Use Appendix to calculate$\left[{\text{H}}_{\text{2}}\text{S}\right]\mathbf{\text{,}}\left[{\text{HS}}^{\text{-}}\right]\mathbf{\text{,}}\left[{\text{S}}^{\text{2-}}\right]\mathbf{\text{,}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\mathbf{\text{,pH,}}\left[{\text{OH}}^{\text{-}}\right]$ and$\mathbf{\text{pOH}}$ in a $\mathbf{\text{0.10M}}$solution of the diprotic acid hydro sulfuric acid.

Chemical agents that release hydrogen ions when introduced to water are referred to as acids in chemistry.

Write the reaction equation for the dissociation of${\text{H}}_{\text{2}}\text{S}$in water using the${\text{K}}_{\text{a}}$from Appendix as a reference.

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{S+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+HS}}^{\text{-}}\\ {\text{K}}_{\text{a1}}{\text{=9×10}}^{\text{-8}}\\ {\text{HS}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+S}}^{\text{2-}}\\ {\text{K}}_{\text{a2}}{\text{=1×10}}^{\text{-17}}\end{array}$

After that, create the ICE table to get the ${\text{K}}_{\text{a1}}$equation.

$\begin{array}{l}\begin{array}{cccccc}& \left[{\text{H}}_{\text{2}}\text{S}\right]\text{+}& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{+}& \left[{\text{HS}}^{\text{-}}\right]\\ \text{I}& \text{0.10M}& & & \text{0}& \text{0}\\ \text{C}& \text{-x}& & & \text{+x}& \text{+x}\\ \text{E}& \text{0.10M-x}& & & \text{+x}& \text{+x}\end{array}\\ \\ {\text{K}}_{\text{a1}}\text{=}\frac{\left[{\text{HS}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{H}}_{\text{2}}\text{S}\right]}\\ {\text{K}}_{\text{a1}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.10-x}}\end{array}$

As$\text{x=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\left[{\text{HS}}^{\text{-}}\right]$ is well-known. As ${\text{H}}_{\text{2}}\text{S}$ is a weak acid, ${\text{K}}_{\text{a}}$its value is quite low. Assume that the has no effect on the denominator's$\text{0.10M}$ .

Then, to solve for $\text{x}$, substitute the ${\text{K}}_{\text{a}}$.

$\begin{array}{c}{\text{K}}_{\text{a1}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.10}}\\ {\text{x}}^{\text{2}}\text{=}\left({\text{K}}_{\text{a1}}\right)\text{(0.10)}\\ \text{x=}\sqrt{\left({\text{K}}_{\text{a1}}\right)\text{(0.10)}}\\ \text{=}\sqrt{\left({\text{9.0×10}}^{\text{-8}}\right)\text{(0.10)}}\\ {\text{x=9.5×10}}^{\text{-5}}\text{M}\end{array}$

Let us now solve for $\left[{\text{S}}^{\text{2-}}\right]$. Since ${\text{K}}_{\text{a2}}$ is so little, the ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$produced will have no discernible effect on the present ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$concentration. As a result, we can apply the equation immediately.

$\begin{array}{c}{\text{K}}_{\text{a2}}\text{=}\frac{\left[{\text{S}}^{\text{2-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{HS}}^{\text{-}}\right]}\\ \left[{\text{S}}^{\text{2-}}\right]\text{=}\frac{\left[{\text{K}}_{\text{a2}}\right]\left[{\text{HS}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}\\ \left[{\text{S}}^{\text{2-}}\right]\text{=}\frac{\left({\text{1×10}}^{\text{-17}}\right)\left({\text{9.5×10}}^{\text{-5}}\right)}{{\text{9.5×10}}^{\text{-5}}}\\ {\text{=1×10}}^{\text{-17}}\text{M}\end{array}$

After that, use the ${\text{K}}_{\text{w}}$of water to find the $\left[{\text{OH}}^{\text{-}}\right]$.

$\begin{array}{c}{\text{K}}_{\text{w}}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}\\ \text{=}\frac{{\text{1.0×10}}^{\text{-14}}}{{\text{9.5×10}}^{\text{-5}}}\\ \left[{\text{OH}}^{\text{-}}\right]{\text{=1.05×10}}^{\text{-10}}\text{M}\end{array}$

Calculate the final ${\text{H}}_{\text{2}}\text{S}$concentration.

$\begin{array}{c}\left[{\text{H}}_{\text{2}}\text{S}\right]{\text{=0.10M-9.5×10}}^{\text{-5}}\\ \text{=0.10M}\end{array}$

Calculate for $\text{pH}$using $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$.

$\begin{array}{c}\text{pH=-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{=-log}\left({\text{9.5×10}}^{\text{-5}}\right)\\ \text{pH=4.02}\end{array}$

At last, calculate the $\text{pOH}$using equation $\text{pH+pOH=14}$.

$\begin{array}{c}\text{pH+pOH=14}\\ \text{pOH=14-pH}\\ \text{=14-4.02}\\ \text{pOH=9.98}\end{array}$

Therefore,

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\left[{\text{HS}}^{\text{-}}\right]{\text{=9.5×10}}^{\text{-5}}\text{M}\\ \left[{\text{S}}^{\text{2-}}\right]{\text{=1.0×10}}^{\text{-17}}\text{M}\\ \left[{\text{OH}}^{\text{-}}\right]{\text{=1.05×10}}^{\text{-10}}\text{M}\\ \left[{\text{H}}_{\text{2}}\text{S}\right]\text{=0.10MpH=4.02pOH=9.98}\end{array}$