Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\begin{array}{ccccc}\mathrm{HCl}& \to & {\mathrm{H}}^{+}& +& {\mathrm{Cl}}^{-}\\ & & \left(0.2+\mathrm{x}+\mathrm{y}\right)& & 0.2\end{array}$

$\begin{array}{ccccc}{\mathrm{H}}_2\mathrm{S}& \rightleftharpoons & {\mathrm{H}}^{+}& +& {\mathrm{HS}}^{-}\\ \left(0.1-\mathrm{x}\right)& & \left(0.2+\mathrm{x}+\mathrm{y}\right)& & \left(\mathrm{x}-\mathrm{y}\right)\end{array}; {\mathrm{K}}_{{\mathrm{a}}_1}={10}^{-7}$

$\begin{array}{ccccc}{\mathrm{HS}}^{-}& \to & {\mathrm{H}}^{+}& +& {\mathrm{S}}^{2-}\\ \left(\mathrm{x}-\mathrm{y}\right)& & \left(0.2+\mathrm{x}+\mathrm{y}\right)& & \left(\mathrm{y}\right)\end{array}; {\mathrm{K}}_{{\mathrm{a}}_2}=1.2\times {10}^{-13}$

$\left[{\mathrm{H}}^{+}\right]=\left(0.2+\mathrm{x}+\mathrm{y}\right)\mathrm{M}=0.2\mathrm{M}$

${\mathrm{K}}_{{\mathrm{a}}_1}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{HS}}^{-}\right]}{\left[{\mathrm{H}}_2\mathrm{S}\right]}=\frac{0.2\times \left[{\mathrm{HS}}^{-}\right]}{0.1-\mathrm{x}}=\frac{0.2\left[{\mathrm{HS}}^{-}\right]}{0.1}$           ($\mathrm{x}=$negligible)

$\left[{\mathrm{HS}}^{-}\right]=\frac{{\mathrm{K}}_{{\mathrm{a}}_1}\times 0.1}{0.2}=\frac{1}{2}\times {10}^{-7}$

${\mathrm{K}}_{{\mathrm{a}}_2}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{S}}^{2-}\right]}{\left[{\mathrm{HS}}^{-}\right]}=\frac{\left[0.2 \mathrm{M}\right]\left[{\mathrm{S}}^{2-}\right]}{{\displaystyle \frac{1}{2}}\times {10}^{-7}}$

$\left[{\mathrm{S}}^{2-}\right]=\frac{{\mathrm{K}}_{{\mathrm{a}}_2}\times {\displaystyle \frac{1}{2}}\times {10}^{-7}}{0.2}=\frac{1.2\times {10}^{-13}\times {\displaystyle \frac{1}{2}}\times {10}^{-7}}{0.2}=3\times {10}^{-20} \mathrm{M}$