Calculate $[S^{2-}]$ and $[HS^{-}]$ of the solution which contains $0.1 \ M \ H_2S$ and $0.3 \ M \ HCl$.
[$K_{a1} = 1.0 \times 10^{-7}$ and $K_{a2} = 1.3 \times 10^{-13}$ for $H_2S$]

(N/A) The dissociation of $H_2S$ is suppressed by the presence of $0.3 \ M \ HCl$ (a strong acid).
$1$. For the first dissociation: $H_2S \rightleftharpoons H^+ + HS^-$
$K_{a1} = \frac{[H^+][HS^-]}{[H_2S]} = 1.0 \times 10^{-7}$
Since $[H^+] \approx 0.3 \ M$ and $[H_2S] \approx 0.1 \ M$,we have:
$[HS^-] = \frac{K_{a1} \times [H_2S]}{[H^+]} = \frac{1.0 \times 10^{-7} \times 0.1}{0.3} = 3.33 \times 10^{-8} \ M$
$2$. For the second dissociation: $HS^- \rightleftharpoons H^+ + S^{2-}$
$K_{a2} = \frac{[H^+][S^{2-}]}{[HS^-]} = 1.3 \times 10^{-13}$
$[S^{2-}] = \frac{K_{a2} \times [HS^-]}{[H^+]} = \frac{1.3 \times 10^{-13} \times 3.33 \times 10^{-8}}{0.3} = 1.44 \times 10^{-20} \ M$