Chapter 14, Problem 119E

Calculate the pH of each of the following solutions.

a. 0.10 M CH₃NH₃Cl

b. 0.050 M NaCN

Interpretation Introduction

Interpretation: The $\text{pH}$ value for each of the given solutions 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 $\text{pOH}$ of a solution is calculated by the formula, $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

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

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

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

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

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

To determine: The $\text{pH}$ value for each of the given solution of $0.10\text{M}$ ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}$.

Answer to Problem 119E

Answer

The $\text{pH}$ of the given solution of $0.10\text{M}$ ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}$ is $\underset{\_}{5.82}$.

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, $K_{\text{a}}=\frac{\left[{\text{CH}}_3{\text{NH}}_2\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CH}}_{\text{3}}{\text{NH}}_3^{+}\right]}$

${\text{CH}}_{\text{3}}{\text{NH}}_3^{+}$ is a stronger acid than ${\text{H}}_{\text{2}}\text{O}$.

The dominant equilibrium reaction is,

${\text{CH}}_{\text{3}}{\text{NH}}_3^{+}\left(aq\right)\rightleftharpoons {\text{CH}}_3{\text{NH}}_2\left(aq\right)+{\text{H}}^{+}\left(aq\right)$

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,

$K_{\text{a}}=\frac{\left[{\text{CH}}_3{\text{NH}}_2\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CH}}_{\text{3}}{\text{NH}}_3^{+}\right]}$ (1)

The $K_{\text{a}}$ value is $\underset{\_}{2.28\times 10^{-11}M}$.

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

The value of $K_{\text{b}}$ for methyl amine is $4.38\times {10}^{-4}$.

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.

$\begin{array}{c}{K}_{\text{a}}=\frac{1.0\times {10}^{-14}}{4.38\times {10}^{-4}}\\ =\underset{\_}{2.28\times 10^{-11}}\end{array}$

The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{1.50\times 10^{-6}M}$.

The change in concentration of ${\text{CH}}_{\text{3}}{\text{NH}}_3^{+}$ 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,

$\begin{array}{cccccc}& {\text{CH}}_{\text{3}}{\text{NH}}_3^{+}\left(aq\right)& \rightleftharpoons & {\text{CH}}_3{\text{NH}}_2\left(aq\right)& +& {\text{H}}^{+}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.1& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.1-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{CH}}_{\text{3}}{\text{NH}}_3^{+}\right]$ is $\left(0.1-x\right)\text{M}$.

The equilibrium concentration of $\left[{\text{CH}}_3{\text{NH}}_2\right]$ is $x\text{M}$.

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

The calculated value of $K_{\text{a}}$ is $2.28\times {10}^{-11}$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{CH}}_{\text{3}}{\text{NH}}_3^{+}\right]$, $\left[{\text{CH}}_3{\text{NH}}_2\right]$ and $\left[{\text{H}}^{+}\right]$ in equation (1).

$\begin{array}{c}2.28\times {10}^{-11}=\frac{\left[x\right]\left[x\right]}{\left[0.1-x\right]}\\ 2.28\times {10}^{-11}=\frac{{\left[x\right]}^2}{\left[0.1-x\right]}\end{array}$

Solve the above expression.

$\begin{array}{c}2.28\times {10}^{-11}=\frac{{\left[x\right]}^2}{\left[0.1-x\right]}\\ \left[x\right]=\underset{\_}{1.50\times 10^{-6}M}\end{array}$

Therefore, the $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{1.50\times 10^{-6}M}$.

The $\text{pH}$ of the given solution is $\underset{\_}{5.82}$.

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.

$\begin{array}{c}\text{pH}=-\log \left[1.50\times {10}^{-6}\right]\\ =\underset{\_}{5.82}\end{array}$

Interpretation Introduction

Interpretation: The $\text{pH}$ value for each of the given solutions 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 $\text{pOH}$ of a solution is calculated by the formula, $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

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

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

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

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

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

To determine: The $\text{pH}$ value for each of the given solution of $0.050\text{M}$ $\text{NaCN}$.

Answer to Problem 119E

Answer

The $\text{pH}$ of the given solution of $0.050\text{M}$ $\text{NaCN}$ is $\underset{\_}{10.945}$.

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is,

$K_{\text{b}}=\frac{\left[\text{HCN}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CN}}^{-}\right]}$

The dominant equilibrium reaction is,

${\text{CN}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons \text{HCN}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

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,

$K_{\text{b}}=\frac{\left[\text{HCN}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CN}}^{-}\right]}$ (1)

The $K_{\text{b}}$ value is $\underset{\_}{1.61\times 10^{-5}}$.

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

The value of $K_{\text{a}}$ for $\text{HCN}$ is $6.2\times {10}^{-10}$.

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

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

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

$\begin{array}{c}{K}_{\text{b}}=\frac{1.0\times {10}^{-14}}{6.2\times {10}^{-10}}\\ =\underset{\_}{1.61\times 10^{-5}}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{8.8\times 10^{-4}M}$.

The change in concentration of ${\text{CN}}^{-}$ 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,

$\begin{array}{cccccc}& {\text{CN}}^{-}\left(aq\right)& \rightleftharpoons & \text{HCN}\left(aq\right)& +& {\text{OH}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.05& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.05-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{CN}}^{-}\right]$ is $\left(0.05-x\right)\text{M}$.

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

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

The calculated value of $K_{\text{b}}$ is $1.61\times {10}^{-5}$.

Substitute the value of $K_{\text{b}}$, $\left[{\text{CN}}^{-}\right]$, $\left[{\text{CN}}^{-}\right]$ and $\left[{\text{OH}}^{-}\right]$ in equation (1).

$\begin{array}{c}1.61\times {10}^{-5}=\frac{\left[x\right]\left[x\right]}{\left[0.05-x\right]}\\ 1.61\times {10}^{-5}=\frac{{\left[x\right]}^2}{\left[0.05-x\right]}\end{array}$

Solve the above expression.

$\begin{array}{c}1.61\times {10}^{-5}=\frac{{\left[x\right]}^2}{\left[0.05-x\right]}\\ \left[x\right]=\underset{\_}{8.8\times 10^{-4}M}\end{array}$

Therefore, the $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{8.8\times 10^{-4}M}$.

The $\text{pOH}$ of the given solution is $\underset{\_}{3.055}$.

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

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

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

$\begin{array}{c}\text{pOH}=-\log \left[8.8\times {10}^{-4}\right]\\ =\underset{\_}{3.055}\end{array}$

The $\text{pH}$ of the given solution is $\underset{\_}{10.945}$.

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

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

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

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

$\begin{array}{c}\text{pH}=14-\text{3}\text{.055}\\ =\underset{\_}{10.945}\end{array}$