Chemistry - Ionic Equilibrium Question with Solution | TestHub

Consider the following equations:

$$\begin{array}{rl}{\mathrm{H}}_2{\mathrm{SO}}_4& \to {\mathrm{H}}^{+}+{\mathrm{HSO}}_4^{-}\\ {\mathrm{HSO}}_4^{-}& \rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{SO}}_4^{2-}\\ {\mathrm{Na}}_2{\mathrm{SO}}_4& \to 2{\mathrm{Na}}^{+}+{\mathrm{SO}}_4^{2-}\\ {\mathrm{PbSO}}_4(\mathrm{s})& \rightleftharpoons {\mathrm{Pb}}^{2+}(\mathrm{aq})+{\mathrm{SO}}_4^{2-}(\mathrm{aq})\\ K_{\mathrm{a}2}& =1.2\times 10^{-4}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{SO}}_4^{2-}\right]}{\left[{\mathrm{HSO}}_4^{-}\right]}\\ K_{\mathrm{sp}}& =1.6\times 10^{-8}=\left[{\mathrm{Pb}}^{2+}\right]\left[{\mathrm{SO}}_4^{2-}\right]\\ \frac{1.2\times 10^{-4}}{1.6\times 10^{-8}}& =\frac{\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{HSO}}_4^{-}\right]\left[{\mathrm{Pb}}^{2+}\right]}\\ \left[{\mathrm{Pb}}^{2+}\right]& =\frac{1.6\times 10^{-8}}{1.2\times 10^{-4}}\times \frac{\left[{\mathrm{HSO}}_4^{-}\right]}{\left[{\mathrm{H}}^{+}\right]}\\ As\left[{\mathrm{H}}^{+}\right]& \approx \left[{\mathrm{HSO}}_4^{-}\right]\\ \left[{\mathrm{Pb}}^{2+}\right]& =1.33\times 10^{-4}\end{array}$$