Learning Objectives 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 We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The reaction of an acid with water is given by the general expression: \[\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq)\] Water is the base that reacts with the acid \(\ce{HA}\), \(\ce{A^{−}}\) is the conjugate base of the acid \(\ce{HA}\), and the hydronium ion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of \(\ce{H3O+}\) and \(\ce{A^{−}}\) when the acid ionizes in water; Figure \(\PageIndex{1}\) lists several strong acids. A weak acid gives small amounts of \(\ce{H3O+}\) and \(\ce{A^{−}}\). Six Strong Acids Six Strong Acids.1 Six Strong Bases Six Strong Bases.1 \(\ce{HClO4}\) perchloric acid \(\ce{LiOH}\) lithium hydroxide \(\ce{HCl}\) hydrochloric acid \(\ce{NaOH}\) sodium hydroxide \(\ce{HBr}\) hydrobromic acid \(\ce{KOH}\) potassium hydroxide \(\ce{HI}\) hydroiodic acid \(\ce{Ca(OH)2}\) calcium hydroxide \(\ce{HNO3}\) nitric acid \(\ce{Sr(OH)2}\) strontium hydroxide \(\ce{H2SO4}\) sulfuric acid \(\ce{Ba(OH)2}\) barium hydroxide The relative strengths of acids may be determined 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, K a . For the reaction of an acid \(\ce{HA}\): \[\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq)\] we write the equation for the ionization constant as: \[K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\] 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 [H 2 O] in the equation. The larger the \(K_a\) of an acid, the larger the concentration of \(\ce{H3O+}\) and \(\ce{A^{−}}\) relative to the concentration of the nonionized acid, \(\ce{HA}\). Thus a stronger acid has a larger ionization constant than does a weaker acid. The ionization constants increase as the strengths of the acids increase. The following data on acid-ionization constants indicate the order of acid strength: \(\ce{CH3CO2H} < \ce{HNO2} < \ce{HSO4-}\) \[ \begin{aligned} \ce{CH3CO2H}(aq) + \ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \quad &K_\ce{a}=1.8×10^{−5} \\[4pt] \ce{HNO2}(aq)+\ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) &K_\ce{a}=4.6×10^{-4} \\[4pt] \ce{HSO4-}(aq)+\ce{H2O}(aq) &⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) & K_\ce{a}=1.2×10^{−2} \end{aligned}\] Another measure of the strength of an acid is its percent ionization . The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100: \[\% \:\ce{ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\% \label{PercentIon} \] Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration. Example \(\PageIndex{1}\): Calculation of Percent Ionization from pH Calculate the percent ionization of a 0.125- M solution of nitrous acid (a weak acid), with a pH of 2.09. Solution The percent ionization for an acid is: \[\ce{\dfrac{[H3O+]_{eq}}{[HNO2]_0}}×100 \] The chemical equation for the dissociation of the nitrous acid is: \[\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{NO2-}(aq)+\ce{H3O+}(aq). \] Since 10 −pH = \(\ce{[H3O+]}\), we find that \(10^{−2.09} = 8.1 \times 10^{−3}\, M\), so that percent ionization (Equation \ref{PercentIon}) is: \[\dfrac{8.1×10^{−3}}{0.125}×100=6.5\% \] Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures. Exercise \(\PageIndex{1}\) Calculate the percent ionization of a 0.10 M solution of acetic acid with a pH of 2.89. Answer 1.3% ionized We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. The reaction of a Brønsted-Lowry base with water is given by: \[\ce{B}(aq)+\ce{H2O}(l)⇌\ce{HB+}(aq)+\ce{OH-}(aq)\] Water is the acid that reacts with the base, \(\ce{HB^{+}}\) is the conjugate acid of the base \(\ce{B}\), and the hydroxide ion is the conjugate base of water. A strong base yields 100% (or very nearly so) of OH − and HB + when it reacts with water; Figure \(\PageIndex{1}\) lists several strong bases. A weak base yields a small proportion of hydroxide ions. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completely when dissolved in water. As we did with acids, we can measure the relative strengths of bases by measuring their base-ionization constant ( K b ) 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, \(\ce{B}\): \[\ce{B}(aq)+\ce{H2O}(l)⇌\ce{HB+}(aq)+\ce{OH-}(aq),\] we write the equation for the ionization constant as: \[K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}}\] where the concentrations are those at equilibrium. Again, we do not include [H 2 O] in the equation because water is the solvent. The chemical reactions and ionization constants of the three bases shown are: \[ \begin{aligned} \ce{NO2-}(aq)+\ce{H2O}(l) &⇌\ce{HNO2}(aq)+\ce{OH-}(aq) \quad &K_\ce{b}=2.17×10^{−11} \\[4pt] \ce{CH3CO2-}(aq)+\ce{H2O}(l) &⇌\ce{CH3CO2H}(aq)+\ce{OH-}(aq) &K_\ce{b}=5.6×10^{−10} \\[4pt] \ce{NH3}(aq)+\ce{H2O}(l) &⇌\ce{NH4+}(aq)+\ce{OH-}(aq) &K_\ce{b}=1.8×10^{−5} \end{aligned}\] A table of ionization constants of weak bases appears in Table E2. As with acids, percent ionization can be measured for basic solutions, but will vary depending on the base ionization constant and the initial concentration of the solution. Consider the ionization reactions for a conjugate acid-base pair, HA − A − : \[\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq) \hspace{20px} K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\] \[\ce{A-}(aq)+\ce{H2O}(l)⇌\ce{OH-}(aq)+\ce{HA}(aq) \hspace{20px} K_\ce{b}=\ce{\dfrac{[HA][OH]}{[A- ]}}\] Adding these two chemical equations yields the equation for the autoionization for water: \[\cancel{\ce{HA}(aq)}+\ce{H2O}(l)+\cancel{\ce{A-}(aq)}+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\cancel{\ce{A-}(aq)}+\ce{OH-}(aq)+\cancel{\ce{HA}(aq)}\] \[\ce{2H2O}(l)⇌\ce{H3O+}(aq)+\ce{OH-}(aq)\] As shown in the previous chapter on equilibrium, the K expression for a chemical equation derived from adding two or more other equations is the mathematical product of the other equations’ K expressions. Multiplying the mass-action expressions together and cancelling common terms, we see that: \[K_\ce{a}×K_\ce{b}=\ce{\dfrac{[H3O+][A- ]}{[HA]}×\dfrac{[HA][OH- ]}{[A- ]}}=\ce{[H3O+][OH- ]}=K_\ce{w}\] For example, the acid ionization constant of acetic acid (CH 3 COOH) is 1.8 × 10 −5 , and the base ionization constant of its conjugate base, acetate ion (\(\ce{CH3COO-}\)), is 5.6 × 10 −10 . The product of these two constants is indeed equal to K w : \[K_\ce{a}×K_\ce{b}=(1.8×10^{−5})×(5.6×10^{−10})=1.0×10^{−14}=K_\ce{w}\] The extent to which an acid, HA, donates protons to water molecules depends on the strength of the conjugate base, A − , of the acid. If A − is a strong base, any protons that are donated to water molecules are recaptured by A − . Thus there is relatively little A − and \(\ce{H3O+}\) in solution, and the acid, HA, is weak. If A − is a weak base, water binds the protons more strongly, and the solution contains primarily A − and H 3 O + —the acid is strong. Strong acids form very weak conjugate bases, and weak acids form stronger conjugate bases (Figure \(\PageIndex{2}\)). Figure \(\PageIndex{3}\) lists a series of acids and bases in order of the decreasing strengths of the acids and the corresponding increasing strengths of the bases. The acid and base in a given row are conjugate to each other. The first six acids in Figure \(\PageIndex{3}\) are the most common strong acids. These acids are completely dissociated in aqueous solution. The conjugate bases of these acids are weaker bases than water. When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base. Those acids that lie between the hydronium ion and water in Figure \(\PageIndex{3}\) form conjugate bases that can compete with water for possession of a proton. Both hydronium ions and nonionized acid molecules are present in equilibrium in a solution of one of these acids. Compounds that are weaker acids than water (those found below water in the column of acids) in Figure \(\PageIndex{3}\) exhibit no observable acidic behavior when dissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate base were formed, it would react with water to re-form the acid. The extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the base relative to that of the hydroxide ion, as shown in the last column in Figure \(\PageIndex{3}\). A strong base, such as one of those lying below hydroxide ion, accepts protons from water to yield 100% of the conjugate acid and hydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution. Example \(\PageIndex{2}\): The Product K a × K b = K w Use the K b for the nitrite ion, \(\ce{NO2-}\), to calculate the K a for its conjugate acid. Solution K b for \(\ce{NO2-}\) is given in this section as 2.17 × 10 −11 . The conjugate acid of \(\ce{NO2-}\) is HNO 2 ; K a for HNO 2 can be calculated using the relationship: \[K_\ce{a}×K_\ce{b}=1.0×10^{−14}=K_\ce{w} \] Solving for K a , we get: \[K_\ce{a}=\dfrac{K_\ce{w}}{K_\ce{b}}=\dfrac{1.0×10^{−14}}{2.17×10^{−11}}=4.6×10^{−4} \] This answer can be verified by finding the K a for HNO 2 in Table E1 Exercise \(\PageIndex{2}\) We can determine the relative acid strengths of \(\ce{NH4+}\) and HCN by comparing their ionization constants. The ionization constant of HCN is given in Table E1 as 4.9 × 10 −10 . The ionization constant of \(\ce{NH4+}\) is not listed, but the ionization constant of its conjugate base, NH 3 , is listed as 1.8 × 10 −5 . Determine the ionization constant of \(\ce{NH4+}\), and decide which is the stronger acid, HCN or \(\ce{NH4+}\). Answer \(\ce{NH4+}\) is the slightly stronger acid ( K a for \(\ce{NH4+}\) = 5.6 × 10 −10 ). The Ionization of Weak Acids and Weak Bases Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid). Acetic acid (\(\ce{CH3CO2H}\)) is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation: \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq)\] giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. This equilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure \(\PageIndex{4}\)). The remaining weak acid is present in the nonionized form. For acetic acid, at equilibrium: \[K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}=1.8 \times 10^{−5}\] Ionization Reaction Ka at 25 °C \(\ce{HSO4- + H2O ⇌ H3O+ + SO4^2-}\) 1.2 × 10−2 \(\ce{HF + H2O ⇌ H3O+ + F-}\) 3.5 × 10−4 \(\ce{HNO2 + H2O ⇌ H3O+ + NO2-}\) 4.6 × 10−4 \(\ce{HNCO + H2O ⇌ H3O+ + NCO-}\) 2 × 10−4 \(\ce{HCO2H + H2O ⇌ H3O+ + HCO2-}\) 1.8 × 10−4 \(\ce{CH3CO2H + H2O ⇌ H3O+ + CH3CO2-}\) 1.8 × 10−5 \(\ce{HCIO + H2O ⇌ H3O+ + CIO-}\) 2.9 × 10−8 \(\ce{HBrO + H2O ⇌ H3O+ + BrO-}\) 2.8 × 10−9 \(\ce{HCN + H2O ⇌ H3O+ + CN-}\) 4.9 × 10−10 Table \(\PageIndex{1}\) gives the ionization constants for several weak acids; additional ionization constants can be found in Table E1. At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base). For example, a solution of the weak base trimethylamine, (CH 3 ) 3 N, in water reacts according to the equation: \[\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq)\] This gives an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids. We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure 14.4.5). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, K b , is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium: \[K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}}\] The ionization constants of several weak bases are given in Table \(\PageIndex{2}\) and Table E2. Ionization Reaction Kb at 25 °C \(\ce{(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-}\) 5.9 × 10−4 \(\ce{CH3NH2 + H2O ⇌ CH3NH3+ + OH-}\) 4.4 × 10−4 \(\ce{(CH3)3N + H2O ⇌ (CH3)3NH+ + OH-}\) 6.3 × 10−5 \(\ce{NH3 + H2O ⇌ NH4+ + OH-}\) 1.8 × 10−5 \(\ce{C6H5NH2 + H2O ⇌ C6N5NH3+ + OH-}\) 4.3 × 10−10 Example \(\PageIndex{3}\): Determination of K a from Equilibrium Concentrations Acetic acid is the principal ingredient in vinegar; that's why it tastes sour. At equilibrium, a solution contains [CH 3 CO 2 H] = 0.0787 M and \(\ce{[H3O+]}=\ce{[CH3CO2- ]}=0.00118\:M\). What is the value of K a for acetic acid? Solution We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction: \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \] \[K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}=\dfrac{(0.00118)(0.00118)}{0.0787}=1.77×10^{−5} \] Exercise \(\PageIndex{3}\) What is the equilibrium constant for the ionization of the \(\ce{HSO4-}\) ion, the weak acid used in some household cleansers: \[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \] In one mixture of NaHSO 4 and Na 2 SO 4 at equilibrium, \(\ce{[H3O+]}\) = 0.027 M ; \(\ce{[HSO4- ]}=0.29\:M\); and \(\ce{[SO4^2- ]}=0.13\:M\). Answer \(K_a\) for \(\ce{HSO_4^-}= 1.2 ×\times 10^{−2}\) Example \(\PageIndex{4}\): Determination of K b from Equilibrium Concentrations Caffeine, C 8 H 10 N 4 O 2 is a weak base. What is the value of K b for caffeine if a solution at equilibrium has [C 8 H 10 N 4 O 2 ] = 0.050 M , \(\ce{[C8H10N4O2H+]}\) = 5.0 × 10 −3 M , and [OH − ] = 2.5 × 10 −3 M ? Solution At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction: \[\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \] \[K_\ce{b}=\ce{\dfrac{[C8H10N4O2H+][OH- ]}{[C8H10N4O2]}}=\dfrac{(5.0×10^{−3})(2.5×10^{−3})}{0.050}=2.5×10^{−4} \] Exercise \(\PageIndex{4}\) What is the equilibrium constant for the ionization of the \(\ce{HPO4^2-}\) ion, a weak base: \[\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H2PO4-}(aq)+\ce{OH-}(aq) \] In a solution containing a mixture of NaH 2 PO 4 and Na 2 HPO 4 at equilibrium, [OH − ] = 1.3 × 10 −6 M ; \(\ce{[H2PO4- ]}=0.042\:M\) ; and \(\ce{[HPO4^2- ]}=0.341\:M\). Answer K b for \(\ce{HPO4^2-}=1.6×10^{−7} \) Example \(\PageIndex{5}\): Determination of K a or K b from pH The pH of a 0.0516- M solution of nitrous acid, \(\ce{HNO2}\), is 2.34. What is its K a ? \[\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) \] Solution We determine an equilibrium constant starting with the initial concentrations of HNO 2 , \(\ce{H3O+}\), and \(\ce{NO2-}\) as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.) We can solve this problem with the following steps in which x is a change in concentration of a species in the reaction: We can summarize the various concentrations and changes as shown here (the concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its concentration): To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate \(\ce{[H3O+]}\), the equilibrium concentration of \(\ce{H3O+}\), from the pH: \[\ce{[H3O+]}=10^{−2.34}=0.0046\:M \] The change in concentration of \(\ce{H3O+}\), \(x_{\ce{[H3O+]}}\), is the difference between the equilibrium concentration of H 3 O + , which we determined from the pH, and the initial concentration, \(\mathrm{[H_3O^+]_i}\). The initial concentration of \(\ce{H3O+}\) is its concentration in pure water, which is so much less than the final concentration that we approximate it as zero (~0). The change in concentration of \(\ce{NO2-}\) is equal to the change in concentration of \(\ce{[H3O+]}\). For each 1 mol of \(\ce{H3O+}\) that forms, 1 mol of \(\ce{NO2-}\) forms. The equilibrium concentration of HNO 2 is equal to its initial concentration plus the change in its concentration. Now we can fill in the ICE table with the concentrations at equilibrium, as shown here: Finally, we calculate the value of the equilibrium constant using the data in the table: \[K_\ce{a}=\ce{\dfrac{[H3O+][NO2- ]}{[HNO2]}}=\dfrac{(0.0046)(0.0046)}{(0.0470)}=4.5×10^{−4} \] Exercise \(\PageIndex{5}\) The pH of a solution of household ammonia, a 0.950- M solution of NH 3, is 11.612. What is K b for NH 3 . Answer \(K_b = 1.8 × 10^{−5}\) Example \(\PageIndex{6}\): Equilibrium Concentrations in a Solution of a Weak Acid Formic acid, HCO 2 H, is the irritant that causes the body’s reaction to ant stings. What is the concentration of hydronium ion and the pH in a 0.534- M solution of formic acid? \[\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−4} \] Soluti on 1. Determine x and equilibrium concentrations . The equilibrium expression is: \[\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \] The concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its change in concentration when setting up the ICE table. The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color): 2. Solve for x and the equilibrium concentrations. At equilibrium: \[\begin{align*} K_\ce{a} &=1.8×10^{−4}=\ce{\dfrac{[H3O+][HCO2- ]}{[HCO2H]}} \\[4pt] &=\dfrac{(x)(x)}{0.534−x}=1.8×10^{−4} \end{align*}\] Now solve for x . Because the initial concentration of acid is reasonably large and K a is very small, we assume that x << 0.534, which permits us to simplify the denominator term as (0.534 − x ) = 0.534. This gives: \[K_\ce{a}=1.8×10^{−4}=\dfrac{x^{2}}{0.534} \] Solve for x as follows: \[ x^2 =0.534×(1.8×10^{−4}) =9.6×10^{−5} \\[4pt] x =\sqrt{9.6×10^{−5}} \\[4pt] =9.8×10^{−3} \] To check the assumption that \(x\) is small compared to 0.534, we calculate: \[\begin{align*} \dfrac{x}{0.534} &=\dfrac{9.8×10^{−3}}{0.534} \\[4pt] &=1.8×10^{−2} \, \textrm{(1.8% of 0.534)} \end{align*}\] x is less than 5% of the initial concentration; the assumption is valid. We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table: \[\begin{align*} \ce{[H3O+]} &=~0+x=0+9.8×10^{−3}\:M. \\[4pt] &=9.8×10^{−3}\:M \end{align*}\] The pH of the solution can be found by taking the negative log of the \(\ce{[H3O+]}\), so: \(pH = −\log(9.8×10^{−3})=2.01\) Exercise \(\PageIndex{6}\): acetic acid Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100- M solution of acetic acid, CH 3 CO 2 H? \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−5} \] Hint Determine \(\ce{[CH3CO2- ]}\) at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or \(\ce{\dfrac{[CH3CO2- ]}{[CH3CO2H]_{initial}}}×100\). Answer percent ionization = 1.3% The following example shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid. Example \(\PageIndex{7}\): Equilibrium Concentrations in a Solution of a Weak Base Find the concentration of hydroxide ion in a 0.25- M solution of trimethylamine, a weak base: \[\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=6.3×10^{−5} \] Solution This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in Example \(\PageIndex{6}\). The reactants and products will be different and the numbers will be different, but the logic will be the same: 1. Determine x and equilibrium concentrations . The table shows the changes and concentrations: 2. Solve for x and the equilibrium concentrations . At equilibrium: \[K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}}=\dfrac{(x)(x)}{0.25−x=}6.3×10^{−5} \] If we assume that x is small relative to 0.25, then we can replace (0.25 − x ) in the preceding equation with 0.25. Solving the simplified equation gives: \[x=4.0×10^{−3} \] This change is less than 5% of the initial concentration (0.25), so the assumption is justified. Recall that, for this computation, x is equal to the equilibrium concentration of hydroxide ion in the solution (see earlier tabulation): \[(\ce{[OH- ]}=~0+x=x=4.0×10^{−3}\:M \] \[=4.0×10^{−3}\:M \] Then calculate pOH as follows: \[\ce{pOH}=−\log(4.3×10^{−3})=2.40 \] Using the relation introduced in the previous section of this chapter: \[\mathrm{pH + pOH=p\mathit{K}_w=14.00} \] permits the computation of pH: \[\mathrm{pH=14.00−pOH=14.00−2.37=11.60} \] Check the work . A check of our arithmetic shows that K b = 6.3 × 10 −5 . Exercise \(\PageIndex{7}\) Show that the calculation in Step 2 of this example gives an x of 4.3 × 10 −3 and the calculation in Step 3 shows K b = 6.3 × 10 −5 . Find the concentration of hydroxide ion in a 0.0325- M solution of ammonia, a weak base with a K b of 1.76 × 10 −5 . Calculate the percent ionization of ammonia, the fraction ionized × 100, or \(\ce{\dfrac{[NH4+]}{[NH3]}}×100 \%\) Answer a \(7.56 × 10^{−4}\, M\), 2.33% Answer b 2.33% Some weak acids and weak bases ionize to such an extent that the simplifying assumption that x is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation. Example \(\PageIndex{8}\): Equilibrium Concentrations in a Solution of a Weak Acid Sodium bisulfate, NaHSO 4 , is used in some household cleansers because it contains the \(\ce{HSO4-}\) ion, a weak acid. What is the pH of a 0.50- M solution of \(\ce{HSO4-}\)? \[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{a}=1.2×10^{−2} \] Soluti on We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of \(\ce{HSO4-}\) so that we can use \(\ce{[H3O+]}\) to determine the pH. As in the previous examples, we can approach the solution by the following steps: 1. Determine x and equilibrium concentrations . This table shows the changes and concentrations: 2. Solve for x and the concentrations . As we begin solving for x , we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of x . At equilibrium: \[K_\ce{a}=1.2×10^{−2}=\ce{\dfrac{[H3O+][SO4^2- ]}{[HSO4- ]}}=\dfrac{(x)(x)}{0.50−x} \] If we assume that x is small and approximate (0.50 − x ) as 0.50, we find: \[x=7.7×10^{−2} \] When we check the assumption, we confirm: \[\dfrac{x}{\mathrm{[HSO_4^- ]_i}} \overset{?}{\le} 0.05 \] which for this system is \[\dfrac{x}{0.50}=\dfrac{7.7×10^{−2}}{0.50}=0.15(15\%) \] The value of \(x\) is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find \(x\). The equation: \[K_\ce{a}=1.2×10^{−2}=\dfrac{(x)(x)}{0.50−x} \] gives \[6.0×10^{−3}−1.2×10^{−2}x=x^{2+} \] or \[x^{2+}+1.2×10^{−2}x−6.0×10^{−3}=0\ \] This equation can be solved using the quadratic formula. For an equation of the form \[ax^{2+} + bx + c=0, \] x is given by the equation: \[x=\dfrac{−b±\sqrt{b^{2+}−4ac}}{2a} \] In this problem, a = 1, b = 1.2 × 10 −3 , and c = −6.0 × 10 −3 . Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root: \[x=7.2×10^{−2} \] Now determine the hydronium ion concentration and the pH: \[\begin{align*} \ce{[H3O+]} &=~0+x=0+7.2×10^{−2}\:M \\[4pt] &=7.2×10^{−2}\:M \end{align*}\] The pH of this solution is: \[\mathrm{pH=−log[H_3O^+]=−log7.2×10^{−2}=1.14} \] Exercise \(\PageIndex{8}\) Show that the quadratic formula gives \(x = 7.2 × 10^{−2}\). Calculate the pH in a 0.010- M solution of caffeine, a weak base: \[\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=2.5×10^{−4} \] Hint It will be necessary to convert [OH − ] to \(\ce{[H3O+]}\) or pOH to pH toward the end of the calculation. Answer pH 11.16 The Relative Strengths of Strong Acids and Bases Strong acids, such as HCl, HBr, and HI, all exhibit the same strength in water. The water molecule is such a strong base compared to the conjugate bases Cl − , Br − , and I − that ionization of these strong acids is essentially complete in aqueous solutions. In solvents less basic than water, we find HCl, HBr, and HI differ markedly in their tendency to give up a proton to the solvent. For example, when dissolved in ethanol (a weaker base than water), the extent of ionization increases in the order HCl < HBr < HI, and so HI is demonstrated to be the strongest of these acids. The inability to discern differences in strength among strong acids dissolved in water is known as the leveling effect of water . Water also exerts a leveling effect on the strengths of strong bases. For example, the oxide ion, O 2− , and the amide ion, \(\ce{NH2-}\), are such strong bases that they react completely with water: \[\ce{O^2-}(aq)+\ce{H2O}(l)⟶\ce{OH-}(aq)+\ce{OH-}(aq)\] \[\ce{NH2-}(aq)+\ce{H2O}(l)⟶\ce{NH3}(aq)+\ce{OH-}(aq)\] Thus, O 2− and \(\ce{NH2-}\) appear to have the same base strength in water; they both give a 100% yield of hydroxide ion. 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 H 2 O < H 2 S < H 2 Se < H 2 Te. 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 CH 4 < NH 3 < H 2 O < HF; across the third row, it is SiH 4 < PH 3 < H 2 S < HCl (see Figure \(\PageIndex{6}\)). Figure \(\PageIndex{6}\): As you move from left to right and down the periodic table, the acid strength increases. As you move from right to left and up, the base strength increases. Compounds containing oxygen and one or more hydroxyl (OH) groups can be acidic, basic, or amphoteric, depending on the position in the periodic table of the central atom E, the atom bonded to the hydroxyl group. Such compounds have the general formula O n E(OH) m , and include sulfuric acid, O 2 S(OH) 2 , sulfurous acid, OS(OH) 2 , nitric acid, O 2 NOH, perchloric acid, O 3 ClOH, 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, H 2 SO 4 , or O 2 S(OH) 2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H 2 SO 3 , or OS(OH) 2 (with a sulfur oxidation number of +4). Likewise nitric acid, HNO 3 , or O 2 NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO 2 , 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 \(\PageIndex{7}\)). 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(H 2 O) 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(H 2 O) 3 (OH) 3 , is converted into the soluble ion, \(\ce{[Al(H2O)2(OH)4]-}\), by reaction with hydroxide ion: \[[\ce{Al(H2O)3(OH)3}(aq)+\ce{OH-}(aq)⇌\ce{H2O}(l)+\ce{[Al(H2O)2(OH)4]-}(aq)\] In this reaction, a proton is transferred from one of the aluminum-bound H 2 O molecules to a hydroxide ion in solution. The Al(H 2 O) 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 \(\ce{[Al(H2O)6]^3+}\) by reaction with hydronium ion: \[\ce{3H3O+}(aq)+\ce{Al(H2O)3(OH)3}(aq)⇌\ce{Al(H2O)6^3+}(aq)+\ce{3H2O}(l)\] In this case, protons are transferred from hydronium ions in solution to Al(H 2 O) 3 (OH) 3 , and the compound functions as a base. Summary The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. Strong bases react with water to quantitatively form hydroxide ions. Weak bases give only small amounts of hydroxide ion. The strengths of the binary acids increase from left to right across a period of the periodic table (CH 4 < NH 3 < H 2 O < 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 (H 2 SO 3 < H 2 SO 4 ). The strengths of oxyacids also increase as the electronegativity of the central element increases [H 2 SeO 4 < H 2 SO 4 ]. Key Equations \(K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\) \(K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}}\) \(K_a \times K_b = 1.0 \times 10^{−14} = K_w \,(\text{at room temperature})\) \(\textrm{Percent ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\) Glossary acid ionization constant ( K a ) equilibrium constant for the ionization of a weak acid base ionization constant ( K b ) equilibrium constant for the ionization of a weak base leveling effect of water any acid stronger than \(\ce{H3O+}\), or any base stronger than OH − will react with water to form \(\ce{H3O+}\), or OH − , respectively; water acts as a base to make all strong acids appear equally strong, and it acts as an acid to make all strong bases appear equally strong oxyacid compound containing a nonmetal and one or more hydroxyl groups percent ionization ratio of the concentration of the ionized acid to the initial acid concentration, times 100