# 14.1: Ionization of Water

In the section on amphiprotic species, we saw that water can act as a very weak acid and a very weak base, 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 equilibrium law to this reaction, we obtain

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

However, as can be seen in the section on the law of chemical equilibrium, the concentration of water has a constant value of 55.5 mol/L, and so its square can be multiplied by Kc to give a new constant Kw, called the ion-product constant of water:

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

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

\begin{align}K_{w}={1.00}\times{10}^{-7}\text{ mol L}^{-1}\times{1.00}\times{10}^{-7}\text{ mol L}^{-1} \nonumber \\\text{ }\\\text{ }={1.00}\times{10}^{-14}\text{ mol}^2\text{L}^{-2} \nonumber \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/L2 is sufficiently accurate for all such calculations.)

The equilibrium constant 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 acid HNO3 to H2O to make a total volume of 1 L, 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/L is obtained. Although this solution is very acidic, there are still hydroxide ions present. We can calculate their concentration by rearranging Eq. $$\ref{3}$$:

\begin{align}\text{ }[\text{OH}^{-}]=\dfrac{K_{w}}{[\text{ H}_{3}\text{O}^{+}]}=\dfrac{\text{1.00 }\times \text{ 10}^{-14}\text{ mol}^{2}\text{ L}^{-2}}{\text{1.00 mol L}^{-1}}\\\text{ }\\\text{ }=\text{1.00 }\times \text{ 10}^{-14}\text{ mol L}^{-1}\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/L to an even more minute 10–14 mol/L.

Example $$\PageIndex{2}$$: Ion Concentration

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

$[OH^-] = 2 \times 0.306 \dfrac{mol}{L} = 0.612 \dfrac{mol}{L}$

Then

\begin{align}\text{ }[\text{ H}_{3}\text{O}^{+}]=\dfrac{K_{w}}{[\text{OH}^{-}]}=\dfrac{\text{1.00 }\times \text{ 10}^{-14}\text{ mol}^{2}\text{ L}^{-2}}{\text{0.612 mol L}^{-1}} \nonumber \\\text{ }\\\text{ }=\text{1.63 }\times \text{ 10}^{-14}\text{ mol L}^{-1} \nonumber \end{align}

Note

Note that since strong acids like HNO3 are completely converted to H3O+ in aqueous solution, it is a simple matter to determine [H3O+], and from it, [OH]. Similarly, when a strong base dissolves in H2O it is entirely converted to OH, so that [OH], and from it [H3O+] are easily obtained.