Hydrogen sulfide $(H_2(S))$ is a weak diprotic acid and has the following acid $-$

dissociation $(K_a)$ values:

$K_{a,1}=1\mathrm{.}0{10}^{-7}$

$K_{a,2}=1\mathrm{.}3{10}^{-13}$

The forms of $H_{2}S$ are as follows:

$H_{2}S$ is $H_{2}L,($acidic form$)$

$H{S}^{-}$is $H{L}^{-}($intermediate form$)$

$S^{2-}$ is $L^{2-},($basic form$)$

When the formal concentration of $H_{2}S$ is $0\mathrm{.}5650M,$ what would be the $H^{+}$and

$pH$ of the solution? For full credit, in your calculation please use the $H^{+}$

equation for the intermediate form of diprotic acid$/$base systems.