An aqueous solution contains 0.10 M $H_{2}S$ and 0.20 M $HCl$ . If the equilibrium constants for the formation of  $H{S}^{-}$ from  $H_{2}S$ is  $1.0\times {10}^{-7}$ and that of  $S^{2-}$ from  $H{S}^{-}$ ions is  $1.2\times {10}^{-13}$ that the concentration of $s^{2-}$  ions in aqueous solutions is $y\times {10}^{-20}M$. Then y = 

Formation of  $H\overline{S}$ from  $H_2\overline{S}$

$H_{2}S{H}^{+}\left(aq\right)+H{S}^{-}\left(aq\right)-------\left(i\right)$

                                  from of  $S^{2-}from{H}^{s-}$

$H{S}^{-}aq{H}^{+}\left(aq\right)+S^2\left(aq\right)------\left(ii\right)$

Where in  $K{a}_1=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}$

     $K{a}_1=\frac{\left[H^{+}\right].\left[S^{2-}\right]}{\left[H{S}^{-1}\right]}$

Add   $e{q}^n\left(i\right)\&\left(2\right),$    we get – 

$H_{2}S\left(aq\right)2H^{+}\left(aq\right)+S^{2-}\left(aq\right),K=K{a}_1\times K{a}_2$

$$\begin{gathered} K=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}\times \frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]} \\ K=\frac{{\left[H^{+}\right]}^2\left[S^{2-}\right]}{\left[H_{2}S\right]} \\ ={10}^{-7}\times 1.2\times {10}^{-20} \\ at t=00.10,0.20m \\ K=\frac{{\left(0.2\right)}^2\left(x\right)}{0.1} \\ 1.2\times {10}^{-20}=\frac{0.2\times 0.2\times x}{0.1} \\ x=3\times {10}^{-20} \end{gathered}$$