The acid dissociation constants of $H_{2}S$  and  $H{S}^{-}$ are  ${10}^{-7}$ and ${10}^{-13}$  respectively. The $pH$ of 0.1M  $H_{2}S$ aqueous solution will be _______

As  $K_{a_2}<<K_{a_1}$,  $H^{+}$ are mainly obtained from the first dissociation , second dissociation constant is extremely low when compare to first, therefore we can neglect ${K_{a_2}}_{ }$for calculation
 $H_{2}S\rightleftharpoons H^{+}+H{S}^{-}$ for which 
$$\begin{gathered} \left[H^{+}\right]=\sqrt{KC}=\sqrt{{10}^{-7}\times {10}^{-1}}={10}^{-4} \\ pH=4 \end{gathered}$$