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

(i) To calculate $[H{S}^{-}]$ in absence of HCl:
Let, $[H{S}^{-}]=x$ M.
$H_{2}S\rightleftharpoons H^{+}+H{S}^{-}$
The initial concentrations of $H_{2}S,H^{+}$ and $H{S}^{-}$ are 0.1 M, 0 M and 0 M respectively.
Their final concentrations are 0.1-x M, x M and x M respectively.
$K_a=\frac{[H^{+}][H{S}^{-}]}{[H_{2}S]}$
$9.1\times 10^{-8}=\frac{x\times x}{0.1-x}$
In the denominator, 0.1-x can be approximated to 0.1 as x is very small.
$9.1\times 10^{-8}=\frac{x\times x}{0.1}$
$x=9.54\times 10^{-5}M=[H{S}^{-}]$

(ii) To calculate $[H{S}^{-}]$ in presence of HCl:
Let $[H{S}^{-}]=y$ M.
$H_{2}S\rightleftharpoons H^{+}+H{S}^{-}$
The initial concentrations of $H_{2}S,H^{+}$ and $H{S}^{-}$ are 0.1 M, 0 M and 0 M respectively.
Their final concentrations are 0.1-y M, y M and y M respectively.
Also, $HCl\rightleftharpoons H^{+}+C{l}^{-}$
For HCl $[H^{+}]=[C{l}^{-}]=0.1M$
$K_a=\frac{[H{S}^{-}][H^{+}]}{[H_{2}S]}$
$K_a=\frac{y(0.1+y)}{0.1-y}$
In the denominator, 0.1-y can be approximated to 0.1 and in the numerator, 0.1+y can be approximated to 0.1 as y is very small.
$9.1\times 10^{-8}=\frac{y\times 0.1}{0.1}$
$y=9.1\times 10^{-8}M=[H{S}^{-}]$