Learning Objectives To write chemical equations for water acting as an acid and as a base. Define and use the ion product constant for water, \(K_w\), to calculate concentrations of \(\ce{H3O+}\) and \(\ce{OH-}\) in aqueous solutions. Water (H 2 O) is an interesting compound in many respects. Here, we will consider its ability to behave as an acid or a base. In some circumstances, a water molecule will accept a proton and thus act as a Brønsted-Lowry base . We saw an example in the dissolving of HCl in H 2 O: \[HCl + H_2O(ℓ) \rightarrow H_3O^+(aq)+ Cl^−(aq) \label{Eq1} \] In other circumstances, a water molecule can donate a proton and thus act as a Brønsted-Lowry acid . For example, in the presence of the amide ion, a water molecule donates a proton, making ammonia as a product: \[H_2O(ℓ) + NH^−_2(aq) \rightarrow OH^−(aq) + NH_3(aq) \label{Eq2} \] In this case, \(NH^−_2\) is a Brønsted-Lowry base (the proton acceptor). So, depending on the circumstances, H 2 O can act as either a Brønsted-Lowry acid or a Brønsted-Lowry base. Water is not the only substance that can react as an acid in some cases or a base in others, but it is certainly the most common example—and the most important one. A substance that can either donate or accept a proton, depending on the circumstances, is called an amphiprotic compound. A water molecule can act as an acid or a base even in a sample of pure water. About 6 in every 100 million (6 in 10 8 ) water molecules undergo the following reaction: \[H_2O(ℓ) + H_2O(ℓ) \rightarrow H_3O^+(aq) + OH^−(aq) \label{Eq3} \] This process is called the autoionization of water (Figure \(\PageIndex{1}\)) and occurs in every sample of water, whether it is pure or part of a solution. Autoionization occurs to some extent in any amphiprotic liquid. (For comparison, liquid ammonia undergoes autoionization as well, but only about 1 molecule in a million billion (1 in 10 15 ) reacts with another ammonia molecule.) Example \(\PageIndex{1}\) Identify water as either a Brønsted-Lowry acid or a Brønsted-Lowry base. H 2 O(ℓ) + NO 2 − (aq) → HNO 2 (aq) + OH − (aq) HC 2 H 3 O 2 (aq) + H 2 O(ℓ) → H 3 O + (aq) + C 2 H 3 O 2 − (aq) Solution In this reaction, the water molecule donates a proton to the NO 2 − ion, making OH − (aq). As the proton donor, H 2 O acts as a Brønsted-Lowry acid. In this reaction, the water molecule accepts a proton from HC 2 H 3 O 2 , becoming H 3 O + (aq). As the proton acceptor, H 2 O is a Brønsted-Lowry base. Exercise \(\PageIndex{2}\) Identify water as either a Brønsted-Lowry acid or a Brønsted-Lowry base. HCOOH(aq) + H 2 O(ℓ) → H 3 O + (aq) + HCOO − (aq) H 2 O(ℓ) + PO 4 3 − (aq) → OH − (aq) + HPO 4 2 − (aq) Answer 1. H 2 O acts as the proton acceptor (Brønsted-Lowry base) 2. H 2 O acts as the proton donor (Brønsted-Lowry acid) Dissociation of Water As we have already seen, H 2 O can act as an acid or a base. Within any given sample of water, some \(\ce{H2O}\) molecules are acting as acids, and other \(\ce{H2O}\) molecules are acting as bases. The chemical equation is as follows: \[\color{red}{\underbrace{\ce{H2O}}_{\text{acid}}} + \color{blue}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{<=> H3O^{+} + OH^{−}} \label{Auto} \] Similar to a weak acid, the autoionization of water is an equilibrium process, and is more properly written as follows: \[\ce{H2O(ℓ) + H2O(ℓ) <=> H3O^{+}(aq) + OH^{-}(aq)} \label{Eq7} \] We often use the simplified form of the reaction: \[\ce{H2O(l) <=> H+(aq) + OH−(aq)} \nonumber \] The equilibrium constant for the autoionization of water is referred to as the ion-product for water and is given the symbol \(K_w\). \[K_w = [\ce{H^{+}}][\ce{OH^{-}}] \nonumber \] The ion-product constant for water (\(K_w\)) is the mathematical product of the concentration of hydrogen ions and hydroxide ions. Note that H 2 O is not included in the ion-product expression because it is a pure liquid. The value of \(K_w\) is very small, in accordance with a reaction that favors the reactants. At 25 o C, the experimentally determined value of \(K_w\) in pure water is 1.0×10 −14 . \[K_w = [\ce{H^{+}}][\ce{OH^{−}}] = 1.0 \times 10^{−14} \nonumber \] In a sample of pure water, the concentrations of hydrogen and hydroxide ions are equal to one another. Pure water or any other aqueous solution in which this ratio holds is said to be neutral . To find the molarity of each ion, the square root of \(K_w\) is taken. [H + ] = [OH − ] = 1.0×10 −7 The product of these two concentrations is 1.0×10 −14 \[\color{red}{\ce{[H^{+}]}} \color{black}{\times} \color{blue}{\ce{[OH^{-}]}} \color{black} = (1.0 \times 10^{-7})( 1.0 \times 10^{-7}) = 1.0 \times 10^{-14} \nonumber \] For acids, the concentration of H + or [H + ] is greater than 1.0×10 −7 M For bases, the concentration of OH − or [OH − ] is greater than 1.0×10 −7 M. Aqueous HCl is an example of acidic solution. Hydrogen chloride (HCl) ionizes to produce H + and Cl − ions upon dissolving in water. This increases the concentration of H + ions in the solution. According to Le Chatelier's principle, the equilibrium represented by \[\ce{H2O(l) <=> H^{+}(aq) + OH^{−}(aq)} \nonumber \] \[\ce{HCl(g) -> H^{+}(aq) + Cl^{−}(aq)} \nonumber \] is f orced to the left, towards the reactant. As a result, the concentration of the hydroxide ion decreases. Now, consider KOH (aq), a basic solution . Solid potassium hydroxide (KOH) dissociates in water to yield potassium ions and hydroxide ions. KOH(s) → K + (aq) + OH − (aq) The increase in concentration of the OH − ions will cause a decrease in the concentration of the H + ions. No matter whether the aqueous solution is an acid, a base, or neutral:and the ion-product of [H + ][OH − ] remains constant. For acidic solutions, [H + ] is greater than [OH − ]. For basic solutions, [OH−] is greater than [H + ]. For neutral solutions, [H 3 O + ] = [OH−] = 1.0×10 −7 M This means that if you know \(\ce{[H^{+}]}\) for a solution, you can calculate what \(\ce{[OH^{−}]}\)) has to be for the product to equal \(1.0 \times 10^{−14}\), or if you know \(\ce{[OH^{−}]}\)), you can calculate \(\ce{[H^{+}]}\). This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of \(K_w\). \[K_w = \color{red}{\ce{[H_3O^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14} \label{eq10} \] Example \(\PageIndex{2}\) Hydrochloric acid (HCl) is a strong acid, meaning it is 100% ionized in solution. What is the [H + ] and the [OH − ] in a solution of 2.0×10 −3 M HCl? Solution Step 1: List the known values and plan the problem. Known [HCl] = 2.0×10 −3 M Kw = 1.0×10 −14 Unknown [H + ]=?M [OH − ]=?M Because HCl is 100% ionized, the concentration of H+ ions in solution will be equal to the original concentration of HCl. Each HCl molecule that was originally present ionizes into one H+ ion and one Cl− ion. The concentration of OH− can then be determined from the [H+] and Kw. Step 2: Solve. [H + ]=2.0×10 −3 M Kw = [H + ][OH − ] = 1.0×10 −14 [OH − ] = Kw/[H + ] = 1.0×10 −14 /2.0×10 −3 = 5.0×10 −12 M Step 3: Think about your result. The [H + ] is much higher than the [OH − ] because the solution is acidic. As with other equilibrium constants, the unit for Kw is customarily omitted. Exercise \(\PageIndex{2}\) Sodium hydroxide (NaOH) is a strong base. What is the [H + ] and the [OH − ] in a 0.001 M NaOH solution at 25 °C? Answer [OH − ] = 0.001M or 1 x 10 -3 M; [H + ]=1×10 − 11 M. Concept Review Exercises Explain how water can act as an acid. Explain how water can act as a base. Answers Under the right conditions, H 2 O can donate a proton, making it a Brønsted-Lowry acid. Under the right conditions, H 2 O can accept a proton, making it a Brønsted-Lowry base. Key Takeaway Water molecules can act as both an acid and a base, depending on the conditions. Exercises Is H 2 O(ℓ) acting as an acid or a base? H 2 O(ℓ) + NH 4 + (aq) → H 3 O + (aq) + NH 3 (aq) Is H 2 O(ℓ) acting as an acid or a base? CH 3 − (aq) + H 2 O(ℓ) → CH 4 (aq) + OH − (aq) In the aqueous solutions of some salts, one of the ions from the salt can react with water molecules. In some C 2 H 3 O 2 − solutions, the following reaction can occur: C 2 H 3 O 2 − (aq) + H 2 O(ℓ) → HC 2 H 3 O 2 (aq) + OH − (aq) Is H 2 O acting as an acid or a base in this reaction? In the aqueous solutions of some salts, one of the ions from the salt can react with water molecules. In some NH 4 + solutions, the following reaction can occur: NH 4 + (aq) + H 2 O → NH 3 (aq) + H 3 O + (aq) Is H 2 O acting as an acid or a base in this reaction? Why is pure water considered neutral? Answers base acid acid base 5. When water ionizes, equal amounts of H + (acid) and OH − (base) are formed, so the solution is neither acidic nor basic: H 2 O(ℓ) → H + (aq) + OH − (aq) [SIDE NOTE: It is rare to truly have pure water. Water exposed to air will usually be slightly acidic because dissolved carbon dioxide gas, or carbonic acid, decreases the pH slightly below 7. Alternatively, dissolved minerals, like calcium carbonate (limestone), can make water slightly basic.]