Question

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.

Short answer

The values are,

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

Step-by-step solution

Define acids

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

Explanation

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 ${\text{K}}_{\text{a}}$.

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

Evaluating the values

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,

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