Chapter 13, Problem 126E

Interpretation Introduction

Interpretation: A $0.20\text{M}$ solution of the salt ${\text{NaC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_{\text{2}}$ is given to have a $\text{pH}$ value of $8.65$. The $\text{pH}$ of a $0.20\text{M}$ solution of ${\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_{\text{2}}$ is to be calculated.

Concept introduction: The $\text{pH}$ of a solution is define 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, $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

The sum, $\text{pH}+\text{pOH}=14$

The equilibrium constant for water is denoted by $K_{\text{w}}$ and is expressed as,

$K_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$

The value of $K_{\text{w}}$ is calculated by the formula,

$K_{\text{w}}=K_{\text{a}}\cdot K_{\text{b}}$

Answer to Problem 126E

Answer

The $\text{pH}$ of the given $0.20\text{M}$ solution of ${\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_{\text{2}}$ is $\underset{\_}{2.65}$.

Explanation of Solution

Explanation

To determine: The $\text{pH}$ of a $0.20\text{M}$ solution of ${\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_{\text{2}}$.

The $\text{pOH}$ of the given ${\text{NaC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_{\text{2}}$ salt is $\underset{\_}{5.35}$.

The given value of $\text{pH}$ is $8.65$.

The sum, $\text{pH}+\text{pOH}=14$

The value of $\text{pOH}$ is calculated by the formula,

$\text{pOH}=14-\text{pH}$

Substitute the value of $\text{pH}$ in the above expression.

$\begin{array}{c}\text{pOH}=14-\text{8}\text{.65}\\ =\underset{\_}{5.35}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{4.467\times 10^{-6}M}$.

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

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Rearrange the above expression to calculate the value of $\left[{\text{OH}}^{-}\right]$.

$\left[{\text{OH}}^{-}\right]={10}^{-\text{pOH}}$

Substitute the value of $\text{pOH}$ in the above expression.

$\begin{array}{c}\left[{\text{OH}}^{-}\right]={10}^{-\text{5}\text{.35}}\\ =\underset{\_}{4.467\times 10^{-6}M}\end{array}$

The equilibrium constant expression for the given reaction is, $K_{\text{b}}=\frac{\left[{\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_{\text{2}}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2^{-}\right]}$

The dominant equilibrium reaction is,

${\text{NaC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2^{-}\left(aq\right)\rightleftharpoons {\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_{\text{2}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

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

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

Where,

• $K_{\text{b}}$ is the base ionization constant.

The equilibrium constant expression for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_{\text{2}}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2^{-}\right]}$ (1)

The $K_{\text{b}}$ value is $\underset{\_}{3.99\times 10^{-10}}$.

The change in concentration of ${\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2^{-}$ is assumed to be $x$.

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2^{-}\left(aq\right)& \rightleftharpoons & {\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2\left(aq\right)& +& {\text{OH}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.20& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.20-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2^{-}\right]$ is $\left(0.20-x\right)\text{M}$.

The equilibrium concentration of $\left[{\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2\right]$ is $x\text{M}$.

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

Substitute the value of $\left[{\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2^{-}\right]$, $\left[{\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2\right]$ and $\left[{\text{OH}}^{-}\right]$ in equation (1).

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[x\right]\left[x\right]}{\left[0.20-x\right]}\\ K_{\text{b}}=\frac{{\left[x\right]}^2}{\left[0.20-x\right]}\end{array}$

The value of $\left[{\text{OH}}^{-}\right]$ that is equal to the value of $x$ is $\text{4}{\text{.467×10}}^{-6}\text{M}$.

Substitute the value of $x$ in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{{\left[\text{4}{\text{.467×10}}^{-6}\right]}^2}{\left[0.050-\left(\text{4}{\text{.467×10}}^{-6}\right)\right]}\\ K_{\text{b}}=\underset{\_}{3.99\times 10^{-10}}\end{array}$

The $K_{\text{a}}$ value is $\underset{\_}{2.51\times 10^{-5}}$.

The value, $K_{\text{w}}=K_{\text{a}}\cdot K_{\text{b}}$

The calculated value of $K_{\text{b}}$ is $3.99\times {10}^{-10}$.

The value of $K_{\text{a}}$ is calculated by the formula,

$K_{\text{a}}=\frac{K_{\text{w}}}{K_{\text{b}}}$

Substitute the value of $K_{\text{b}}$ in the above expression.

$\begin{array}{c}{K}_{\text{a}}=\frac{1.0\times {10}^{-14}}{3.99\times {10}^{-10}}\\ =\underset{\_}{2.51\times 10^{-5}}\end{array}$

The equilibrium constant expression for the second reaction is, $K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2^{-}\right]}{\left[{\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2\right]}$

The dominant equilibrium reaction is,

${\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2^{-}\left(aq\right)$

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,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2^{-}\right]}{\left[{\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2\right]}$ (2)

The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{2.24\times 10^{-3}M}$.

The change in concentration of ${\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2$ is assumed to be $x$.

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2\left(aq\right)& \rightleftharpoons & {\text{H}}^{+}\left(aq\right)& +& {\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.20& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.20-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2\right]$ is $\left(0.20-x\right)\text{M}$.

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

The equilibrium concentration of $\left[{\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2^{-}\right]$ is $x\text{M}$.

The calculated value of $K_{\text{a}}$ is $2.51\times {10}^{-5}$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2\right]$, $\left[{\text{H}}^{+}\right]$ and $\left[{\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_2^{-}\right]$ in equation (2).

$\begin{array}{c}2.51\times {10}^{-5}=\frac{\left[x\right]\left[x\right]}{\left[0.20-x\right]}\\ 2.51\times {10}^{-5}=\frac{{\left[x\right]}^2}{\left[0.20-x\right]}\end{array}$

The value of $\left[0.20-x\right]$ is approximated to be equal to $0.20\text{M}$.

Simplify the above expression.

$\begin{array}{c}2.51\times {10}^{-5}=\frac{{\left[x\right]}^2}{\left[0.20\right]}\\ \left[x\right]=\underset{\_}{2.24\times 10^{-3}M}\end{array}$

Therefore, the $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{2.24\times 10^{-3}M}$.

The $\text{pH}$ of the given solution is $\underset{\_}{2.65}$.

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

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

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

$\begin{array}{c}\text{pH}=-\log \left[2.24\times {10}^{-3}\right]\\ =\underset{\_}{2.65}\end{array}$

Conclusion

Conclusion

The $\text{pH}$ of the given $0.20\text{M}$ solution of ${\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_{\text{2}}$ is $\underset{\_}{2.65}$.