The first ionization constant of $H_2S$ is $9.1 \times 10^{-8}$. Calculate the concentration of $HS^{-}$ ion in its $0.1 \ M$ solution. How will this concentration be affected if the solution is $0.1 \ M$ in $HCl$ also? If the second dissociation constant of $H_2S$ is $1.2 \times 10^{-13}$,calculate the concentration of $S^{2-}$ under both conditions.

(N/A) $(i)$ To calculate the concentration of $HS^{-}$ ion:
Case $I$ (in the absence of $HCl$):
Let the concentration of $HS^{-}$ be $x \ M$.
$H_2S \leftrightarrow H^{+} + HS^{-}$
$K_{a_1} = \frac{[H^{+}][HS^{-}]}{[H_2S]} = \frac{x^2}{0.1-x} \approx \frac{x^2}{0.1} = 9.1 \times 10^{-8}$
$x^2 = 9.1 \times 10^{-9} \Rightarrow x = 9.54 \times 10^{-5} \ M$.
So,$[HS^{-}] = 9.54 \times 10^{-5} \ M$.
Case $II$ (in the presence of $0.1 \ M \ HCl$):
$HCl$ is a strong acid,so $[H^{+}] \approx 0.1 \ M$.
$K_{a_1} = \frac{[H^{+}][HS^{-}]}{[H_2S]} = \frac{(0.1)[HS^{-}]}{0.1} = 9.1 \times 10^{-8}$
$[HS^{-}] = 9.1 \times 10^{-8} \ M$.
$(ii)$ To calculate the concentration of $[S^{2-}]$:
Case $I$ (in the absence of $HCl$):
$K_{a_2} = \frac{[H^{+}][S^{2-}]}{[HS^{-}]} = 1.2 \times 10^{-13}$
Since $[H^{+}] = [HS^{-}] = 9.54 \times 10^{-5} \ M$,we get $[S^{2-}] = K_{a_2} = 1.2 \times 10^{-13} \ M$.
Case $II$ (in the presence of $0.1 \ M \ HCl$):
$[H^{+}] = 0.1 \ M$ and $[HS^{-}] = 9.1 \times 10^{-8} \ M$.
$1.2 \times 10^{-13} = \frac{(0.1)[S^{2-}]}{9.1 \times 10^{-8}}$
$[S^{2-}] = \frac{1.2 \times 10^{-13} \times 9.1 \times 10^{-8}}{0.1} = 1.092 \times 10^{-19} \ M$.