The \(pH\) of a \(0.15 - M\) solution of \(HS{O_4}\) − is \(1.43. \)Determine \({K_a}\)for \( HS{O_4}\) − from these data.

The reaction in the question is \({\bf{HSO}}_{\bf{4}}^{\bf{ - }}{\bf{(aq)}} \to {{\bf{H}}^{\bf{ + }}}{\bf{(aq) + SO}}_{\bf{4}}^{{\bf{2 - }}}{\bf{(aq)}}.\)

Another possible reaction is the reaction of \({\bf{HSO}}_{\bf{4}}^{\bf{ - }}\) and water, which produces \({{\bf{H}}_{\bf{2}}}{\bf{SO}}_{\bf{4}}^{\bf{ - }}\) and \({\bf{O}}{{\bf{H}}^{\bf{ - }}}\)ions. However, because \({{\bf{H}}_{\bf{2}}}{\bf{SO}}_{\bf{4}}^{\bf{ - }}\) is a strong acid that totally ionizes, this reverse reaction is minor and we do not need to account for it in our calculations.

We can readily compute the concentration of \({H^ + }\)ions because we know the \(pH\):

\(\begin{aligned}\;\;\;pH &= - log\left( {{H^ + }} \right)\\\left( {{H^ + }} \right) &= 10 - pH M\\\;\;\;\;\;\;\;\; &= 1{0^{ - 1.43}} M\\\;\;\;\;\;\;\;\; &= 0.04 M.\end{aligned}\)

We can assume that the concentrations of \({H^ + }\) and \(SO_4^{2 - }\) are the same. Only \(0.15 M\) of \(HSO_4^ - \) is available at the start of the reaction. The concentration of \(HSO_4^ - \) lowers by \(x\) and the concentrations of both \({H^ + }\) and \(SO_4^{2 - }\) rise by \(x\) until the reaction achieves equilibrium. The equilibrium concentrations of \({H^ + }\) and \(SO_4^{2 - }\) ions are known, therefore:

\(\begin{aligned}\left( {HSO_4^ - } \right) &= 0.15 M - x\\\;\;\;\;\;\;\;\;\;\;\;\; &= (0.15 - 0.04) M\\\;\;\;\;\;\;\;\;\;\;\;\; &= 0.11 M\end{aligned}\)

Now that we know all the equilibrium concentrations, we can calculate the \({K_a}\) value as:

\(\begin{aligned}{K_a} &= \frac{{\left( {{H^ + }} \right) \times \left( {SO_4^{2 - }} \right)}}{{\left( {HSO_4^ - } \right)}}\\\;\;\;\;\; &= \frac{{0.04 \times 0.04}}{{0.11}}\\\;\;\;\;\; &= 0.01.\end{aligned}\)

The \({K_a}\) value of of \(HSO_4^ - \) is \(0.01.\)

The final solution is \({K_a} = 0.01.\)