# Ionization of Water

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 14:51

In the section on amphiprotic species, we saw that water can act as a very weak acidAn acid that ionizes only partially in a given solvent. and a very weak baseAn base that ionizes only partially in a given solvent., donating protons to itself to a limited extent:

$\text{2H}_{2}\text{O}\text{ }({l}) \rightleftharpoons \text{H}_{3}\text{O}^{+} ({aq}) + \text{OH}^{-} ({aq})$

Applying the equilibriumA state in which no net change is occurring, that is, in which the concentrations of reactants and products remain constant; chemical equilibrium is characterized by forward and reverse reactions occurring at the same rate. law to this reaction, we obtain

$K_{c}=\frac{[\text{ H}_{3}\text{O}^{+}][\text{OH}^{-}]}{[\text{ H}_{2}\text{O }]^{2}}$

However, as can be seen in the section on the law of chemical equilibrium, the concentrationA measure of the ratio of the quantity of a substance to the quantity of solvent, solution, or ore. Also, the process of making something more concentrated. of water has a constant value of 55.5 mol dm–3, and so its square can be multiplied by Kc to give a new constant Kw, called the ion-productA substance produced by a chemical reaction. constant of water:

$K_{w}={K_{c}}\times{({55.5}\text{ mol dm}^{-3})^{2}}={[\text{ H}_{3}\text{O}^{+}] [\text{OH}^{-}]}$      (1)

Measurements of the electrical conductivity of carefully purified water indicate that at 25°C [H3O+] = [OH] = 1.00 × 10–7 mol dm–3, so that

\begin{align}K_{w}&={1.00}\times{10}^{-7}\text{ mol dm}^{-3}\times{1.00}\times{10}^{-7}\text{ mol dm}^{-3}\ \text{ }\ \text{ }&={1.00}\times{10}^{-14}\text{ mol}^2\text{dm}^{-6}\end{align}

(Since the equilibrium law is not obeyed exactly, even in dilute solutions, results of most equilibrium calculations are rounded to three significant figures. Hence the value of Kw = 1.00 × 10–14 mol2 dm–6 is sufficiently accurate for all such calculations.)

The equilibrium constantThe value of the equilibrium constant expression when equilibrium concentrations are substituted; a value greater than one indicates the position of equilibrium lies toward products (product-favored), and a value less than one indicates the position of equilibrium lies toward reactants (reactant-favored). Kw applies not only to pure water but to any aqueous solution at 25°C. Thus, for example, if we add 1.00 mol of the strong acidAn acid that ionizes completely in a particular solvent. HNO3 to H2O to make a total volume of 1 dm3, essentially all the HNO3 molecules donate their protons to H2O:

$\text{HNO}_{3} + \text{H}_{2}\text{O} \rightarrow \text{NO}_{3}^{-} + \text{H}_{3}\text{O}^{+}$

and a solution in which [H3O+] = 1.00 mol dm–3 is obtained. Although this solution is very acidic, there are still hydroxide ions present. We can calculate their concentration by rearranging Eq. (1):

\begin{align}\text{ }[\text{OH}^{-}]&=\frac{K_{w}}{[\text{ H}_{3}\text{O}^{+}]}=\frac{\text{1.00 }\times \text{ 10}^{-14}\text{ mol}^{2}\text{ dm}^{-6}}{\text{1.00 mol dm}^{-3}}\ \text{ }\ \text{ }&=\text{1.00 }\times \text{ 10}^{-14}\text{ mol dm}^{-3}\end{align}

The addition of the HNO3 to H2O not only increases the hydronium-ion concentration but also reduces the hydroxide-ion concentration from an initially minute 10–7 mol dm–3 to an even more minute 10–14 mol dm–3.

EXAMPLE Calculate the hydronium-ion concentration in a solution of 0.306 M Ba(OH)2.

Solution Since 1 mol Ba(OH)2 produces 2 mol OH in solution, we have

[OH] = 2 × 0.306 mol dm–3 = 0.612 mol dm–3

Then

\begin{align}\text{ }[\text{ H}_{3}\text{O}^{+}]&=\frac{K_{w}}{[\text{OH}^{-}]}=\frac{\text{1.00 }\times \text{ 10}^{-14}\text{ mol}^{2}\text{ dm}^{-6}}{\text{0.612 mol dm}^{-3}}\ \text{ }\ \text{ }&=\text{1.63 }\times \text{ 10}^{-14}\text{ mol dm}^{-3}\end{align}

Note that since strong acids like HNO3 are completely converted to H3O+ in aqueous solution, it is a simple matterAnything that occupies space and has mass; contrasted with energy. to determine [H3O+], and from it, [OH]. Similarly, when a strong baseA base that dissociates completely or ionizes completely in a particular solvent. dissolves in H2O it is entirely converted to OH, so that [OH], and from it [H3O+] are easily obtained.