Chapter 13, Problem 98E

Calculate [OH⁻], [H⁺], and the pH of 0.40 M solutions of each of the following amines (the K_b values are found in Table 13-3).

a. aniline

b. methylamine

Interpretation Introduction

Interpretation: The $\text{pH,}\left[{\text{H}}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ of the given amines 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 $\text{pOH}$ 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,

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

To determine: The $\text{pH,}\left[{\text{H}}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ of ${\text{C}}_6{\text{H}}_5{\text{NH}}_2$.

Answer to Problem 98E

Answer

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.23\times 10^{-5}M}$. The $\text{pH}$ value is $\underset{\_}{9.09}$. The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{8.13\times 10^{-10}}$.

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_3^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_2\right]}$

The reaction involved is,

${\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{C}}_6{\text{H}}_5{\text{NH}}_3^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

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,

$K_{\text{b}}=\frac{\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_3^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_2\right]}$ (1)

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.23\times 10^{-5}M}$.

The change in concentration of ${\text{C}}_6{\text{H}}_5{\text{NH}}_2$ 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,

$\begin{array}{cccccc}& {\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(aq\right)& \rightleftharpoons & {\text{C}}_6{\text{H}}_5{\text{NH}}_3^{+}\left(aq\right)& +& {\text{OH}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.40& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.40-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_2\right]$ is $\left(0.40-x\right)\text{M}$.

The equilibrium concentration of $\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_3^{+}\right]$ is $x\text{M}$.

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

The $K_{\text{b}}$ value is given to be $3.8\times {10}^{-10}$.

Substitute the value of $K_{\text{b}}$, $\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_2\right]$, $\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_3^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ in equation (1).

$\begin{array}{c}3.8\times {10}^{-10}=\frac{\left[x\right]\left[x\right]}{\left[0.40-x\right]}\\ 3.8\times {10}^{-10}=\frac{{\left[x\right]}^2}{\left[0.40-x\right]}\end{array}$

The value of $x$ will be very small as compared to $0.40$. Hence, it is ignored from the term $\left[0.40-x\right]$.

Simplify the above expression.

$\begin{array}{c}3.8\times {10}^{-10}=\frac{{\left[x\right]}^2}{\left[0.40\right]}\\ {\left[x\right]}^2=\left(1.52\times {10}^{-10}\right)\\ \left[x\right]=\underset{\_}{1.23\times 10^{-5}M}\end{array}$

Therefore, the $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.23\times 10^{-5}M}$.

The $\text{pOH}$ value is $\underset{\_}{4.91}$.

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.

$\begin{array}{c}\text{pOH}=-\log \left[1.23\times {10}^{-5}\right]\\ =\underset{\_}{4.91}\end{array}$

The $\text{pH}$ value is $\underset{\_}{9.09}$.

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

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

$\begin{array}{c}\text{pH}+\text{4}\text{.91}=14\\ \text{pH}=\underset{\_}{9.09}\end{array}$

The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{8.13\times 10^{-10}}$.

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

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

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

$\left[{\text{H}}^{+}\right]={10}^{-\text{pH}}$

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

$\begin{array}{c}\left[{\text{H}}^{+}\right]={10}^{-9.09}\\ =\underset{\_}{8.13\times 10^{-10}}\end{array}$

Interpretation Introduction

Interpretation: The $\text{pH,}\left[{\text{H}}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ of the given amines 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 $\text{pOH}$ 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,

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

To determine: The $\text{pH,}\left[{\text{H}}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ of ${\text{CH}}_3{\text{NH}}_2$.

Answer to Problem 98E

Answer

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.3\times 10^{-2}M}$. The $\text{pH}$ value is $\underset{\_}{12.11}$. The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{7.76\times 10^{-13}}$.

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{CH}}_3{\text{NH}}_3^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_3{\text{NH}}_2\right]}$

The reaction involved is,

${\text{CH}}_3{\text{NH}}_2\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{CH}}_3{\text{NH}}_3^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

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 dissociation constant.

The equilibrium constant expression for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{CH}}_3{\text{NH}}_3^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_3{\text{NH}}_2\right]}$ (1)

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.3\times 10^{-2}M}$.

The change in concentration of ${\text{C}}_6{\text{H}}_5{\text{NH}}_2$ 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,

$\begin{array}{cccccc}& {\text{CH}}_3{\text{NH}}_2\left(aq\right)& \rightleftharpoons & {\text{CH}}_3{\text{NH}}_3^{+}\left(aq\right)& +& {\text{OH}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.40& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.40-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{CH}}_3{\text{NH}}_2\right]$ is $\left(0.40-x\right)\text{M}$.

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

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

The given $K_{\text{b}}$ value is given to be $4.38\times {10}^{-4}$.

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

$\begin{array}{c}4.38\times {10}^{-4}=\frac{\left[x\right]\left[x\right]}{\left[0.40-x\right]}\\ 4.38\times {10}^{-4}=\frac{{\left[x\right]}^2}{\left[0.40-x\right]}\end{array}$

The value of $x$ will be very small as compared to $0.40$. Hence, it is ignored from the term $\left[0.40-x\right]$.

Simplify the above expression.

$\begin{array}{c}4.38\times {10}^{-4}=\frac{{\left[x\right]}^2}{\left[0.40\right]}\\ {\left[x\right]}^2=\left(1.752\times {10}^{-4}\right)\\ \left[x\right]=\underset{\_}{1.3\times 10^{-2}M}\end{array}$

Therefore, the $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.3\times 10^{-2}M}$.

The $\text{pOH}$ value is $\underset{\_}{1.89}$.

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.

$\begin{array}{c}\text{pOH}=-\log \left[1.3\times {10}^{-2}\right]\\ =\underset{\_}{1.89}\end{array}$

The $\text{pH}$ value is $\underset{\_}{12.11}$.

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

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

$\begin{array}{c}\text{pH}+\text{1}\text{.89}=14\\ \text{pH}=\underset{\_}{12.11}\end{array}$

The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{7.76\times 10^{-13}}$.

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

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

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

$\left[{\text{H}}^{+}\right]={10}^{-\text{pH}}$

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

$\begin{array}{c}\left[{\text{H}}^{+}\right]={10}^{-12.11}\\ =\underset{\_}{7.76\times 10^{-13}}\end{array}$