Chapter 13, Problem 109E

Interpretation Introduction

Interpretation: The $\text{pH}$ of the given ${\text{H}}_{\text{2}}\text{S}$ solution and the $\left[{\text{S}}^{2-}\right]$ is to be calculated.

Concept introduction: The $\text{pH}$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $\text{pH}$ of a solution is calculated by the formula, $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$K_{\text{a}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Answer to Problem 109E

Answer

The $\text{pH}$ of the given solution is $\underset{\_}{4.0}$ and the $\left[{\text{S}}^{2-}\right]$ is $\underset{\_}{1.0\times 10^{-19}M}$.

Explanation of Solution

Explanation

To determine: T The $\text{pH}$ of the given ${\text{H}}_{\text{2}}\text{S}$ solution and the $\left[{\text{S}}^{2-}\right]$.

The equilibrium constant expression for the initial dissociation reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{HS}}^{-}\right]}{\left[{\text{H}}_2\text{S}\right]}$

The initial dissociation reaction is,

${\text{H}}_2\text{S}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{HS}}^{-}\left(aq\right)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$K_{\text{a}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K_{\text{a}}$ is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{HS}}^{-}\right]}{\left[{\text{H}}_2\text{S}\right]}$ (1)

The $\left[{\text{H}}^{\text{+}}\right]$ from first reaction is $\underset{\_}{1.0\times 10^{-4}M}$.

The change in concentration of ${\text{H}}_{\text{2}}\text{S}$ is assumed to be $x$.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{H}}_{\text{2}}\text{S}\left(aq\right)& \rightleftharpoons & {\text{H}}^{+}\left(aq\right)& +& {\text{HS}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.10& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.10-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{H}}_{\text{2}}\text{S}\right]$ is $\left(0.10-x\right)\text{M}$.

The equilibrium concentration of $\left[{\text{H}}^{+}\right]$ is $x\text{M}$.

The equilibrium concentration of $\left[{\text{HS}}^{-}\right]$ is $x\text{M}$.

The $K_{\text{a}}$ for the first reaction is given to be is $1.0\times {10}^{-7}$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{H}}_{\text{2}}\text{S}\right]$, $\left[{\text{H}}^{+}\right]$ and $\left[{\text{HS}}^{-}\right]$ in equation (1).

$\begin{array}{c}1.0\times {10}^{-7}=\frac{\left[x\right]\left[x\right]}{\left[0.10-x\right]}\\ 1.0\times {10}^{-7}=\frac{{\left[x\right]}^2}{\left[0.10-x\right]}\end{array}$

The value of $x$ will be very small as compared to $0.10$. Hence, it is ignored from the term $\left[0.10-x\right]$.

Simplify the above expression.

$\begin{array}{c}1.0\times {10}^{-7}=\frac{{\left[x\right]}^2}{\left[0.10\right]}\\ {\left[x\right]}^2=\left(1.0\times {10}^{-8}\right)\\ \left[x\right]=\underset{\_}{1.0\times 10^{-4}M}\end{array}$

Therefore, the $\left[{\text{H}}^{\text{+}}\right]$ from the first reaction is $\underset{\_}{1.0\times 10^{-4}M}$.

The initial $\left[{\text{HS}}^{-}\right]$ is $\underset{\_}{1.0\times 10^{-4}M}$.

According to the ICE table formed,

The $\left[{\text{HS}}^{-}\right]$ is equal to the $\left[{\text{H}}^{\text{+}}\right]$,that is $1.0\times {10}^{-4}\text{M}$.

This is the initial $\left[{\text{HS}}^{-}\right]$ for the further reaction.

The $\left[{\text{H}}^{\text{+}}\right]$ from second reaction is $\underset{\_}{1.0\times 10^{-19}M}$.

The change in concentration of ${\text{HS}}^{-}$ is assumed to be $y$.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{HS}}^{-}\left(aq\right)& \rightleftharpoons & {\text{H}}^{+}\left(aq\right)& +& {\text{S}}^{2-}\left(aq\right)\\ \text{Inititial}& 1.0\times {10}^{-4}& & 1.0\times {10}^{-4}& & 0\\ \text{Change}& -y& & +y& & +y\\ \text{Equilibrium}& \left(1.0\times {10}^{-4}\right)-y& & \left(1.0\times {10}^{-4}\right)+y& & y\end{array}$

The equilibrium concentration of $\left[{\text{HS}}^{-}\right]$ is $\left(\left(1.0\times {10}^{-4}\right)-y\right)\text{M}$.

The equilibrium concentration of $\left[{\text{H}}^{+}\right]$ is $\left(1.0\times {10}^{-4}\right)+y\text{M}$.

The equilibrium concentration of $\left[{\text{S}}^{2-}\right]$ is $y\text{M}$.

The equilibrium constant expression for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{S}}^{2-}\right]}{\left[{\text{HS}}^{-}\right]}$ (2)

The $K_{\text{a}}$ for this reaction is $1.0\times {10}^{-19}$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{HS}}^{-}\right]$, $\left[{\text{H}}^{+}\right]$ and $\left[{\text{S}}^{2-}\right]$ in equation (2).

$1.0\times {10}^{-19}=\frac{\left[\left(1.0\times {10}^{-4}\right)+y\right]\left[y\right]}{\left[\left(\left(1.0\times {10}^{-4}\right)-y\right)\right]}$

The value of $y$ will be very small as compared to $1.0\times {10}^{-4}$. Hence, it is ignored from the term $\left[\left(1.0\times {10}^{-4}\right)-y\right]$ and from $\left[\left(1.0\times {10}^{-4}\right)+y\right]$

Simplify the above expression.

$\begin{array}{c}1.0\times {10}^{-19}=\frac{\left[\left(1.0\times {10}^{-4}\right)\right]\left[y\right]}{\left[1.0\times {10}^{-4}\right]}\\ \left[y\right]=\underset{\_}{1.0\times 10^{-19}M}\end{array}$

Therefore, the $\left[{\text{H}}^{\text{+}}\right]$ from the second reaction is $\underset{\_}{1.0\times 10^{-19}M}$.

The $\left[{\text{S}}^{2-}\right]$ is $\underset{\_}{1.0\times 10^{-19}M}$.

According to the ICE table formed,

The $\left[{\text{S}}^{2-}\right]$ is equal to the $\left[{\text{H}}^{\text{+}}\right]$ from the second reaction, that is $\underset{\_}{1.0\times 10^{-19}M}$.

The total $\left[{\text{H}}^{\text{+}}\right]$ is $\underset{\_}{1.0\times 10^{-4}M}$.

The total $\left[{\text{H}}^{\text{+}}\right]$ is calculated by the formula,

$\text{Total}\left[{\text{H}}^{\text{+}}\right]=\left[{\text{H}}^{\text{+}}\right]\text{from}\text{first}\text{reaction}+\left[{\text{H}}^{\text{+}}\right]\text{from}\text{second}\text{reaction}$

Substitute the value of the $\left[{\text{H}}^{\text{+}}\right]$ from the first and second reaction in the above expression.

$\begin{array}{c}\text{Total}\left[{\text{H}}^{\text{+}}\right]=\left(\left(1.0\times {10}^{-4}\right)+\left(1.0\times {10}^{-19}\right)\right)\text{M}\\ =\underset{\_}{1.0\times 10^{-4}M}\end{array}$

The required $\text{pH}$ value is $\underset{\_}{4.0}$.

The $\text{pH}$ of a solution is calculated by the formula,

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{c}\text{pH}=-\log \left[1.0\times {10}^{-4}\right]\\ =\underset{\_}{4.0}\end{array}$

Conclusion

Conclusion

The $\text{pH}$ of the given solution is $\underset{\_}{4.0}$ and the $\left[{\text{S}}^{2-}\right]$ is $\underset{\_}{1.0\times 10^{-19}M}.$