The saturation solubility of ${\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)$ in water is $0.1 \mathrm{M}$. If${{\mathrm{K}}_{\mathrm{a}}}_1 \left({\mathrm{H}}_2\mathrm{S}\right) = 1.0 \mathrm{x} {10}^{-7} \mathrm{M} \mathrm{and} {\mathrm{K}}_{\mathrm{a}2}\left({\mathrm{H}}_2\mathrm{S}\right) = 1.0 \mathrm{x} {10}^{-14} \mathrm{M}$, then the number of ${\mathrm{S}}^{2-}$ ions in 1 L of $0.2 \mathrm{M} \mathrm{HCI}$ solution will be about

$\begin{array}{l}{\mathrm{H}}_2\mathrm{S}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{HS}}^{-}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em};\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{K}_{\mathrm{a}1}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{HS}}^{-}\right]/\left[{\mathrm{H}}_2\mathrm{S}\right]\\ {\mathrm{HS}}^{-}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{S}}^{2-}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em};\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{K}_{\mathrm{a}2}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{S}}^{2-}\right]/\left[{\mathrm{HS}}^{-}\right]\\ {\mathrm{H}}_2\mathrm{S}\rightleftharpoons 2{\mathrm{H}}^{+}+{\mathrm{S}}^{2-}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em};\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{K}_{\mathrm{a}}={\left[{\mathrm{H}}^{+}\right]}^2\left[{\mathrm{S}}^{2-}\right]/\left[{\mathrm{H}}_2\mathrm{S}\right]\end{array}$

Obviuosly , $K_{\mathrm{a}}=K_{\mathrm{a}1}{K}_{\mathrm{a}2}=\left({10}^{-7}\mathrm{M}\right)\left({10}^{-14}\mathrm{M}\right)={10}^{-21}{\mathrm{M}}^2$

$\left[{\mathrm{S}}^{2-}\right]=\frac{K_{\mathrm{a}}\left[{\mathrm{H}}_2\mathrm{S}\right]}{{\left[{\mathrm{H}}^{+}\right]}^2}=\frac{\left({10}^{-21}{\mathrm{M}}^2\right)\left(0.1\mathrm{M}\right)}{(0.2\mathrm{M}{)}^2}=2.5\times {10}^{-21}\mathrm{M}$

Number of ${\mathrm{S}}^{2-}$ ions $=\left[{\mathrm{S}}^{2-}\right]N_{\mathrm{A}}=\left(2.5\times {10}^{-21}{\mathrm{molL}}^{-1}\right)\left(6.022\times {10}^{23}{\mathrm{mol}}^{-1}\right)=1506{\mathrm{L}}^{-1}$