Example 13.7: 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

\(\mathrm{pH}\) \(= 2.09\)


\([\ce{HNO2}]_{\mathrm{0}}\) \(= 0.125\)


The percent ionization for an acid is:

\(\mathrm{fraction}_{\mathrm{ionized}} = \dfrac{[\ce{H3O+}]_{\mathrm{eq}}}{[\ce{HNO2}]_{\mathrm{0}}} \cdot 100\ \mathrm{%}\)     

The chemical equation for the dissociation of the nitrous acid is

\(\ce{HNO2(aq)}\)\(\ce{ + }\)\(\ce{H2O(l)}\)\(\ce{<=>}\)\(\ce{NO2-(aq)}\)\(\ce{ + }\)\(\ce{H3O+(aq)}\)\(\ce{ }\)

Since \(\ce{10^-pH}\) \( = [\ce{H3O+}]\), we find:

\([\ce{H3O+}]_{\mathrm{eq}}\) \(= {10}^{(-\mathrm{pH})}\)

\(\ \ \ ={10}^{-2.09}\)

\(\ \ \ =8.1\times 10^{-3}\)


\(\mathrm{fraction}_{\mathrm{ionized}}\) \(= \dfrac{[\ce{H3O+}]_{\mathrm{eq}}}{[\ce{HNO2}]_{\mathrm{0}}} \cdot 100\ \mathrm{%}\)

\(\ \ \ =\dfrac{8.1\times 10^{-3}}{0.125} \cdot 100\ \mathrm{%}\)

\(\ \ \ =0.0650 \cdot 100\ \mathrm{%}\)

\(\ \ \ =6.5\ \mathrm{%}\)


Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures.