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Raising the temperature of the system is akin to increasing the amount of a reactant, and so the equilibrium will shift to the right. Lowering the system temperature will likewise cause the equilibrium to shift left. For exothermic processes, heat is viewed as a product of the reaction and so the opposite temperature dependence is observed.
Effect of a Catalyst
The kinetics chapter of this text identifies a catalyst as a substance that enables a reaction to proceed via a different mechanism with an accelerated rate. The catalyzed reaction mechanism involves a lower energy transition state than the uncatalyzed reaction, resulting in a lower activation energy, Ea, and a correspondingly greater rate constant.
To discern the effect of catalysis on an equilibrium system, consider the reaction diagram for a simple one- step (elementary) reaction shown in . The lowered transition state energy of the catalyzed reaction results in lowered activation energies for both the forward and the reverse reactions. Consequently, both forward and reverse reactions are accelerated, and equilibrium is achieved more quickly but without a change in the equilibrium constant.
FIGURE 13.8 Reaction diagrams for an elementary process in the absence (red) and presence (blue) of a catalyst. The presence of catalyst lowers the activation energies of both the forward and reverse reactions but does not affect the value of the equilibrium constant.
An interesting case study highlighting these equilibrium concepts is the industrial production of ammonia, NH3. This substance is among the “top 10” industrial chemicals with regard to production, with roughly two billion pounds produced annually in the US. Ammonia is used as a chemical feedstock to synthesize a wide range of commercially useful compounds, including fertilizers, plastics, dyes, and explosives.
Most industrial production of ammonia uses the Haber-Bosch process based on the following equilibrium reaction:
The traits of this reaction present challenges to its use in an efficient industrial process. The equilibrium constant is relatively small (Kp on the order of 10−5 at 25 °C), meaning very little ammonia is present in an equilibrium mixture. Also, the rate of this reaction is relatively slow at low temperatures. To raise the yield of ammonia, the industrial process is designed to operate under conditions favoring product formation:
High pressures (concentrations) of reactants are used, ~150−250 atm, to shift the equilibrium right, favoring product formation.
Ammonia is continually removed (collected) from the equilibrium mixture during the process, lowering its concentration and also shifting the equilibrium right.
Although low temperatures favor product formation for this exothermic process, the reaction rate at low temperatures is inefficiently slow. A catalyst is used to accelerate the reaction to reasonable rates at relatively moderate temperatures (400−500 °C).
A diagram illustrating a typical industrial setup for production of ammonia via the Haber-Bosch process is shown in .
FIGURE 13.9 The figure shows a typical industrial setup for the commercial production of ammonia by the Haber- Bosch process. The process operates under conditions that stress the chemical equilibrium to favor product formation.
Equilibrium Calculations
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Identify the changes in concentration or pressure that occur for chemical species in equilibrium systems
Calculate equilibrium concentrations or pressures and equilibrium constants, using various algebraic approaches
Having covered the essential concepts of chemical equilibria in the preceding sections of this chapter, this final section will demonstrate the more practical aspect of using these concepts and appropriate mathematical strategies to perform various equilibrium calculations. These types of computations are essential to many areas of science and technology—for example, in the formulation and dosing of pharmaceutical products. After a drug is ingested or injected, it is typically involved in several chemical equilibria that affect its ultimate concentration in the body system of interest. Knowledge of the quantitative aspects of these equilibria is required to compute a dosage amount that will solicit the desired therapeutic effect.
Many of the useful equilibrium calculations that will be demonstrated here require terms representing changes in reactant and product concentrations. These terms are derived from the stoichiometry of the reaction, as illustrated by decomposition of ammonia:
As shown earlier in this chapter, this equilibrium may be established within a sealed container that initially contains either NH3 only, or a mixture of any two of the three chemical species involved in the equilibrium. Regardless of its initial composition, a reaction mixture will show the same relationships between changes in the concentrations of the three species involved, as dictated by the reaction stoichiometry (see also the related content on expressing reaction rates in the chapter on kinetics). For example, if the nitrogen concentration increases by an amount x: |
Calculation of an Equilibrium Constant
The equilibrium constant for a reaction is calculated from the equilibrium concentrations (or pressures) of its reactants and products. If these concentrations are known, the calculation simply involves their substitution into the K expression, as was illustrated by . A slightly more challenging example is provided next, in which the reaction stoichiometry is used to derive equilibrium concentrations from the information
provided. The basic strategy of this computation is helpful for many types of equilibrium computations and relies on the use of terms for the reactant and product concentrations initially present, for how they change as the reaction proceeds, and for what they are when the system reaches equilibrium. The acronym ICE is commonly used to refer to this mathematical approach, and the concentrations terms are usually gathered in a tabular format called an ICE table. |
If a solution with the concentrations of I2 and I− both equal to 1.000 10−3 M before reaction gives an equilibrium concentration of I2 of 6.61 10−4 M, what is the equilibrium constant for the reaction?
Solution
To calculate the equilibrium constants, equilibrium concentrations are needed for all the reactants and products:
Provided are the initial concentrations of the reactants and the equilibrium concentration of the product. Use this information to derive terms for the equilibrium concentrations of the reactants, presenting all the information in an ICE table. |
When 1 mol each of C2H5OH and CH3CO2H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is a solute in this reaction.)
Answer:
Kc = 4
Calculation of a Missing Equilibrium Concentration
When the equilibrium constant and all but one equilibrium concentration are provided, the other equilibrium concentration(s) may be calculated. A computation of this sort is illustrated in the next example exercise.
Calculation of a Missing Equilibrium Concentration
Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the Kc for the reaction, is 4.1 10−4. Calculate the equilibrium concentration of NO(g) in air at 1 atm pressure and 2000 °C. The equilibrium concentrations of N2 and O2 at this pressure and temperature are 0.036 M and 0.0089 M, respectively.
Solution
Substitute the provided quantities into the equilibrium constant expression and solve for [NO]: |
This result is consistent with the provided value for K within nominal uncertainty, differing by just 1 in the least significant digit’s place.
Check Your Learning
The equilibrium constant Kc for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 10−2. Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M, respectively.
Answer:
1.53 mol/L
Calculation of Equilibrium Concentrations from Initial Concentrations
Perhaps the most challenging type of equilibrium calculation can be one in which equilibrium concentrations are derived from initial concentrations and an equilibrium constant. For these calculations, a four-step approach is typically useful:
Identify the direction in which the reaction will proceed to reach equilibrium.
Develop an ICE table.
Calculate the concentration changes and, subsequently, the equilibrium concentrations.
Confirm the calculated equilibrium concentrations.
The last two example exercises of this chapter demonstrate the application of this strategy.
Calculation of Equilibrium Concentrations
Under certain conditions, the equilibrium constant Kc for the decomposition of PCl5(g) into PCl3(g) and Cl2(g) is 0.0211. What are the equilibrium concentrations of PCl5, PCl3, and Cl2 in a mixture that initially contained only PCl5 at a concentration of 1.00 M?
Solution
Use the stepwise process described earlier.
Step 1.
Determine the direction the reaction proceeds.
The balanced equation for the decomposition of PCl5 is |
The equilibrium constant calculated from the equilibrium concentrations is equal to the value of Kc given in the problem (when rounded to the proper number of significant figures).
Check Your Learning
Acetic acid, CH3CO2H, reacts with ethanol, C2H5OH, to form water and ethyl acetate, CH3CO2C2H5. |
Thus [H+] = [CN–] = x = 8.6 10–6 M and [HCN] = 0.15 – x = 0.15 M.
Note in this case that the change in concentration is significantly less than the initial concentration (a consequence of the small K), and so the initial concentration experiences a negligible change: |
and so the approximation was justified. If this simplified approach were to yield a value for x that did not
justify the approximation, the calculation would need to be repeated without making the approximation.
Check Your Learning
What are the equilibrium concentrations in a 0.25 M NH3 solution? |
Key Terms
equilibrium state of a reversible reaction in which the forward and reverse processes occur at equal rates
equilibrium constant (K) value of the reaction
quotient for a system at equilibrium; may be expressed using concentrations (Kc) or partial pressures (Kp)
heterogeneous equilibria equilibria in which reactants and products occupy two or more different phases
homogeneous equilibria equilibria in which all
reactants and products occupy the same phase
law of mass action when a reversible reaction has |
attained equilibrium at a given temperature, the reaction quotient remains constant
Le Châtelier’s principle an equilibrium subjected
to stress will shift in a way to counter the stress and re-establish equilibrium
reaction quotient (Q) mathematical function
describing the relative amounts of reactants and products in a reaction mixture; may be expressed in terms of concentrations (Qc) or pressures (Qp)
reversible reaction chemical reaction that can
proceed in both the forward and reverse directions under given conditions |
A reversible reaction is at equilibrium when the forward and reverse processes occur at equal rates. Chemical equilibria are dynamic processes characterized by constant amounts of reactant and product species.
The composition of a reaction mixture may be represented by a mathematical function known as the reaction quotient, Q. For a reaction at equilibrium, the composition is constant, and Q is called the equilibrium constant, K.
A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases. |
to temperature, concentration, and, in some cases, volume and pressure. The system’s response to these disturbances is described by Le Châtelier’s principle: An equilibrium system subjected to a disturbance will shift in a way that counters the disturbance and re-establishes equilibrium. A catalyst will increase the rate of both the forward and reverse reactions of a reversible process, increasing the rate at which equilibrium is reached but not altering the equilibrium mixture’s composition (K does not change).
Calculating values for equilibrium constants and/or equilibrium concentrations is of practical benefit to many applications. A mathematical strategy that uses initial concentrations, changes in concentrations, and equilibrium concentrations (and goes by the acronym ICE) is useful for several types of equilibrium calculations. |
What does it mean to describe a reaction as “reversible”?
When writing an equation, how is a reversible reaction distinguished from a nonreversible reaction?
If a reaction is reversible, when can it be said to have reached equilibrium?
Is a system at equilibrium if the rate constants of the forward and reverse reactions are equal?
If the concentrations of products and reactants are equal, is the system at equilibrium?
Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.
Explain why an equilibrium between Br2(l) and Br2(g) would not be established if the container were not a closed vessel shown in .
If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO2 or with pure N2O4?
Among the solubility rules previously discussed is the statement: All chlorides are soluble except Hg2Cl2, AgCl, PbCl2, and CuCl.
Write the expression for the equilibrium constant for the reaction represented by the equation
Is Kc > 1, < 1, or ≈ 1? Explain your answer.
Write the expression for the equilibrium constant for the reaction represented by the equation
Is Kc > 1, < 1, or ≈ 1? Explain your answer.
Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates—except those of the ammonium ion and the alkali metals—are insoluble.
Write the expression for the equilibrium constant for the reaction represented by the equation
Is Kc > 1, < 1, or ≈ 1? Explain your answer.
Write the expression for the equilibrium constant for the reaction represented by the equation
Is Kc > 1, < 1, or ≈ 1? Explain your answer.
Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: Which value of Kc would make this reaction most useful commercially? Kc ≈ 0.01, Kc ≈ 1, or Kc ≈ 10. Explain your answer.
Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation give the same expression for the reaction quotient. KI3 is composed of the ions K+ and
For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is Kc > 1, < 1, or ≈ 1 for a titration reaction?
For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is Kc > 1, < 1, or ≈ 1 for a useful precipitation reaction?
Write the mathematical expression for the reaction quotient, Qc, for each of the following reactions: (a)
(b)
(c)
(d)
(e)
(f )
(g)
(h) |
Convert the values of Kc to values of KP or the values of KP to values of Kc. (a)
(b)
(c)
(d)
What is the value of the equilibrium constant expression for the change at 30 °C? (See .)
Write the expression of the reaction quotient for the ionization of HOCN in water.
Write the reaction quotient expression for the ionization of NH3 in water.
What is the approximate value of the equilibrium constant KP for the change
at 25 °C. (The equilibrium vapor pressure for this substance is 570
torr at 25 °C.)
The following equation represents a reversible decomposition:
Under what conditions will decomposition in a closed container proceed to completion so that no CaCO3 remains?
Explain how to recognize the conditions under which changes in volume will affect gas-phase systems at
equilibrium.
What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?
The following reaction occurs when a burner on a gas stove is lit:
Is an equilibrium among CH4, O2, CO2, and H2O established under these conditions? Explain your answer.
A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, SO3, from sulfur dioxide, SO2, and oxygen, O2, shown here. At high temperatures, the rate of formation of SO3 is higher, but the equilibrium amount (concentration or partial pressure) of SO3 is lower than it would be at lower temperatures.
Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?
Is the reaction endothermic or exothermic?
Suggest four ways in which the concentration of hydrazine, N2H4, could be increased in an equilibrium described by the following equation:
Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:
How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
(a)
(b)
(c)
(d)
How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
(a)
(b)
(c)
(d)
Methanol can be prepared from carbon monoxide and hydrogen at high temperature and pressure in the presence of a suitable catalyst.
Write the expression for the equilibrium constant (Kc) for the reversible reaction
What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more H2 is added?
What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if CO is removed?
What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if CH3OH is added?
What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if the temperature of the system is increased?
What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more catalyst is added?
Nitrogen and oxygen react at high temperatures.
Write the expression for the equilibrium constant (Kc) for the reversible reaction
What will happen to the concentrations of N2, O2, and NO at equilibrium if more O2 is added?
What will happen to the concentrations of N2, O2, and NO at equilibrium if N2 is removed?
What will happen to the concentrations of N2, O2, and NO at equilibrium if NO is added?
What will happen to the concentrations of N2, O2, and NO at equilibrium if the volume of the reaction vessel is decreased?
What will happen to the concentrations of N2, O2, and NO at equilibrium if the temperature of the system is increased?
What will happen to the concentrations of N2, O2, and NO at equilibrium if a catalyst is added?
Water gas, a mixture of H2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon.
Write the expression for the equilibrium constant for the reversible reaction
What will happen to the concentration of each reactant and product at equilibrium if more C is added?
What will happen to the concentration of each reactant and product at equilibrium if H2O is removed?
What will happen to the concentration of each reactant and product at equilibrium if CO is added?
What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.
Write the expression for the equilibrium constant (Kc) for the reversible reaction
What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?
What will happen to the concentration of each reactant and product at equilibrium if H2O is removed?
What will happen to the concentration of each reactant and product at equilibrium if H2 is added?
What will happen to the concentration of each reactant and product at equilibrium if the volume of the reaction vessel is decreased?
What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
Ammonia is a weak base that reacts with water according to this equation:
Will any of the following increase the percent of ammonia that is converted to the ammonium ion in water?
Addition of NaOH
Addition of HCl
Addition of NH4Cl
Acetic acid is a weak acid that reacts with water according to this equation:
Will any of the following increase the percent of acetic acid that reacts and produces ion?
Addition of HCl
Addition of NaOH
Addition of NaCH3CO2 |
A small amount of solid silver sulfate containing a radioactive isotope of silver is added to this solution. Within a few minutes, a portion of the solution phase is sampled and tests positive for radioactive Ag+ ions. Explain this observation.
When equal molar amounts of HCl and HOCl are dissolved separately in equal amounts of water, the solution of HCl freezes at a lower temperature. Which compound has the larger equilibrium constant for acid ionization?
HCl
H+ + Cl–
HOCl
H+ + OCl–
A reaction is represented by this equation:
Write the mathematical expression for the equilibrium constant.
Using concentrations ≤1 M, identify two sets of concentrations that describe a mixture of A, B, and C at equilibrium.
A reaction is represented by this equation:
Write the mathematical expression for the equilibrium constant.
Using concentrations of ≤1 M, identify two sets of concentrations that describe a mixture of W, X, and Y at equilibrium.
What is the value of the equilibrium constant at 500 °C for the formation of NH3 according to the following equation?
An equilibrium mixture of NH3(g), H2(g), and N2(g) at 500 °C was found to contain 1.35 M H2, 1.15 M N2, and 4.12 10−1 M NH3.
Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures.
What is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: CH4, 0.126 M; H2O, 0.242 M; CO, 0.126 M; H2 1.15 M, at a temperature of 760 °C?
A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.40 mol of Cl2(g). Calculate the value of the equilibrium constant for the decomposition of PCl5 to PCl3 and Cl2 at this temperature.
At 1 atm and 25 °C, NO2 with an initial concentration of 1.00 M is 0.0033% decomposed into NO and O2. Calculate the value of the equilibrium constant for the reaction.
Calculate the value of the equilibrium constant KP for the reaction from these equilibrium pressures: NO, 0.050 atm; Cl2, 0.30 atm; NOCl, 1.2 atm.
When heated, iodine vapor dissociates according to this equation:
At 1274 K, a sample exhibits a partial pressure of I2 of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, KP, for the decomposition at 1274 K.
A sample of ammonium chloride was heated in a closed container.
At equilibrium, the pressure of NH3(g) was found to be 1.75 atm. What is the value of the equilibrium constant KP for the decomposition at this temperature?
At a temperature of 60 °C, the vapor pressure of water is 0.196 atm. What is the value of the equilibrium constant KP for the vaporization equilibrium at 60 °C?
Complete the following partial ICE tables. (a) |
Analysis of the gases in a sealed reaction vessel containing NH3, N2, and H2 at equilibrium at 400 °C established the concentration of N2 to be 1.2 M and the concentration of H2 to be 0.24 M.
Calculate the equilibrium molar concentration of NH3.
Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H2 and 1.25 mol of I2 in a 5.00−L flask at 448 °C.
What is the pressure of BrCl in an equilibrium mixture of Cl2, Br2, and BrCl if the pressure of Cl2 in the mixture is 0.115 atm and the pressure of Br2 in the mixture is 0.450 atm?
What is the pressure of CO2 in a mixture at equilibrium that contains 0.50 atm H2, 2.0 atm of H2O, and 1.0 atm of CO at 990 °C?
Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.
What concentration of CO remains in an equilibrium mixture with [CO2] = 0.100 M?
Carbon reacts with water vapor at elevated temperatures.
Assuming a reaction mixture initially contains only reactants, what is the concentration of CO in an equilibrium mixture with [H2O] = 0.500 M at 1000 °C?
Sodium sulfate 10−hydrate, Na2SO4·10H2O, dehydrates according to the equation
What is the pressure of water vapor at equilibrium with a mixture of Na2SO4·10H2O and NaSO4?
Calcium chloride 6−hydrate, CaCl2·6H2O, dehydrates according to the equation
What is the pressure of water vapor at equilibrium with a mixture of CaCl2·6H2O and CaCl2 at 25 °C?
A student solved the following problem and found the equilibrium concentrations to be [SO2] = 0.590 M, [O2] = 0.0450 M, and [SO3] = 0.260 M. How could this student check the work without reworking the problem? The problem was: For the following reaction at 600 °C:
A student solved the following problem and found [N2O4] = 0.16 M at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of N2O4 in a mixture formed from a sample of NO2 with a concentration of 0.10 M?
Assume that the change in concentration of N2O4 is small enough to be neglected in the following problem.
Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N2O4 with chloroform as the solvent.
in chloroform
Confirm that the change is small enough to be neglected.
Assume that the change in concentration of COCl2 is small enough to be neglected in the following problem.
Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCl2 with an initial concentration of 0.3166 M.
Confirm that the change is small enough to be neglected.
Assume that the change in pressure of H2S is small enough to be neglected in the following problem.
Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H2S with an initial pressure of 0.824 atm. |
What are all concentrations after a mixture that contains [H2O] = 1.00 M and [Cl2O] = 1.00 M comes to equilibrium at 25 °C?
What are the concentrations of PCl5, PCl3, and Cl2 in an equilibrium mixture produced by the decomposition of a sample of pure PCl5 with [PCl5] = 2.00 M? |
KP = 2.5 at 25 °C
What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of
atm?
What is the minimum mass of CaCO3 required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant (Kc) is 0.50 for the decomposition reaction of CaCO3 at that temperature?
The equilibrium constant (Kc) for this reaction is 1.60 at 990 °C:
Calculate the number of moles of each component in the final equilibrium mixture obtained from adding
1.00 mol of H2, 2.00 mol of CO2, 0.750 mol of H2O, and 1.00 mol of CO to a 5.00-L container at 990 °C.
In a 3.0-L vessel, the following equilibrium partial pressures are measured: N2, 190 torr; H2, 317 torr; NH3, 1.00 103 torr.
How will the partial pressures of H2, N2, and NH3 change if H2 is removed from the system? Will they increase, decrease, or remain the same?
Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.
The equilibrium constant (Kc) for this reaction is 5.0 at a given temperature.
On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.90 mol of H2 in a liter. How many moles of CO2 were there in the equilibrium mixture?
Maintaining the same temperature, additional H2 was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.30 mol of water vapor, and 1.2 mol of H2 in a liter. How many moles of CO2 were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.
Antimony pentachloride decomposes according to this equation:
An equilibrium mixture in a 5.00-L flask at 448 °C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2. How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature?
Consider the equilibrium
What is the expression for the equilibrium constant (Kc) of the reaction?
How must the concentration of NH3 change to reach equilibrium if the reaction quotient is less than the equilibrium constant?
If the reaction were at equilibrium, how would an increase in the volume of the reaction vessel affect the pressure of NO2?
If the change in the pressure of NO2 is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O2 change?
The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO2), is partially regulated by the
concentration of H3O+ and dissolved CO2 in the blood. Although the equilibrium is complicated, it can be summarized as
Write the equilibrium constant expression for this reaction.
Explain why the production of lactic acid and CO2 in a muscle during exertion stimulates release of O2 from the oxyhemoglobin in the blood passing through the muscle.
Liquid N2O3 is dark blue at low temperatures, but the color fades and becomes greenish at higher
temperatures as the compound decomposes to NO and NO2. At 25 °C, a value of KP = 1.91 has been established for this decomposition. If 0.236 moles of N2O3 are placed in a 1.52-L vessel at 25 °C, calculate the equilibrium partial pressures of N2O3(g), NO2(g), and NO(g).
A 1.00-L vessel at 400 °C contains the following equilibrium concentrations: N2, 1.00 M; H2, 0.50 M; and NH3, 0.25 M. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 M? The equilibrium reaction is |
INTRODUCTION Liquid water is essential to life on our planet, and chemistry involving the characteristic ions of water, H+ and OH–, is widely encountered in nature and society. As introduced in another chapter of this text, acid-base chemistry involves the transfer of hydrogen ions from donors (acids) to acceptors (bases).
These H+ transfer reactions are reversible, and the equilibria established by acid-base systems are essential aspects of phenomena ranging from sinkhole formation () to oxygen transport in the human body. This chapter will further explore acid-base chemistry with an emphasis on the equilibrium aspects of this important reaction class.
Brønsted-Lowry Acids and Bases
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Identify acids, bases, and conjugate acid-base pairs according to the Brønsted-Lowry definition
Write equations for acid and base ionization reactions
Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations
Describe the acid-base behavior of amphiprotic substances
The acid-base reaction class has been studied for quite some time. In 1680, Robert Boyle reported traits of acid solutions that included their ability to dissolve many substances, to change the colors of certain natural dyes, and to lose these traits after coming in contact with alkali (base) solutions. In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO2), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.
Johannes Brønsted and Thomas Lowry proposed a more general description in 1923 in which acids and bases were defined in terms of the transfer of hydrogen ions, H+. (Note that these hydrogen ions are often referred to simply as protons, since that subatomic particle is the only component of cations derived from the most abundant hydrogen isotope, 1H.) A compound that donates a proton to another compound is called a Brønsted-Lowry acid, and a compound that accepts a proton is called a Brønsted-Lowry base. An acid-base reaction is, thus, the transfer of a proton from a donor (acid) to an acceptor (base).
The concept of conjugate pairs is useful in describing Brønsted-Lowry acid-base reactions (and other reversible reactions, as well). When an acid donates H+, the species that remains is called the conjugate base of the acid because it reacts as a proton acceptor in the reverse reaction. Likewise, when a base accepts H+, it is converted to its conjugate acid. The reaction between water and ammonia illustrates this idea. In the forward direction, water acts as an acid by donating a proton to ammonia and subsequently becoming a hydroxide ion, OH−, the conjugate base of water. The ammonia acts as a base in accepting this proton, becoming an ammonium ion, the conjugate acid of ammonia. In the reverse direction, a hydroxide ion acts as a base in accepting a proton from ammonium ion, which acts as an acid.
The reaction between a Brønsted-Lowry acid and water is called acid ionization. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:
Base ionization of a species occurs when it accepts protons from water molecules. In the example below, pyridine molecules, C5NH5, undergo base ionization when dissolved in water, yielding hydroxide and pyridinium ions:
The preceding ionization reactions suggest that water may function as both a base (as in its reaction with hydrogen fluoride) and an acid (as in its reaction with ammonia). Species capable of either donating or accepting protons are called amphiprotic, or more generally, amphoteric, a term that may be used for acids and bases per definitions other than the Brønsted-Lowry one. The equations below show the two possible acid- base reactions for two amphiprotic species, bicarbonate ion and water:
The first equation represents the reaction of bicarbonate as an acid with water as a base, whereas the second represents reaction of bicarbonate as a base with water as an acid. When bicarbonate is added to water, both these equilibria are established simultaneously and the composition of the resulting solution may be determined through appropriate equilibrium calculations, as described later in this chapter.
In the liquid state, molecules of an amphiprotic substance can react with one another as illustrated for water in the equations below:
The process in which like molecules react to yield ions is called autoionization. Liquid water undergoes autoionization to a very slight extent; at 25 °C, approximately two out of every billion water molecules are ionized. The extent of the water autoionization process is reflected in the value of its equilibrium constant, the ion-product constant for water, Kw:
The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, Kw has a value of 1.0 10−14. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 °C, the value for Kw is about 5.6 10−13, roughly 50 times larger than the value at 25 °C.
Ion Concentrations in Pure Water
What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 °C?
Solution
The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, [H3O+] = [OH−] = x. At 25 °C: |
The hydronium ion concentration and the hydroxide ion concentration are the same, 1.0 10−7 M.
Check Your Learning
The ion product of water at 80 °C is 2.4 10−13. What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C?
Answer:
[H3O+] = [OH−] = 4.9 10−7 M
The Inverse Relation between [H3O+] and [OH−]
A solution of an acid in water has a hydronium ion concentration of 2.0 10−6 M. What is the concentration of hydroxide ion at 25 °C?
Solution
Use the value of the ion-product constant for water at 25 °C |
Compared with pure water, a solution of acid exhibits a higher concentration of hydronium ions (due to ionization of the acid) and a proportionally lower concentration of hydroxide ions. This may be explained via Le Châtelier’s principle as a left shift in the water autoionization equilibrium resulting from the stress of increased hydronium ion concentration.
Substituting the ion concentrations into the Kw expression confirms this calculation, resulting in the expected value: |
pH and pOH
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Explain the characterization of aqueous solutions as acidic, basic, or neutral
Express hydronium and hydroxide ion concentrations on the pH and pOH scales
Perform calculations relating pH and pOH
As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water (Kw). The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions.
A common means of expressing quantities that may span many orders of magnitude is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where “X” is the quantity of interest and “log” is the base-10 logarithm: |
And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than 1.0 10−7 M and hydroxide ion molarities less than 1.0 10−7 M (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than 1.0 10−7 M and hydroxide ion molarities greater than 1.0 10−7 M (corresponding to pH values greater than 7.00 and pOH values less than 7.00).
Since the autoionization constant Kw is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, the “Check Your Learning” exercise accompanying showed the hydronium molarity of pure water at 80 °C is
4.9 10−7 M, which corresponds to pH and pOH values of:
At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at other temperatures, such as enzyme reactions in warm-blooded organisms at a temperature around 36–40 °C. Unless otherwise noted, references to pH values are presumed to be those at 25 °C (). |
(The use of logarithms is explained in . When taking the log of a value, keep as many decimal places in the result as there are significant figures in the value.)
Check Your Learning
Water exposed to air contains carbonic acid, H2CO3, due to the reaction between carbon dioxide and water: |
Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO2, SO2, SO3, NO, and NO2 being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here:
Carbon dioxide is naturally present in the atmosphere because most organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or fossil fuels.
Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also originates from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal- refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine.
Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until
the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India.
For further information on acid rain, visit this hosted by the US Environmental Protection Agency.
FIGURE 14.3 (a) Acid rain makes trees more susceptible to drought and insect infestation, and depletes nutrients in the soil. (b) It also is corrodes statues that are carved from marble or limestone. (credit a: modification of work by Chris M Morris; credit b: modification of work by “Eden, Janine and Jim”/Flickr)
Calculation of pOH
What are the pOH and the pH of a 0.0125-M solution of potassium hydroxide, KOH?
Solution
Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [OH−] = 0.0125 M: |
The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter ().
FIGURE 14.4 (a) A research-grade pH meter used in a laboratory can have a resolution of 0.001 pH units, an accuracy of ± 0.002 pH units, and may cost in excess of $1000. (b) A portable pH meter has lower resolution (0.01 pH units), lower accuracy (± 0.2 pH units), and a far lower price tag. (credit b: modification of work by Jacopo Werther)
The pH of a solution may also be visually estimated using colored indicators (). The acid-base equilibria that enable use of these indicator dyes for pH measurements are described in a later section of this chapter.
FIGURE 14.5 (a) A solution containing a dye mixture, called universal indicator, takes on different colors depending upon its pH. (b) Convenient test strips, called pH paper, contain embedded indicator dyes that yield pH- dependent color changes on contact with aqueous solutions.(credit: modification of work by Sahar Atwa)
Relative Strengths of Acids and Bases
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Assess the relative strengths of acids and bases according to their ionization constants
Rationalize trends in acid–base strength in relation to molecular structure
Carry out equilibrium calculations for weak acid–base systems
Acid and Base Ionization Constants
The relative strength of an acid or base is the extent to which it ionizes when dissolved in water. If the ionization reaction is essentially complete, the acid or base is termed strong; if relatively little ionization occurs, the acid or base is weak. As will be evident throughout the remainder of this chapter, there are many more weak acids and bases than strong ones. The most common strong acids and bases are listed in .
FIGURE 14.6 Some of the common strong acids and bases are listed here.
The relative strengths of acids may be quantified by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, Ka. For the reaction of an acid HA: |
where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include [H2O] in the equation. The larger the Ka of an acid, the larger the concentration of
and A− relative to the concentration of the nonionized acid, HA, in an equilibrium mixture, and the stronger the acid. An acid is classified as “strong” when it undergoes complete ionization, in which case the concentration of HA is zero and the acid ionization constant is immeasurably large (Ka ≈ ∞). Acids that are partially ionized are called “weak,” and their acid ionization constants may be experimentally measured. A table of ionization constants for weak acids is provided in .
To illustrate this idea, three acid ionization equations and Ka values are shown below. The ionization constants increase from first to last of the listed equations, indicating the relative acid strength increases in the order CH3CO2H < HNO2 < |
[H3O+]). Unlike the Ka value, the percent ionization of a weak acid varies with the initial concentration of acid, typically decreasing as concentration increases. Equilibrium calculations of the sort described later in this chapter can be used to confirm this behavior. |
(Recall the provided pH value of 2.09 is logarithmic, and so it contains just two significant digits, limiting the certainty of the computed percent ionization.)
Check Your Learning
Calculate the percent ionization of a 0.10-M solution of acetic acid with a pH of 2.89.
Answer:
1.3% ionized |
Just as for acids, the relative strength of a base is reflected in the magnitude of its base-ionization constant (Kb) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, B: |
but will vary depending on the base ionization constant and the initial concentration of the solution.
Relative Strengths of Conjugate Acid-Base Pairs
Brønsted-Lowry acid-base chemistry is the transfer of protons; thus, logic suggests a relation between the relative strengths of conjugate acid-base pairs. The strength of an acid or base is quantified in its ionization constant, Ka or Kb, which represents the extent of the acid or base ionization reaction. For the conjugate acid- base pair HA / A−, ionization equilibrium equations and ionization constant expressions are |
This equation states the relation between ionization constants for any conjugate acid-base pair, namely, their mathematical product is equal to the ion product of water, Kw. By rearranging this equation, a reciprocal relation between the strengths of a conjugate acid-base pair becomes evident: |
FIGURE 14.8 This figure shows strengths of conjugate acid-base pairs relative to the strength of water as the reference substance.
The listing of conjugate acid–base pairs shown in is arranged to show the relative strength of each species as compared with water, whose entries are highlighted in each of the table’s columns. In the acid column, those species listed below water are weaker acids than water. These species do not undergo acid ionization in water; they are not Bronsted-Lowry acids. All the species listed above water are stronger acids, transferring protons to water to some extent when dissolved in an aqueous solution to generate hydronium ions. Species above water but below hydronium ion are weak acids, undergoing partial acid ionization, wheres those above hydronium ion are strong acids that are completely ionized in aqueous solution.
If all these strong acids are completely ionized in water, why does the column indicate they vary in strength, with nitric acid being the weakest and perchloric acid the strongest? Notice that the sole acid species present in an aqueous solution of any strong acid is H3O+(aq), meaning that hydronium ion is the strongest acid that may exist in water; any stronger acid will react completely with water to generate hydronium ions. This limit on the acid strength of solutes in a solution is called a leveling effect. To measure the differences in acid strength for “strong” acids, the acids must be dissolved in a solvent that is less basic than water. In such solvents, the acids will be “weak,” and so any differences in the extent of their ionization can be determined. For example, the binary hydrogen halides HCl, HBr, and HI are strong acids in water but weak acids in ethanol (strength increasing HCl < HBr < HI).
The right column of lists a number of substances in order of increasing base strength from top to bottom. Following the same logic as for the left column, species listed above water are weaker bases and so they don’t undergo base ionization when dissolved in water. Species listed between water and its conjugate base, hydroxide ion, are weak bases that partially ionize. Species listed below hydroxide ion are strong bases that completely ionize in water to yield hydroxide ions (i.e., they are leveled to hydroxide). A comparison of the acid and base columns in this table supports the reciprocal relation between the strengths of conjugate acid- base pairs. For example, the conjugate bases of the strong acids (top of table) are all of negligible strength. A strong acid exhibits an immeasurably large Ka, and so its conjugate base will exhibit a Kb that is essentially zero:
∞
∞ |
Calculating Ionization Constants for Conjugate Acid-Base Pairs
Use the Kb for the nitrite ion, to calculate the Ka for its conjugate acid.
Solution
Kb for is given in this section as 2.17 10−11. The conjugate acid of is HNO2; Ka for HNO2 can be calculated using the relationship: |
This answer can be verified by finding the Ka for HNO2 in .
Check Your Learning
Determine the relative acid strengths of and HCN by comparing their ionization constants. The ionization constant of HCN is given in as 4.9 10−10. The ionization constant of is not listed, but the ionization constant of its conjugate base, NH3, is listed as 1.8 10−5.
Answer:
is the slightly stronger acid (Ka for = 5.6 10−10).
Acid-Base Equilibrium Calculations
The chapter on chemical equilibria introduced several types of equilibrium calculations and the various mathematical strategies that are helpful in performing them. These strategies are generally useful for equilibrium systems regardless of chemical reaction class, and so they may be effectively applied to acid-base equilibrium problems. This section presents several example exercises involving equilibrium calculations for acid-base systems.
Determination of Ka from Equilibrium Concentrations
Acetic acid is the principal ingredient in vinegar () that provides its sour taste. At equilibrium, a solution contains [CH3CO2H] = 0.0787 M and What is the value of Ka for acetic acid?
FIGURE 14.9 Vinegar contains acetic acid, a weak acid. (credit: modification of work by “HomeSpot HQ”/Flickr)
Solution
The relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the Ka for acetic acid. |
Determination of Kb from Equilibrium Concentrations
Caffeine, C8H10N4O2 is a weak base. What is the value of Kb for caffeine if a solution at equilibrium has [C8H10N4O2] = 0.050 M, = 5.0 10−3 M, and [OH−] = 2.5 10−3 M?
Solution
The relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the Kb for caffeine. |
Solution
The nitrous acid concentration provided is a formal concentration, one that does not account for any chemical equilibria that may be established in solution. Such concentrations are treated as “initial” values for equilibrium calculations using the ICE table approach. Notice the initial value of hydronium ion is listed as approximately zero because a small concentration of H3O+ is present (1 × 10−7 M) due to the autoprotolysis of water. In many cases, such as all the ones presented in this chapter, this concentration is much less than that generated by ionization of the acid (or base) in question and may be neglected.
The pH provided is a logarithmic measure of the hydronium ion concentration resulting from the acid ionization of the nitrous acid, and so it represents an “equilibrium” value for the ICE table: |
In some cases, the strength of the weak acid or base and its formal (initial) concentration result in an appreciable ionization. Though the ICE strategy remains effective for these systems, the algebra is a bit more involved because the simplifying assumption that x is negligible cannot be made. Calculations of this sort are demonstrated in below.
Calculating Equilibrium Concentrations without Simplifying Assumptions
Sodium bisulfate, NaHSO4, is used in some household cleansers as a source of the ion, a weak acid. What is the pH of a 0.50-M solution of |
Solving for the two roots of this quadratic equation results in a negative value that may be discarded as physically irrelevant and a positive value equal to x. As defined in the ICE table, x is equal to the hydronium concentration. |
Effect of Molecular Structure on Acid-Base Strength
Binary Acids and Bases
In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is HF < HCl < HBr < HI. Likewise, for group 16, the order of increasing acid strength is H2O < H2S < H2Se < H2Te.
Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is CH4 < NH3 < H2O < HF; across the third row, it is SiH4 < PH3 < H2S < HCl (see ).
FIGURE 14.11 The figure shows trends in the strengths of binary acids and bases.
Ternary Acids and Bases
Ternary compounds composed of hydrogen, oxygen, and some third element (“E”) may be structured as depicted in the image below. In these compounds, the central E atom is bonded to one or more O atoms, and at least one of the O atoms is also bonded to an H atom, corresponding to the general molecular formula OmE(OH)n. These compounds may be acidic, basic, or amphoteric depending on the properties of the central E atom. Examples of such compounds include sulfuric acid, O2S(OH)2, sulfurous acid, OS(OH)2, nitric acid, O2NOH, perchloric acid, O3ClOH, aluminum hydroxide, Al(OH)3, calcium hydroxide, Ca(OH)2, and potassium hydroxide, KOH:
If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond a between the element and oxygen is more readily broken than bond b between oxygen and hydrogen. Hence bond a is ionic, hydroxide ions are released to the solution, and the material behaves as a base—this is the case with Ca(OH)2 and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds.
If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond a relatively strongly covalent. The oxygen-hydrogen bond, bond b, is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic −OH groups that are called oxyacids.
Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond.
Sulfuric acid, H2SO4, or O2S(OH)2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid,
H2SO3, or OS(OH)2 (with a sulfur oxidation number of +4). Likewise nitric acid, HNO3, or O2NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO2, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid ().
FIGURE 14.12 As the oxidation number of the central atom E increases, the acidity also increases.
Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate Al(H2O)3(OH)3, is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, Al(H2O)3(OH)3, is converted into the soluble ion, by reaction with hydroxide ion:
In this reaction, a proton is transferred from one of the aluminum-bound H2O molecules to a hydroxide ion in solution. The Al(H2O)3(OH)3 compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion by reaction with hydronium ion:
In this case, protons are transferred from hydronium ions in solution to Al(H2O)3(OH)3, and the compound functions as a base.
Hydrolysis of Salts
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Predict whether a salt solution will be acidic, basic, or neutral
Calculate the concentrations of the various species in a salt solution
Describe the acid ionization of hydrated metal ions
Salts with Acidic Ions
Salts are ionic compounds composed of cations and anions, either of which may be capable of undergoing an acid or base ionization reaction with water. Aqueous salt solutions, therefore, may be acidic, basic, or neutral, depending on the relative acid-base strengths of the salt's constituent ions. For example, dissolving ammonium chloride in water results in its dissociation, as described by the equation |
Since ammonia is a weak base, Kb is measurable and Ka > 0 (ammonium ion is a weak acid).
The chloride ion is the conjugate base of hydrochloric acid, and so its base ionization (or base hydrolysis) reaction is represented by
Since HCl is a strong acid, Ka is immeasurably large and Kb ≈ 0 (chloride ions don’t undergo appreciable hydrolysis).
Thus, dissolving ammonium chloride in water yields a solution of weak acid cations ( ) and inert anions (Cl−), resulting in an acidic solution.
Calculating the pH of an Acidic Salt Solution
Aniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of anilinium chloride |
The sodium ion does not undergo appreciable acid or base ionization and has no effect on the solution pH. This may seem obvious from the ion's formula, which indicates no hydrogen or oxygen atoms, but some dissolved metal ions function as weak acids, as addressed later in this section.
The acetate ion, is the conjugate base of acetic acid, CH3CO2H, and so its base ionization (or base hydrolysis) reaction is represented by
Because acetic acid is a weak acid, its Ka is measurable and Kb > 0 (acetate ion is a weak base).
Dissolving sodium acetate in water yields a solution of inert cations (Na+) and weak base anions resulting in a basic solution. |
Solution
The provided equilibrium concentrations and a value for the equilibrium constant will permit calculation of the missing equilibrium concentration. The process in question is the base ionization of acetate ion, for which |
relative strengths of these two species. Likewise, some salts contain a single ion that is amphiprotic, and so the relative strengths of this ion’s acid and base character will determine its effect on solution pH. For both types of salts, a comparison of the Ka and Kb values allows prediction of the solution’s acid-base status, as illustrated in the following example exercise.
Determining the Acidic or Basic Nature of Salts
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
KBr
NaHCO3
Na2HPO4
NH4F
Solution
Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here:
The K+ cation is inert and will not affect pH. The bromide ion is the conjugate base of a strong acid, and so it is of negligible base strength (no appreciable base ionization). The solution is neutral.
The Na+ cation is inert and will not affect the pH of the solution; while the anion is amphiprotic. The Ka of is 4.7 10−11,and its Kb is
Since Kb >> Ka, the solution is basic.
The Na+ cation is inert and will not affect the pH of the solution, while the anion is amphiprotic. The Ka of is 4.2 10−13,
and its Kb is Because Kb >> Ka, the solution is basic.
The ion is acidic (see above discussion) and the F− ion is basic (conjugate base of the weak acid HF). Comparing the two ionization constants: Ka of is 5.6 10−10 and the Kb of F− is 1.6 10−11, so the solution is acidic, since Ka > Kb.
Check Your Learning
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
K2CO3
CaCl2
KH2PO4
(NH4)2CO3
Answer:
(a) basic; (b) neutral; (c) acidic; (d) basic
The Ionization of Hydrated Metal Ions
Unlike the group 1 and 2 metal ions of the preceding examples (Na+, Ca2+, etc.), some metal ions function as acids in aqueous solutions. These ions are not just loosely solvated by water molecules when dissolved, instead they are covalently bonded to a fixed number of water molecules to yield a complex ion (see chapter on coordination chemistry). As an example, the dissolution of aluminum nitrate in water is typically represented as |
As shown in , the ions involve bonds between a central Al atom and the O atoms of the six water molecules. Consequently, the bonded water molecules' O–H bonds are more polar than in nonbonded water molecules, making the bonded molecules more prone to donation of a hydrogen ion:
The conjugate base produced by this process contains five other bonded water molecules capable of acting as acids, and so the sequential or step-wise transfer of protons is possible as depicted in few equations below: |
FIGURE 14.13 When an aluminum ion reacts with water, the hydrated aluminum ion becomes a weak acid.
Aside from the alkali metals (group 1) and some alkaline earth metals (group 2), most other metal ions will undergo acid ionization to some extent when dissolved in water. The acid strength of these complex ions typically increases with increasing charge and decreasing size of the metal ions. The first-step acid ionization equations for a few other acidic metal ions are shown below:
Hydrolysis of [Al(H2O)6]3+
Calculate the pH of a 0.10-M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion in solution.
Solution
The equation for the reaction and Ka are: |
Polyprotic Acids
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton
Acids are classified by the number of protons per molecule that they can give up in a reaction. Acids such as HCl, HNO3, and HCN that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are: |
Similarly, monoprotic bases are bases that will accept a single proton.
Diprotic acids contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:
This stepwise ionization process occurs for all polyprotic acids. Carbonic acid, H2CO3, is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts. |
is larger than by a factor of 104, so H2CO3 is the dominant producer of hydronium ion in the solution. This means that little of the formed by the ionization of H2CO3 ionizes to give hydronium ions (and carbonate ions), and the concentrations of H3O+ and are practically equal in a pure aqueous solution of H2CO3.
If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This approach is demonstrated in the following example exercise.
Ionization of a Diprotic Acid
“Carbonated water” contains a palatable amount of dissolved carbon dioxide. The solution is acidic because CO2 reacts with water to form carbonic acid, H2CO3. What are and in a saturated solution of CO2 with an initial [H2CO3] = 0.033 M? |
As for the diprotic acid examples, each successive ionization reaction is less extensive than the former, reflected in decreasing values for the stepwise acid ionization constants. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about 105 to 106.
This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H3PO4 complicated. However, because the successive ionization constants differ by a factor of 105 to 106, large differences exist in the small changes in concentration accompanying the ionization reactions. This allows the use of math-simplifying assumptions and processes, as demonstrated in the examples above.
Polyprotic bases are capable of accepting more than one hydrogen ion. The carbonate ion is an example of a diprotic base, because it can accept two protons, as shown below. Similar to the case for polyprotic acids, note the ionization constants decrease with ionization step. Likewise, equilibrium calculations involving polyprotic bases follow the same approaches as those for polyprotic acids.
Buffers
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Describe the composition and function of acid–base buffers
Calculate the pH of a buffer before and after the addition of added acid or base
A solution containing appreciable amounts of a weak conjugate acid-base pair is called a buffer solution, or a buffer. Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (). A solution of acetic acid and sodium acetate (CH3COOH + CH3COONa) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia and ammonium chloride (NH3(aq) + NH4Cl(aq)).
FIGURE 14.14 (a) The unbuffered solution on the left and the buffered solution on the right have the same pH (pH 8); they are basic, showing the yellow color of the indicator methyl orange at this pH. (b) After the addition of 1 mL of a 0.01-M HCl solution, the buffered solution has not detectably changed its pH but the unbuffered solution has become acidic, as indicated by the change in color of the methyl orange, which turns red at a pH of about 4. (credit: modification of work by Mark Ott)
How Buffers Work
To illustrate the function of a buffer solution, consider a mixture of roughly equal amounts of acetic acid and sodium acetate. The presence of a weak conjugate acid-base pair in the solution imparts the ability to neutralize modest amounts of added strong acid or base. For example, strong base added to this solution will neutralize hydronium ion, causing the acetic acid ionization equilibrium to shift to the right and generate additional amounts of the weak conjugate base (acetate ion):
Likewise, strong acid added to this buffer solution will shift the above ionization equilibrium left, producing additional amounts of the weak conjugate acid (acetic acid). provides a graphical illustration of the changes in conjugate-partner concentration that occur in this buffer solution when strong acid and base are added. The buffering action of the solution is essentially a result of the added strong acid and base being converted to the weak acid and base that make up the buffer's conjugate pair. The weaker acid and base undergo only slight ionization, as compared with the complete ionization of the strong acid and base, and the solution pH, therefore, changes much less drastically than it would in an unbuffered solution. |
pH Changes in Buffered and Unbuffered Solutions
Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might affect the biochemical activity of these compounds.
Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.
Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100 mL of this buffer.
For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74.
Solution
Following the ICE approach to this equilibrium calculation yields the following: |
Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer.
Adding strong base will neutralize some of the acetic acid, yielding the conjugate base acetate ion. Compute the new concentrations of these two buffer components, then repeat the equilibrium calculation of part (a) using these new concentrations. |
Using these concentrations, the pH of the solution may be computed as in part (a) above, yielding pH = 4.75 (only slightly different from that prior to adding the strong base).
For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74.
The amount of hydronium ion initially present in the solution is |
Buffer Capacity
Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (). Instead, the ability of a buffer solution to resist changes in pH relies on the presence of appreciable amounts of its conjugate weak acid-base pair. When enough strong acid or base is added to substantially lower the concentration of either member of the buffer pair, the buffering action within the solution is compromised.
FIGURE 14.16 The indicator color (methyl orange) shows that a small amount of acid added to a buffered solution of pH 8 (beaker on the left) has little affect on the buffered system (middle beaker). However, a large amount of acid exhausts the buffering capacity of the solution and the pH changes dramatically (beaker on the right). (credit: modification of work by Mark Ott)
The buffer capacity is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak
acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion.
Selection of Suitable Buffer Mixtures
There are two useful rules of thumb for selecting buffer mixtures:
A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other. shows how pH changes for an acetic acid-acetate ion buffer as base is added. The initial pH is
4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to 11% of the acetate ion concentration.
FIGURE 14.17 Change in pH as an increasing amount of a 0.10-M NaOH solution is added to 100 mL of a buffer solution in which, initially, [CH3CO2H] = 0.10 M and Note the greatly diminished buffering action occurring after the buffer capacity has been reached, resulting in drastic rises in pH on adding more strong base.
Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.
Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, H2CO3, and the bicarbonate ion, When a hydronium ion is introduced to the blood stream, it is removed primarily by the reaction: |
The added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair (H3O+ is converted to H2CO3 and OH- is converted to HCO3-). The pH of human blood thus remains very near the value determined by the buffer pairs pKa, in this case, 7.35. Normal variations in blood pH are usually less than 0.1, and pH changes of 0.4 or greater are likely to be fatal. |
where pKa is the negative of the logarithm of the ionization constant of the weak acid (pKa = −log Ka). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak conjugate acid-base pair in a buffered solution. Scientists often use this expression, called the Henderson-Hasselbalch equation, to calculate the pH of buffer solutions. It is important to note that the “x is small” assumption must be valid to use this equation. |
The concentration of carbonic acid, H2CO3 is approximately 0.0012 M, and the concentration of the hydrogen carbonate ion, is around 0.024 M. Using the Henderson-Hasselbalch equation and the pKa of carbonic acid at body temperature, we can calculate the pH of blood:
The fact that the H2CO3 concentration is significantly lower than that of the ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded.
Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the ion, producing H2CO3. An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO2 from the blood through the lungs driving the equilibrium reaction such that [H3O+] is lowered. If the blood is too alkaline, a lower breath rate increases CO2 concentration in the blood, driving the equilibrium reaction the other way, increasing [H+] and restoring an appropriate pH. |
Acid-Base Titrations
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Interpret titration curves for strong and weak acid-base systems
Compute sample pH at important stages of a titration
Explain the function of acid-base indicators
As seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. In this section, we will explore the underlying chemical equilibria that make acid-base titrimetry a useful analytical technique.
Titration Curves
A titration curve is a plot of some solution property versus the amount of added titrant. For acid-base titrations, solution pH is a useful property to monitor because it varies predictably with the solution composition and, therefore, may be used to monitor the titration’s progress and detect its end point. The following example exercise demonstrates the computation of pH for a titration solution after additions of several specified titrant volumes. The first example involves a strong acid titration that requires only stoichiometric calculations to derive the solution pH. The second example addresses a weak acid titration requiring equilibrium calculations.
Calculating pH for Titration Solutions: Strong Acid/Strong Base
A titration is carried out for 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M of a strong base NaOH (the titration curve is shown in ). Calculate the pH at these volumes of added base solution:
0.00 mL
12.50 mL
25.00 mL
37.50 mL
Solution
Titrant volume = 0 mL. The solution pH is due to the acid ionization of HCl. Because this is a strong acid, the ionization is complete and the hydronium ion molarity is 0.100 M. The pH of the solution is then
Titrant volume = 12.50 mL. Since the acid sample and the base titrant are both monoprotic and equally concentrated, this titrant addition involves less than a stoichiometric amount of base, and so it is completely consumed by reaction with the excess acid in the sample. The concentration of acid remaining is computed by subtracting the consumed amount from the intial amount and then dividing by the solution volume:
Titrant volume = 25.00 mL. This titrant addition involves a stoichiometric amount of base (the equivalence point), and so only products of the neutralization reaction are in solution (water and NaCl). Neither the cation nor the anion of this salt undergo acid-base ionization; the only process generating hydronium ions is the autoprotolysis of water. The solution is neutral, having a pH = 7.00.
Titrant volume = 37.50 mL. This involves the addition of titrant in excess of the equivalence point. The solution pH is then calculated using the concentration of hydroxide ion: |
The acid and titrant are both monoprotic and the sample and titrant solutions are equally concentrated; thus, this volume of titrant represents the equivalence point. Unlike the strong-acid example above, however, the reaction mixture in this case contains a weak conjugate base (acetate ion). The solution pH is computed considering the base ionization of acetate, which is present at a concentration of |
Note that the pH at the equivalence point of this titration is significantly greater than 7, as expected when titrating a weak acid with a strong base.
Titrant volume = 12.50 mL. This volume represents one-half of the stoichiometric amount of titrant, and so one-half of the acetic acid has been neutralized to yield an equivalent amount of acetate ion. The concentrations of these conjugate acid-base partners, therefore, are equal. A convenient approach to computing the pH is use of the Henderson-Hasselbalch equation:
(pH = pKa at the half-equivalence point in a titration of a weak acid)
Titrant volume = 37.50 mL. This volume represents a stoichiometric excess of titrant, and a reaction solution containing both the titration product, acetate ion, and the excess strong titrant. In such solutions, the solution pH is determined primarily by the amount of excess strong base: |
Performing additional calculations similar to those in the preceding example permits a more full assessment of titration curves. A summary of pH/volume data pairs for the strong and weak acid titrations is provided in and plotted as titration curves in . A comparison of these two curves illustrates several important concepts that are best addressed by identifying the four stages of a titration:
initial state (added titrant volume = 0 mL): pH is determined by the acid being titrated; because the two acid samples are equally concentrated, the weak acid will exhibit a greater initial pH
pre-equivalence point (0 mL < V < 25 mL): solution pH increases gradually and the acid is consumed by reaction with added titrant; composition includes unreacted acid and the reaction product, its conjugate base
equivalence point (V = 25 mL): a drastic rise in pH is observed as the solution composition transitions from acidic to either neutral (for the strong acid sample) or basic (for the weak acid sample), with pH determined by ionization of the conjugate base of the acid
postequivalence point (V > 25 mL): pH is determined by the amount of excess strong base titrant added; since both samples are titrated with the same titrant, both titration curves appear similar at this stage. |
FIGURE 14.18 (a) The titration curve for the titration of 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M NaOH (strong base) has an equivalence point of 7.00 pH. (b) The titration curve for the titration of 25.00 mL of 0.100 M acetic acid (weak acid) with 0.100 M NaOH (strong base) has an equivalence point of 8.72 pH.
Acid-Base Indicators
Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than 5.0 10−9 M (pH < 8.3). In more basic solutions where the hydronium ion concentration is less than 5.0 10−9 M (pH > 8.3), it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called acid-base indicators. Acid- base indicators are either weak organic acids or weak organic bases.
The equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use HIn as a simple representation for the complex methyl orange molecule:
The anion of methyl orange, In−, is yellow, and the nonionized form, HIn, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Châtelier’s principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers.
The perceived color of an indicator solution is determined by the ratio of the concentrations of the two species In− and HIn. If most of the indicator (typically about 60−90% or more) is present as In−, the perceived color of the solution is yellow. If most is present as HIn, then the solution color appears red. The Henderson- Hasselbalch equation is useful for understanding the relationship between the pH of an indicator solution and its composition (thus, perceived color):
In solutions where pH > pKa, the logarithmic term must be positive, indicating an excess of the conjugate base form of the indicator (yellow solution). When pH < pKa, the log term must be negative, indicating an excess of the conjugate acid (red solution). When the solution pH is close to the indicator pKa, appreciable amounts of both conjugate partners are present, and the solution color is that of an additive combination of each (yellow and red, yielding orange). The color change interval (or pH interval) for an acid-base indicator is defined as the range of pH values over which a change in color is observed, and for most indicators this range is approximately pKa ± 1.
There are many different acid-base indicators that cover a wide range of pH values and can be used to determine the approximate pH of an unknown solution by a process of elimination. Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pHs. presents several indicators, their colors, and their color-change intervals. |
FIGURE 14.20 Titration curves for strong and weak acids illustrating the proper choice of acid-base indicator. Any of the three indicators will exhibit a reasonably sharp color change at the equivalence point of the strong acid titration, but only phenolphthalein is suitable for use in the weak acid titration.
The titration curves shown in illustrate the choice of a suitable indicator for specific titrations. In the strong acid titration, use of any of the three indicators should yield reasonably sharp color changes and accurate end point determinations. For this titration, the solution pH reaches the lower limit of the methyl orange color change interval after addition of ~24 mL of titrant, at which point the initially red solution would begin to appear orange. When 25 mL of titrant has been added (the equivalence point), the pH is well above the upper limit and the solution will appear yellow. The titration's end point may then be estimated as the volume of titrant that yields a distinct orange-to-yellow color change. This color change would be challenging for most human eyes to precisely discern. More-accurate estimates of the titration end point are possible using either litmus or phenolphthalein, both of which exhibit color change intervals that are encompassed by the steep rise in pH that occurs around the 25.00 mL equivalence point.
The weak acid titration curve in shows that only one of the three indicators is suitable for end point detection. If methyl orange is used in this titration, the solution will undergo a gradual red-to-orange-to- yellow color change over a relatively large volume interval (0–6 mL), completing the color change well before the equivalence point (25 mL) has been reached. Use of litmus would show a color change that begins after adding 7–8 mL of titrant and ends just before the equivalence point. Phenolphthalein, on the other hand, exhibits a color change interval that nicely brackets the abrupt change in pH occurring at the titration's equivalence point. A sharp color change from colorless to pink will be observed within a very small volume interval around the equivalence point.
Key Terms
acid ionization reaction involving the transfer of a proton from an acid to water, yielding hydronium ions and the conjugate base of the acid
acid ionization constant (Ka) equilibrium constant
for an acid ionization reaction
acid-base indicator weak acid or base whose conjugate partner imparts a different solution color; used in visual assessments of solution pH
acidic a solution in which [H3O+] > [OH−]
amphiprotic species that may either donate or accept a proton in a Bronsted-Lowry acid-base reaction
amphoteric species that can act as either an acid
or a base
autoionization reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions
base ionization reaction involving the transfer of a
proton from water to a base, yielding hydroxide ions and the conjugate acid of the base
base ionization constant (Kb) equilibrium
constant for a base ionization reaction basic a solution in which [H3O+] < [OH−] Brønsted-Lowry acid proton donor Brønsted-Lowry base proton acceptor
buffer mixture of appreciable amounts of a weak acid-base pair the pH of a buffer resists change when small amounts of acid or base are added
buffer capacity amount of an acid or base that can
be added to a volume of a buffer solution before its pH changes significantly (usually by one pH unit)
color-change interval range in pH over which the
color change of an indicator is observed
conjugate acid substance formed when a base gains a proton
conjugate base substance formed when an acid |
loses a proton
diprotic acid acid containing two ionizable hydrogen atoms per molecule
diprotic base base capable of accepting two protons
Henderson-Hasselbalch equation logarithmic version of the acid ionization constant expression, conveniently formatted for calculating the pH of buffer solutions
ion-product constant for water (Kw) equilibrium
constant for the autoionization of water
leveling effect observation that acid-base strength of solutes in a given solvent is limited to that of the solvent’s characteristic acid and base species (in water, hydronium and hydroxide ions, respectively)
monoprotic acid acid containing one ionizable
hydrogen atom per molecule
neutral describes a solution in which [H3O+] = [OH−]
oxyacid ternary compound with acidic properties, molecules of which contain a central nonmetallic atom bonded to one or more O atoms, at least one of which is bonded to an ionizable H atom
percent ionization ratio of the concentration of
ionized acid to initial acid concentration expressed as a percentage
pH logarithmic measure of the concentration of
hydronium ions in a solution
pOH logarithmic measure of the concentration of hydroxide ions in a solution
stepwise ionization process in which a polyprotic acid is ionized by losing protons sequentially titration curve plot of some sample property (such
as pH) versus volume of added titrant
triprotic acid acid that contains three ionizable hydrogen atoms per molecule |
A compound that can donate a proton (a hydrogen ion) to another compound is called a Brønsted- Lowry acid. The compound that accepts the proton is called a Brønsted-Lowry base. The species remaining after a Brønsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Brønsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, H3O+, and the hydroxide ion, OH− when it undergoes autoionization: |
Concentrations of hydronium and hydroxide ions in aqueous media are often represented as logarithmic pH and pOH values, respectively. At 25 °C, the autoprotolysis equilibrium for water requires the sum of pH and pOH to equal 14 for any aqueous solution. The relative concentrations of hydronium and hydroxide ion in a solution define its status as acidic ([H3O+] > [OH−]), basic ([H3O+] < [OH−]), or neutral ([H3O+] = [OH−]). At 25 °C, a pH < 7 indicates an acidic solution, a pH > 7 a basic solution, and a pH = 7 a neutral solution.
The relative strengths of acids and bases are reflected in the magnitudes of their ionization constants; the stronger the acid or base, the larger its ionization constant. A reciprocal relation exists
between the strengths of a conjugate acid-base pair: the stronger the acid, the weaker its conjugate base. Water exerts a leveling effect on dissolved acids or bases, reacting completely to generate its characteristic hydronium and hydroxide ions (the strongest acid and base that may exist in water). The strengths of the binary acids increase from left to right across a period of the periodic table (CH4 < NH3
< H2O < HF), and they increase down a group (HF < HCl < HBr < HI). The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases (H2SO3 < H2SO4). The strengths of oxyacids also increase as the electronegativity of the central element increases [H2SeO4 < H2SO4].
The ions composing salts may possess acidic or basic character, ionizing when dissolved in water to yield acidic or basic solutions. Acidic cations are typically the conjugate partners of weak bases, and basic anions are the conjugate partners of weak acids. Many metal ions bond to water molecules when dissolved to yield complex ions that may function as acids.
An acid that contains more than one ionizable proton is a polyprotic acid. These acids undergo stepwise ionization reactions involving the transfer of single protons. The ionization constants for polyprotic acids decrease with each subsequent step; these decreases typically are large enough to permit simple equilibrium calculations that treat each step separately.
Solutions that contain appreciable amounts of a weak conjugate acid-base pair are called buffers. A buffered solution will experience only slight changes in pH when small amounts of acid or base are added. Addition of large amounts of acid or base can exceed the buffer capacity, consuming most of one |
Write equations that show NH3 as both a conjugate acid and a conjugate base.
Write equations that show acting both as an acid and as a base.
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid: (a)
HCl
NH3
CH3CO2H |
H4N2
CH3OH
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
(a)
(b)
(c)
(d)
(e)
(f )
(g)
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
(a)
(b)
(c)
(d)
(e)
(f )
(g)
What are amphiprotic species? Illustrate with suitable equations.
State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species:
H2O |
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
0.200 M HCl
0.0143 M NaOH
3.0 M HNO3
0.0031 M Ca(OH)2
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
0.000259 M HClO4
0.21 M NaOH
0.000071 M Ba(OH)2
2.5 M KOH
What are the pH and pOH of a solution of 2.0 M HCl, which ionizes completely?
What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH. See for useful information.
Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See for useful information.
The hydronium ion concentration in a sample of rainwater is found to be 1.7 10−6 M at 25 °C. What is the
concentration of hydroxide ions in the rainwater?
The hydroxide ion concentration in household ammonia is 3.2 10−3 M at 25 °C. What is the concentration of hydronium ions in the solution?
Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.
Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.
Use this list of important industrial compounds (and ) to answer the following questions regarding: Ca(OH)2, CH3CO2H, HCl, H2CO3, HF, HNO2, HNO3, H3PO4, H2SO4, NH3, NaOH, Na2CO3.
Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases.
Identify the compounds that can behave as Brønsted-Lowry acids with strengths lying between those of H3O+ and H2O.
Identify the compounds that can behave as Brønsted-Lowry bases with strengths lying between those of H2O and OH−.
The odor of vinegar is due to the presence of acetic acid, CH3CO2H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this acid.
Household ammonia is a solution of the weak base NH3 in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this base.
Explain why the ionization constant, Ka, for H2SO4 is larger than the ionization constant for H2SO3.
Explain why the ionization constant, Ka, for HI is larger than the ionization constant for HF.
Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs.
Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, Cu(NO3)2, a compound
that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO3 with CuO.
What is the ionization constant at 25 °C for the weak acid the conjugate acid of the weak base
CH3NH2, Kb = 4.4 10−4.
What is the ionization constant at 25 °C for the weak acid the conjugate acid of the weak base (CH3)2NH, Kb = 5.9 10−4?
Which base, CH3NH2 or (CH3)2NH, is the stronger base? Which conjugate acid, or
, is the stronger acid?
Which is the stronger acid, or HBrO?
Which is the stronger base, (CH3)3N or
Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.
H2O or HF
B(OH)3 or Al(OH)3
or
NH3 or H2S
H2O or H2Te
Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each.
or
NH3 or H2O
PH3 or HI
NH3 or PH3
H2S or HBr
Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
acidity: HCl, HBr, HI
basicity: H2O, OH−, H−, Cl−
basicity: Mg(OH)2, Si(OH)4, ClO3(OH) (Hint: Formula could also be written as HClO4.)
acidity: HF, H2O, NH3, CH4
Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
acidity: NaHSO3, NaHSeO3, NaHSO4
basicity:
acidity: HOCl, HOBr, HOI
acidity: HOCl, HOClO, HOClO2, HOClO3
basicity: HS−, HTe−,
basicity: BrO−,
Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F− or CN−, is the stronger base?
The active ingredient formed by aspirin in the body is salicylic acid, C6H4OH(CO2H). The carboxyl group (−CO2H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001-M aqueous solution of C6H4OH(CO2H).
Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water
directly proportional or inversely proportional? Explain your answer.
What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid or base?
Which of the following will increase the percent of NH3 that is converted to the ammonium ion in water?
addition of NaOH
addition of HCl
addition of NH4Cl
Which of the following will increase the percentage of HF that is converted to the fluoride ion in water?
addition of NaOH
addition of HCl
addition of NaF |
Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
(a) F−
(b)
(c)
(d)
(e)
(f ) (as a base)
Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
HTe− (as a base)
(as a base)
(as a base) (e)
(f ) (as a base)
Using the Ka value of 1.4 10−5, place in the correct location in .
Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected.
0.0092 M HClO, a weak acid
0.0784 M C6H5NH2, a weak base
0.0810 M HCN, a weak acid
0.11 M (CH3)3N, a weak base
0.120 M a weak acid, Ka = 1.6 10−7
Propionic acid, C2H5CO2H (Ka = 1.34 10−5), is used in the manufacture of calcium propionate, a food preservative. What is the pH of a 0.698-M solution of C2H5CO2H?
White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/ cm3, what is the pH?
The ionization constant of lactic acid, CH3CH(OH)CO2H, an acid found in the blood after strenuous exercise, is 1.36 10−4. If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?
Nicotine, C10H14N2, is a base that will accept two protons (Kb1 = 7 10−7, Kb2 = 1.4 10−11). What is the concentration of each species present in a 0.050-M solution of nicotine?
The pH of a 0.23-M solution of HF is 1.92. Determine Ka for HF from these data.
The pH of a 0.15-M solution of is 1.43. Determine Ka for from these data.
The pH of a 0.10-M solution of caffeine is 11.70. Determine Kb for caffeine from these data:
The pH of a solution of household ammonia, a 0.950 M solution of NH3, is 11.612. Determine Kb for NH3 from these data.
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
Al(NO3)3
RbI
KHCO2
CH3NH3Br
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
FeCl3
K2CO3
NH4Br
KClO4
Novocaine, C13H21O2N2Cl, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is 7 10−6. Is a solution of novocaine acidic or basic? What are [H3O+], [OH−], and pH of a 2.0% solution by mass of novocaine, assuming that the density of the solution is 1.0 g/mL.
Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134-M solution of H2CO3, a diprotic acid: [OH−], [H2CO3],
No calculations are needed to answer this question.
Calculate the concentration of each species present in a 0.050-M solution of H2S.
Calculate the concentration of each species present in a 0.010-M solution of phthalic acid, C6H4(CO2H)2.
Salicylic acid, HOC6H4CO2H, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.
Both functional groups of salicylic acid ionize in water, with Ka = 1.0 10−3 for the—CO2H group and
4.2 10−13 for the −OH group. What is the pH of a saturated solution of the acid (solubility = 1.8 g/L).
Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, CH3CO2C6H4CO2H. The −CO2H functional group is still present, but its acidity is reduced, Ka = 3.0 10−4. What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).
The ion HTe− is an amphiprotic species; it can act as either an acid or a base.
What is Ka for the acid reaction of HTe− with H2O?
What is Kb for the reaction in which HTe− functions as a base in water?
Demonstrate whether or not the second ionization of H2Te can be neglected in the calculation of [HTe−] in a 0.10 M solution of H2Te.
Explain why a buffer can be prepared from a mixture of NH4Cl and NaOH but not from NH3 and NaOH.
Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid H3PO4 and a salt of its conjugate base NaH2PO4.
Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base NH3 and a salt of its conjugate acid NH4Cl.
What is [H3O+] in a solution of 0.25 M CH3CO2H and 0.030 M NaCH3CO2? |
What is [OH−] in a solution of 1.25 M NH3 and 0.78 M NH4NO3?
What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:
HCl
KCH3CO2
NaCl
KOH
CH3CO2H
What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:
KI
NH3
HI
NaOH
NH4Cl
What will be the pH of a buffer solution prepared from 0.20 mol NH3, 0.40 mol NH4NO3, and just enough water to give 1.00 L of solution?
Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of KH2PO4, and enough water to make 0.500 L of solution.
How much solid NaCH3CO2•3H2O must be added to 0.300 L of a 0.50-M acetic acid solution to give a buffer with a pH of 5.00? (Hint: Assume a negligible change in volume as the solid is added.)
What mass of NH4Cl must be added to 0.750 L of a 0.100-M solution of NH3 to give a buffer solution with a pH of 9.26? (Hint: Assume a negligible change in volume as the solid is added.)
A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use
1.80 10−5 as Ka for acetic acid.
What is the pH of the solution?
Is the solution acidic or basic?
What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to 0.200 L of the original buffer?
A 5.36–g sample of NH4Cl was added to 25.0 mL of 1.00 M NaOH and the resulting solution diluted to 0.100 L.
What is the pH of this buffer solution?
Is the solution acidic or basic?
What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to the solution?
Explain how to choose the appropriate acid-base indicator for the titration of a weak base with a strong acid.
Explain why an acid-base indicator changes color over a range of pH values rather than at a specific pH.
Calculate the pH at the following points in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (Ka = 9.8 10−5) with 0.100 M KOH.
no KOH added
20 mL of KOH solution added
39 mL of KOH solution added
40 mL of KOH solution added
41 mL of KOH solution added
The indicator dinitrophenol is an acid with a Ka of 1.1 10−4. In a 1.0 10−4-M solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from 10% ionized (colorless) to 90% ionized (yellow). |
INTRODUCTION The mineral fluorite, CaF2 , is commonly used as a semiprecious stone in many types of jewelry because of its striking appearance. Deposits of fluorite are formed through a process called hydrothermal precipitation in which calcium and fluoride ions dissolved in groundwater combine to produce insoluble CaF2 in response to some change in solution conditions. For example, a decrease in temperature may trigger fluorite precipitation if its solubility is exceeded at the lower temperature. Because fluoride ion is a weak base, its solubility is also affected by solution pH, and so geologic or other processes that change groundwater pH will also affect the precipitation of fluorite. This chapter extends the equilibrium discussion of other chapters by addressing some additional reaction classes (including precipitation) and systems involving coupled equilibrium reactions.
Precipitation and Dissolution
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Write chemical equations and equilibrium expressions representing solubility equilibria
Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations
Solubility equilibria are established when the dissolution and precipitation of a solute species occur at equal
rates. These equilibria underlie many natural and technological processes, ranging from tooth decay to water purification. An understanding of the factors affecting compound solubility is, therefore, essential to the effective management of these processes. This section applies previously introduced equilibrium concepts and tools to systems involving dissolution and precipitation.
The Solubility Product
Recall from the chapter on solutions that the solubility of a substance can vary from essentially zero (insoluble or sparingly soluble) to infinity (miscible). A solute with finite solubility can yield a saturated solution when it is added to a solvent in an amount exceeding its solubility, resulting in a heterogeneous mixture of the saturated solution and the excess, undissolved solute. For example, a saturated solution of silver chloride is one in which the equilibrium shown below has been established.
In this solution, an excess of solid AgCl dissolves and dissociates to produce aqueous Ag+ and Cl– ions at the same rate that these aqueous ions combine and precipitate to form solid AgCl (). Because silver chloride is a sparingly soluble salt, the equilibrium concentration of its dissolved ions in the solution is relatively low.
FIGURE 15.2 Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of Ag+ and Cl– ions in equilibrium with undissolved silver chloride.
The equilibrium constant for solubility equilibria such as this one is called the solubility product constant,
Ksp, in this case
Recall that only gases and solutes are represented in equilibrium constant expressions, so the Ksp does not include a term for the undissolved AgCl. A listing of solubility product constants for several sparingly soluble compounds is provided in .
Writing Equations and Solubility Products
Write the dissolution equation and the solubility product expression for each of the following slightly soluble ionic compounds:
AgI, silver iodide, a solid with antiseptic properties
CaCO3, calcium carbonate, the active ingredient in many over-the-counter chewable antacids
Mg(OH)2, magnesium hydroxide, the active ingredient in Milk of Magnesia
Mg(NH4)PO4, magnesium ammonium phosphate, an essentially insoluble substance used in tests for |
Ksp and Solubility
The Ksp of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation, for example:
For cases such as these, one may derive Ksp values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound’s molar solubility, measured as moles of dissolved solute per liter of saturated solution. |
The concentration of Ca2+ in a saturated solution of CaF2 is 2.15 10–4 M. What is the solubility product of fluorite?
Solution
According to the stoichiometry of the dissolution equation, the fluoride ion molarity of a CaF2 solution is equal to twice its calcium ion molarity: |
Determination of Molar Solubility from Ksp
The Ksp of copper(I) bromide, CuBr, is 6.3 10–9. Calculate the molar solubility of copper bromide.
Solution
The dissolution equation and solubility product expression are |
Since the dissolution stoichiometry shows one mole of copper(I) ion and one mole of bromide ion are produced for each moles of Br dissolved, the molar solubility of CuBr is 7.9 10–5 M.
Check Your Learning
The Ksp of AgI is 1.5 10–16. Calculate the molar solubility of silver iodide.
Answer:
10–8 M
Determination of Molar Solubility from Ksp
The Ksp of calcium hydroxide, Ca(OH)2, is 1.3 10–6. Calculate the molar solubility of calcium hydroxide.
Solution
The dissolution equation and solubility product expression are |
As defined in the ICE table, x is the molarity of calcium ion in the saturated solution. The dissolution stoichiometry shows a 1:1 relation between moles of calcium ion in solution and moles of compound dissolved, and so, the molar solubility of Ca(OH)2 is 6.9 10–3 M.
Check Your Learning
The Ksp of PbI2 is 1.4 10–8. Calculate the molar solubility of lead(II) iodide.
Answer:
1.5 10–3 M
Determination of Ksp from Gram Solubility
Many of the pigments used by artists in oil-based paints () are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow, PbCrO4, is 4.6 10–6 g/L. Determine the solubility product for PbCrO4.
FIGURE 15.3 Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow (PbCrO4), examples include Prussian blue (Fe7(CN)18), the reddish-orange color vermilion (HgS), and green color veridian (Cr2O3). (credit: Sonny Abesamis)
Solution
Before calculating the solubility product, the provided solubility must be converted to molarity: |
Calculating the Solubility of Hg2Cl2
Calomel, Hg2Cl2, is a compound composed of the diatomic ion of mercury(I), and chloride ions, Cl–. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel has a very low solubility, as suggested by its very small Ksp:
Calculate the molar solubility of Hg2Cl2.
Solution
The dissolution stoichiometry shows a 1:1 relation between the amount of compound dissolved and the amount of mercury(I) ions, and so the molar solubility of Hg2Cl2 is equal to the concentration of ions
Following the ICE approach results in |
The dissolution stoichiometry shows the molar solubility of Hg2Cl2 is equal to or 6.5 10–7 M.
Check Your Learning
Determine the molar solubility of MgF2 from its solubility product: Ksp = 6.4 10–9.
Answer:
1.2 10–3 M
Using Barium Sulfate for Medical Imaging
Various types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the Ksp of barium sulfate is 2.3 10–8, very little of it dissolves as it coats the lining of the patient’s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray ().
FIGURE 15.4 A suspension of barium sulfate coats the intestinal tract, permitting greater visual detail than a traditional X-ray. (credit modification of work by “glitzy queen00”/Wikimedia Commons)
Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn’s disease, and ulcers in addition to other conditions.
Visit this for more information on how barium is used in medical diagnoses and which conditions it is used to diagnose. |
It is important to realize that this equilibrium is established in any aqueous solution containing Ca2+ and CO32– ions, not just in a solution formed by saturating water with calcium carbonate. Consider, for example, mixing aqueous solutions of the soluble compounds sodium carbonate and calcium nitrate. If the concentrations of calcium and carbonate ions in the mixture do not yield a reaction quotient, Qsp, that exceeds the solubility product, Ksp, then no precipitation will occur. If the ion concentrations yield a reaction quotient greater than the solubility product, then precipitation will occur, lowering those concentrations until equilibrium is established (Qsp = Ksp). The comparison of Qsp to Ksp to predict precipitation is an example of the general approach to predicting the direction of a reaction first introduced in the chapter on equilibrium. For the specific case of solubility equilibria:
Qsp < Ksp: the reaction proceeds in the forward direction (solution is not saturated; no precipitation observed) Qsp > Ksp: the reaction proceeds in the reverse direction (solution is supersaturated; precipitation will occur) This predictive strategy and related calculations are demonstrated in the next few example exercises.
Precipitation of Mg(OH)2
The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of lime, Ca(OH)2, a readily available inexpensive source of OH– ion:
The concentration of Mg2+(aq) in sea water is 0.0537 M. Will Mg(OH)2 precipitate when enough Ca(OH)2 is added to give a [OH–] of 0.0010 M?
Solution
Calculation of the reaction quotient under these conditions is shown here:
Because Q is greater than Ksp (Q = 5.4 10–8 is larger than Ksp = 8.9 10–12), the reverse reaction will proceed, precipitating magnesium hydroxide until the dissolved ion concentrations have been sufficiently lowered, so that Qsp = Ksp.
Check Your Learning
Predict whether CaHPO4 will precipitate from a solution with [Ca2+] = 0.0001 M and = 0.001 M.
Answer:
No precipitation of CaHPO4; Q = 1 10–7, which is less than Ksp (7 × 10–7)
Precipitation of AgCl
Does silver chloride precipitate when equal volumes of a 2.0 10–4-M solution of AgNO3 and a 2.0 10–4-M solution of NaCl are mixed?
Solution
The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:
The solubility product is 1.6 10–10 (see ).
AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO3 and NaCl is greater than Ksp. Because the volume doubles when equal volumes of AgNO3 and NaCl solutions are mixed, each concentration is reduced to half its initial value |
AgCl will precipitate from the mixture until the dissolution equilibrium is established, with Q equal to Ksp.
Check Your Learning
Will KClO4 precipitate when 20 mL of a 0.050-M solution of K+ is added to 80 mL of a 0.50-M solution of (Hint: Use the dilution equation to calculate the concentrations of potassium and perchlorate ions in
the mixture.) |
Precipitation of Calcium Oxalate
Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, for this purpose (). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC2O4·H2O (calcium oxalate monohydrate). The concentration of Ca2+ in a sample of blood serum is 2.2 10–3 M. What concentration of ion must be established before CaC2O4·H2O begins to precipitate?
FIGURE 15.5 Anticoagulants can be added to blood that will combine with the Ca2+ ions in blood serum and prevent the blood from clotting. (credit: modification of work by Neeta Lind)
Solution
The equilibrium expression is: |
conditions.
Check Your Learning
If a solution contains 0.0020 mol of per liter, what concentration of Ag+ ion must be reached by adding solid AgNO3 before Ag2CrO4 begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.
Answer:
6.7 10–5 M
Concentrations Following Precipitation
Clothing washed in water that has a manganese [Mn2+(aq)] concentration exceeding 0.1 mg/L (1.8 10–6 M) may be stained by the manganese upon oxidation, but the amount of Mn2+ in the water can be decreased by adding a base to precipitate Mn(OH)2. What pH is required to keep [Mn2+] equal to 1.8 10–6 M?
Solution
The dissolution of Mn(OH)2 is described by the equation: |
(final result rounded to one significant digit, limited by the certainty of the Ksp)
Check Your Learning
The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of Ca(OH)2. The concentration of Mg2+(aq) in sea water is 5.37 10–2 M. Calculate the pH at which [Mg2+] is decreased to 1.0 10–5 M
Answer:
10.97
In solutions containing two or more ions that may form insoluble compounds with the same counter ion, an experimental strategy called selective precipitation may be used to remove individual ions from solution. By increasing the counter ion concentration in a controlled manner, ions in solution may be precipitated individually, assuming their compound solubilities are adequately different. In solutions with equal concentrations of target ions, the ion forming the least soluble compound will precipitate first (at the lowest concentration of counter ion), with the other ions subsequently precipitating as their compound’s solubilities are reached. As an illustration of this technique, the next example exercise describes separation of a two halide ions via precipitation of one as a silver salt.
Chemistry in Everyday Life
The Role of Precipitation in Wastewater Treatment
Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.
FIGURE 15.6 Wastewater treatment facilities, such as this one, remove contaminants from wastewater before the water is released back into the natural environment. (credit: “eutrophication&hypoxia”/Wikimedia Commons)
One common way to remove phosphates from water is by the addition of calcium hydroxide, or lime, Ca(OH)2. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca5(PO4)3OH, which then precipitates out of the solution:
Because the amount of calcium ion added does not result in exceeding the solubility products for other calcium salts, the anions of those salts remain behind in the wastewater. The precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of CO2 in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.
View this for more information on how phosphorus is removed from wastewater.
Precipitation of Silver Halides
A solution contains 0.00010 mol of KBr and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?
Solution
The two equilibria involved are:
If the solution contained about equal concentrations of Cl– and Br–, then the silver salt with the smaller Ksp (AgBr) would precipitate first. The concentrations are not equal, however, so the [Ag+] at which AgCl begins to precipitate and the [Ag+] at which AgBr begins to precipitate must be calculated. The salt that forms at the lower [Ag+] precipitates first.
AgBr precipitates when Q equals Ksp for AgBr |
AgCl begins to precipitate when [Ag+] is 1.6 10–9 M.
AgCl begins to precipitate at a lower [Ag+] than AgBr, so AgCl begins to precipitate first. Note the chloride ion concentration of the initial mixture was significantly greater than the bromide ion concentration, and so silver chloride precipitated first despite having a Ksp greater than that of silver bromide.
Check Your Learning
If silver nitrate solution is added to a solution which is 0.050 M in both Cl– and Br– ions, at what [Ag+] would precipitation begin, and what would be the formula of the precipitate?
Answer:
[Ag+] = 1.0 10–11 M; AgBr precipitates first
Common Ion Effect
Compared with pure water, the solubility of an ionic compound is less in aqueous solutions containing a common ion (one also produced by dissolution of the ionic compound). This is an example of a phenomenon known as the common ion effect, which is a consequence of the law of mass action that may be explained using Le ChÂtelier’s principle. Consider the dissolution of silver iodide:
This solubility equilibrium may be shifted left by the addition of either silver(I) or iodide ions, resulting in the precipitation of AgI and lowered concentrations of dissolved Ag+ and I–. In solutions that already contain either of these ions, less AgI may be dissolved than in solutions without these ions.
This effect may also be explained in terms of mass action as represented in the solubility product expression:
The mathematical product of silver(I) and iodide ion molarities is constant in an equilibrium mixture regardless of the source of the ions, and so an increase in one ion’s concentration must be balanced by a proportional decrease in the other.
LINK TO LEARNING
View this to explore various aspects of the common ion effect.
Common Ion Effect on Solubility
What is the effect on the amount of solid Mg(OH)2 and the concentrations of Mg2+ and OH– when each of the following are added to a saturated solution of Mg(OH)2?
MgCl2
KOH
NaNO3
Mg(OH)2
Solution
The solubility equilibrium is
Adding a common ion, Mg2+, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of hydroxide ion and increasing the amount of undissolved magnesium hydroxide.
Adding a common ion, OH–, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of magnesium ion and increasing the amount of undissolved magnesium hydroxide.
The added compound does not contain a common ion, and no effect on the magnesium hydroxide solubility equilibrium is expected.
Adding more solid magnesium hydroxide will increase the amount of undissolved compound in the mixture. The solution is already saturated, though, so the concentrations of dissolved magnesium and hydroxide ions will remain the same.
Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of Q, and no shift is required to restore Q to the value of the equilibrium constant.
Check Your Learning
What is the effect on the amount of solid NiCO3 and the concentrations of Ni2+ and when each of the following are added to a saturated solution of NiCO3
Ni(NO3)2
KClO4
NiCO3
K2CO3
Answer:
(a) mass of NiCO3(s) increases, [Ni2+] increases, decreases; (b) no appreciable effect; (c) no effect except to increase the amount of solid NiCO3; (d) mass of NiCO3(s) increases, [Ni2+] decreases,
increases;
Common Ion Effect
Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-M solution of cadmium bromide (CdBr2). The Ksp of CdS is 1.0 10–28.
Solution
This calculation can be performed using the ICE approach: |
Lewis Acids and Bases
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Explain the Lewis model of acid-base chemistry
Write equations for the formation of adducts and complex ions
Perform equilibrium calculations involving formation constants
In 1923, G. N. Lewis proposed a generalized definition of acid-base behavior in which acids and bases are identified by their ability to accept or to donate a pair of electrons and form a coordinate covalent bond.
A coordinate covalent bond (or dative bond) occurs when one of the atoms in the bond provides both bonding electrons. For example, a coordinate covalent bond occurs when a water molecule combines with a hydrogen ion to form a hydronium ion. A coordinate covalent bond also results when an ammonia molecule combines with a hydrogen ion to form an ammonium ion. Both of these equations are shown here.
Reactions involving the formation of coordinate covalent bonds are classified as Lewis acid-base chemistry. The species donating the electron pair that compose the bond is a Lewis base, the species accepting the electron pair is a Lewis acid, and the product of the reaction is a Lewis acid-base adduct. As the two examples above illustrate, Brønsted-Lowry acid-base reactions represent a subcategory of Lewis acid reactions, specifically, those in which the acid species is H+. A few examples involving other Lewis acids and bases are described below.
The boron atom in boron trifluoride, BF3, has only six electrons in its valence shell. Being short of the preferred octet, BF3 is a very good Lewis acid and reacts with many Lewis bases; a fluoride ion is the Lewis base in this reaction, donating one of its lone pairs: |
Another type of Lewis acid-base chemistry involves the formation of a complex ion (or a coordination complex) comprising a central atom, typically a transition metal cation, surrounded by ions or molecules called ligands. These ligands can be neutral molecules like H2O or NH3, or ions such as CN– or OH–. Often, the ligands act as Lewis bases, donating a pair of electrons to the central atom. These types of Lewis acid-base reactions are examples of a broad subdiscipline called coordination chemistry—the topic of another chapter in this text.
The equilibrium constant for the reaction of a metal ion with one or more ligands to form a coordination complex is called a formation constant (Kf) (sometimes called a stability constant). For example, the complex ion |
Alternatively, the reverse reaction (decomposition of the complex ion) can be considered, in which case the equilibrium constant is a dissociation constant (Kd). Per the relation between equilibrium constants for reciprocal reactions described, the dissociation constant is the mathematical inverse of the formation constant, Kd = K –1. A tabulation of formation constants is provided in .
As an example of dissolution by complex ion formation, let us consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a small concentration of Ag+ ([Ag+] = 1.3 10–5 M): |
The large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with NH 3 to form As a consequence, the concentration of silver ions, [Ag + ], is reduced, and the reaction quotient for the dissolution of silver chloride, [Ag + ][Cl – ], falls below the solubility |
Dissociation of a Complex Ion
Calculate the concentration of the silver ion in a solution that initially is 0.10 M with respect to
Solution
Applying the standard ICE approach to this reaction yields the following: |
Because only 1.1% of the dissociates into Ag+ and NH3, the assumption that x is small is justified.
Using this value of x and the relations in the above ICE table allows calculation of all species’ equilibrium concentrations:
The concentration of free silver ion in the solution is 0.0011 M.
Check Your Learning
Calculate the silver ion concentration, [Ag+], of a solution prepared by dissolving 1.00 g of AgNO3 and 10.0 g of KCN in sufficient water to make 1.00 L of solution. (Hint: Because Kf is very large, assume the reaction goes to completion then calculate the [Ag+] produced by dissociation of the complex.) |
Coupled Equilibria
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Describe examples of systems involving two (or more) coupled chemical equilibria
Calculate reactant and product concentrations for coupled equilibrium systems
As discussed in preceding chapters on equilibrium, coupled equilibria involve two or more separate chemical reactions that share one or more reactants or products. This section of this chapter will address solubility equilibria coupled with acid-base and complex-formation reactions.
An environmentally relevant example illustrating the coupling of solubility and acid-base equilibria is the impact of ocean acidification on the health of the ocean’s coral reefs. These reefs are built upon skeletons of sparingly soluble calcium carbonate excreted by colonies of corals (small marine invertebrates). The relevant dissolution equilibrium is |
Inspection of these equilibria shows the carbonate ion is involved in the calcium carbonate dissolution and the acid hydrolysis of bicarbonate ion. Combining the dissolution equation with the reverse of the acid hydrolysis equation yields
The equilibrium constant for this net reaction is much greater than the Ksp for calcium carbonate, indicating its solubility is markedly increased in acidic solutions. As rising carbon dioxide levels in the atmosphere increase the acidity of ocean waters, the calcium carbonate skeletons of coral reefs become more prone to dissolution and subsequently less healthy ().
FIGURE 15.7 Healthy coral reefs (a) support a dense and diverse array of sea life across the ocean food chain. But when coral are unable to adequately build and maintain their calcium carbonate skeletons because of excess ocean acidification, the unhealthy reef (b) is only capable of hosting a small fraction of the species as before, and the local food chain starts to collapse. (credit a: modification of work by NOAA Photo Library; credit b: modification of work by |
The dramatic increase in solubility with increasing acidity described above for calcium carbonate is typical of salts containing basic anions (e.g., carbonate, fluoride, hydroxide, sulfide). Another familiar example is the formation of dental cavities in tooth enamel. The major mineral component of enamel is calcium hydroxyapatite (), a sparingly soluble ionic compound whose dissolution equilibrium is
FIGURE 15.8 Crystal of the mineral hydroxyapatite, Ca5(PO4)3OH, is shown here. The pure compound is white, but like many other minerals, this sample is colored because of the presence of impurities.
This compound dissolved to yield two different basic ions: triprotic phosphate ions |
Of the two basic productions, the hydroxide is, of course, by far the stronger base (it’s the strongest base that can exist in aqueous solution), and so it is the dominant factor providing the compound an acid-dependent solubility. Dental cavities form when the acid waste of bacteria growing on the surface of teeth hastens the dissolution of tooth enamel by reacting completely with the strong base hydroxide, shifting the hydroxyapatite solubility equilibrium to the right. Some toothpastes and mouth rinses contain added NaF or SnF2 that make enamel more acid resistant by replacing the strong base hydroxide with the weak base fluoride:
The weak base fluoride ion reacts only partially with the bacterial acid waste, resulting in a less extensive shift in the solubility equilibrium and an increased resistance to acid dissolution. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information.
Chemistry in Everyday Life
Role of Fluoride in Preventing Tooth Decay
As we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, Ca5(PO4)3F. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride ().
FIGURE 15.9 Fluoride, found in many toothpastes, helps prevent tooth decay (credit: Kerry Ceszyk).
Unfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of 4 ppm (4 mg/L) of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.
The solubility of ionic compounds may also be increased when dissolution is coupled to the formation of a complex ion. For example, aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion
The equations for the dissolution of aluminum hydroxide, the formation of the complex ion, and the combined (net) equation are shown below. As indicated by the relatively large value of K for the net reaction, coupling complex formation with dissolution drastically increases the solubility of Al(OH)3.
Increased Solubility in Acidic Solutions
Compute and compare the molar solublities for aluminum hydroxide, Al(OH)3, dissolved in (a) pure water and
(b) a buffer containing 0.100 M acetic acid and 0.100 M sodium acetate. |
Compared to pure water, the solubility of aluminum hydroxide in this mildly acidic buffer is approximately ten million times greater (though still relatively low).
Check Your Learning
What is the solubility of aluminum hydroxide in a buffer comprised of 0.100 M formic acid and 0.100 M
sodium formate?
Answer:
0.1 M |
What mass of Na2S2O3 is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of
Solution
Two equilibria are involved when silver bromide dissolves in an aqueous thiosulfate solution containing the ion: |
The stoichiometry of the dissolution equilibrium indicates the same concentration of aqueous silver ion will result, 0.00532 M, and the very large value of ensures that essentially all the dissolved silver ion will be complexed by thiosulfate ion: |
The mass of required to dissolve 1.00 g of AgBr in 1.00 L of water is thus 1.68 g + 0.17 g = 1.85 g
Check Your Learning
AgCl(s), silver chloride, has a very low solubility: Ksp = 1.6 10–10. Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed:
Kf = 1.7 107. What mass of NH3 is required to prepare 1.00 L of
solution that will dissolve 2.00 g of AgCl by formation of
Answer:
1.00 L of a solution prepared with 4.81 g NH3 dissolves 2.0 g of AgCl.
Key Terms
common ion effect effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base
complex ion ion consisting of a central atom
surrounding molecules or ions called ligands via coordinate covalent bonds
coordinate covalent bond (also, dative bond)
covalent bond in which both electrons originated from the same atom
coupled equilibria system characterized the
simultaneous establishment of two or more equilibrium reactions sharing one or more reactant or product
dissociation constant (Kd) equilibrium constant
for the decomposition of a complex ion into its components
formation constant (Kf) (also, stability constant)
equilibrium constant for the formation of a |
The equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, Ksp, of the solid. For a heterogeneous equilibrium involving the slightly soluble solid MpXq and its ions Mm+ and Xn–: |
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