Chapter 15, Problem 15.60QP

(a)

Interpretation Introduction

Interpretation: The $K_{\text{b}}$ values of trimethylamine, methylamine and dimethylamine are given to be $6.5\times {10}^{-5}$, $4.4\times {10}^{-4}$ and $5.9\times {10}^{-4}$ respectively. The $\text{pH}$ of $3.00\times {10}^{-4}\text{M}$, $2.88\times {10}^{-3}$ of trimethylamine and methylamine respectively are to be calculated. The concentration of dimethylamine required to have the same $\text{pH}$ as that of methylamine is to be identified.

Concept introduction: The trimethylamine, methylamine and dimethylamine compounds are basic in nature due to presence of methyl group in their structures. The methyl groups possess positive inductive effect that makes the compound more basic in nature.

To determine: The $\text{pH}$ value of $6.5\times {10}^{-5}$ solution of trimethylamine is to be calculated.

Answer to Problem 15.60QP

Solution

The $\text{pH}$ of $6.5\times {10}^{-5}$ solution of trimethylamine is $\underset{\_}{10.14}$.

Explanation of Solution

Explanation

Given

The value of $K_{\text{b}}$ of trimethylamine is $6.5\times {10}^{-5}$.

The concentration of trimethylamine is $3.00\times {10}^{-4}\text{M}$.

The $\left[{\text{OH}}^{-}\right]$ concentration is calculated by the formula,

$\left[{\text{OH}}^{-}\right]=\sqrt{K_{\text{b}}\times \left[\text{base}\right]}$ (1)

Substitute the value of $K_{\text{b}}$ and $\left[\text{base}\right]$ in equation (1).

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\sqrt{\left(6.5\times {10}^{-5}\text{M}\right)\times 3.00\times {10}^{-4}\text{M}}\\ =1.40\times {10}^{-4}\text{M}\end{array}$

The concentration of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is calculated by the formula,

$\left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{K_{\text{w}}}{\left[{\text{OH}}^{-}\right]}$ (2)

The value of $K_{\text{w}}$ is ${10}^{-14}$.

Substitute the value of $K_{\text{w}}$ and $\left[{\text{OH}}^{-}\right]$ in equation (2).

$\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{{10}^{-14}}{1.40\times {10}^{-4}}\\ =7.14\times {10}^{-11}\end{array}$

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

$\text{pH}=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$ (3)

Substitute the value of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in equation 3.

$\begin{array}{c}\text{pH}=-\text{log}\left(7.14\times {10}^{-11}\right)\\ =\underset{\_}{10.14}\end{array}$

Therefore, the $\text{pH}$ of $6.5\times {10}^{-5}$ solution of trimethylamine is $\underset{\_}{10.14}$.

(b)

Interpretation Introduction

To determine: The $\text{pH}$ of $2.88\times {10}^{-3}$ solution of methylamine.

Answer to Problem 15.60QP

Solution

The $\text{pH}$ of $2.88\times {10}^{-3}\text{M}$ solution of methylamine is $\underset{\_}{11.05}$.

Explanation of Solution

Explanation

Given

The value of $K_{\text{b}}$ of methylamine is $4.4\times {10}^{-4}$.

The concentration of methylamine is $2.88\times {10}^{-3}\text{M}$.

The $\left[{\text{OH}}^{-}\right]$ concentration is calculated by the formula,

$\left[{\text{OH}}^{-}\right]=\sqrt{K_{\text{b}}\times \left[\text{base}\right]}$ (1)

Substitute the value of $K_{\text{b}}$ and $\left[\text{base}\right]$ in equation (1).

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\sqrt{\left(4.4\times {10}^{-4}\text{M}\right)\times 2.88\times {10}^{-3}\text{M}}\\ =1.13\times {10}^{-3}\text{M}\end{array}$

The concentration of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is calculated by the formula,

$\left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{K_{\text{w}}}{\left[{\text{OH}}^{-}\right]}$ (2)

The value of $K_{\text{w}}$ is ${10}^{-14}$.

Substitute the value of $K_{\text{w}}$ and $\left[{\text{OH}}^{-}\right]$ in equation (2).

$\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{{10}^{-14}}{1.13\times {10}^{-3}}\\ =8.85\times {10}^{-12}\end{array}$

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

$\text{pH}=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$ (3)

Substitute the value of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in equation 3.

$\begin{array}{c}\text{pH}=-\text{log}\left(8.85\times {10}^{-12}\right)\\ =\underset{\_}{11.05}\end{array}$

Therefore, the $\text{pH}$ of $2.88\times {10}^{-3}\text{M}$ solution of methylamine is $\underset{\_}{11.05}$.

(c)

Interpretation Introduction

To determine: The concentration of dimethylamine required to have the same $\text{pH}$ as that of methylamine.

Answer to Problem 15.60QP

Solution

The required concentration of dimethylamine is $\underset{\_}{2.13\times 10^{-3}M}$.

Explanation of Solution

Explanation

Given

The $K_{\text{b}}$ value of dimethylamine is $5.9\times {10}^{-4}$.

The $\text{pH}$ of dimethylamine is $11.05$.

The concentration of ${\text{H}}^{+}$ is calculated by the formula,

$\left[{\text{H}}^{+}\right]={10}^{-\text{pH}}$ (4)

Substitute the value of $\text{pH}$ in equation (4).

$\begin{array}{c}\left[{\text{H}}^{+}\right]={10}^{-11.05}\\ =8.91\times {10}^{-12}\end{array}$

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

$\begin{array}{c}{K}_{\text{w}}=\left[{\text{H}}^{+}\right]\times \left[{\text{OH}}^{-}\right]\\ ={10}^{-4}\end{array}$

The concentration of $\left[{\text{OH}}^{-}\right]$ is calculated by the formula,

$\left[{\text{OH}}^{-}\right]=\frac{K_{\text{w}}}{\left[{\text{H}}^{+}\right]}$ (5)

Substitute the value of $K_{\text{w}}$ and $\left[{\text{H}}^{+}\right]$ in equation (5).

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{{10}^{-14}}{8.91\times {10}^{-12}}\\ =0.00112\end{array}$

The concentration of base is calculated by the formula,

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\sqrt{K_{\text{b}}\times \left[\text{base}\right]}\\ {\left[{\text{OH}}^{-}\right]}^2=K_{\text{b}}\times \left[\text{base}\right]\end{array}$

Simplify the above expression.

$\begin{array}{c}\left[\text{Base}\right]=\frac{{\left[{\text{OH}}^{-}\right]}^2}{K_{\text{b}}}\\ =\frac{{(0.00112)}^2}{5.9\times {10}^{-4}}\\ \left[\text{Base}\right]=\underset{\_}{2.13\times 10^{-3}M}\end{array}$

Therefore, the required concentration of dimethylamine is $\underset{\_}{2.13\times 10^{-3}M}$ to have the same $\text{pH}$ as the solution in part (b).