Q18.79P - Chemistry: Molecular Nature Of Matter And Change

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 $\mathbf{\text{0.10 M}}$ solution of the diprotic acid hydro sulfuric acid.

Step-by-Step Solution

Answer

The values are,

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

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 $\text{0.10 M}$ .

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.

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

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, 

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