The first ionization constant of $H _{2} S$ 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 $H _{2} S$ is $1.2 \times 10^{-13}$, calculate the concentration of $S^{2-}$ under both conditions.

$(i)$ To calculate the concentration of $HS ^{-}$ ion:

case $I$ (in the absence of $HCl$ ):

Let the concentration of $HS ^{-}$ be $x$ $M$

          ${H_2}S\,\quad  \leftrightarrow \,\quad {H^ + }\quad  + \quad HS$

${C_i}$     $0.1$                       $0$                    $0$

${C_f}$    $0.1-x$                $x$                   $x$

Then, $K_{a_{1}}=\frac{\left|H^{+}\right|\left[H S^{-}\right]}{\left[H_{2} S\right]}$

$9.1 \times 10^{-8}=\frac{(x)(x)}{0.1-x}$

$\left(9.1 \times 10^{-8}\right)(0.1-x)=x^{2}$

Taking $0.1-x \,M ; 0.1 \,M ,$ we have $\left(9.1 \times 10^{-8}\right)(0.1)=x^{2}$

$9.1 \times 10^{-9}=x^{2}$

$x=\sqrt{9.1 \times 10^{-9}}$

$=9.54 \times 10^{-5}\, M$

$\Rightarrow\left[ HS ^{-}\right]=9.54 \times 10^{-5} \,M$

Case $II$ (in the presence of $HCI$ ):

In the presence of $0.1 \,M$ of $HCl$, let $\left[ HS ^{-}\right]$ be $y$ $M$

Then,    ${H_2}S\quad  \leftrightarrow \quad H{S^ - }\quad  + \quad {H^ + }$

${C_i}$            $0.1$                        $0$                $0$

${C_f}$       $0.1-y$                    $y$                 $y$

$Now , \quad K_{a_{1}}=\frac{\left[ HS ^{-}\right]\left[ H ^{+}\right]}{\left[ H _{2} S \right]}$

$K_{a_{1}}=\frac{[y](0.1+y)}{(0.1-y)}$

$9.1 \times 10^{-8}=\frac{y \times 0.1}{0.1}$         $(\because 0.1-y ; 0.1 \,M )$

$9.1 \times 10^{-8}=y$              $(\text { and } 0.1+y ; 0.1 M )$

$\Rightarrow\left[ HS ^{-}\right]=9.1 \times 10^{-8}$

To calculate the concentration of $\left[ S ^{2-}\right]$

Case $I$ (in the absence of $0.1\, M\, HCl$ ):

$HS ^{-} \longleftrightarrow H ^{+}+ S ^{2-}$

$\left[ HS ^{-}\right]=9.54 \times 10^{-5} \,M$  (From first ionization, case $I$)

Let $\left[ S ^{2-}\right]$ be $X$

Also, $\left[ H ^{+}\right]=9.54 \times 10^{-5} \,M$   (From first ionization, case $I$)

$K_{a_{2}}=\frac{\left[ H ^{+}\right]\left[ S ^{2-}\right]}{\left[ HS ^{-}\right]}$

$K_{a_{2}}=\frac{\left(9.54 \times 10^{-5}\right)(X)}{9.54 \times 10^{-5}}$

$1.2 \times 10^{-13}=X=\left[ S ^{2-}\right]$

Case $II$ (in the presence of $0.1 \,M$ $HCI$ )

Again, let the concentration of $HS^{-}$ be $X^{\prime} M$

$\left[ HS ^{-}\right]=9.1 \times 10^{-8} \,M$   (From first ionization, case $II$ )

$\left[ H ^{+}\right]=0.1\, M$     (From $HCl$, case $II$ )

$\left[ S ^{2-}\right]=X^{\prime}$

Then, $K_{a_{2}}=\frac{\left[ H ^{+}\right]\left[ S ^{2-}\right]}{\left[ HS ^{-}\right]}$

$1.2 \times 10^{-13}=\frac{(0.1)\left(X^{\prime}\right)}{9.1 \times 10^{-8}}$

$10.92 \times 10^{-21}=0.1 X^{\prime}$

$\frac{10.92 \times 10^{-21}}{0.1}=X^{\prime}$

$\frac{10.92 \times 10^{-21}}{0.1}=X^{\prime}$

$X^{\prime}=\frac{1.092 \times 10^{-30}}{0.1}$

$=1.092 \times 10^{-14}\, M$

$\Rightarrow K_{a_{1}}=1.74 \times 10^{-5}$