Chapter 16, Problem 16.44QP

Interpretation Introduction

Interpretation: The $\text{pH}$ of the solution of trimethylamine and hydrochloric acid at different volume of acid is to be calculated.

Concept introduction: The $\text{pH}$ of the solution is calculated by using the formula,

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

To determine: The $\text{pH}$ of the solution of trimethylamine and hydrochloric acid at different volume of base.

Answer to Problem 16.44QP

Solution

The $\text{pH}$ of the solution at $10.0\text{mL}$ volume of acid is $\underset{\_}{9.79}$.

The $\text{pH}$ of the solution at $20.0\text{mL}$ volume of acid is $\underset{\_}{5.52}$.

The $\text{pH}$ of the solution at $30.0\text{mL}$ volume of acid is $\underset{\_}{1.64}$.

Explanation of Solution

Explanation

Given

The volume of trimethylamine is $25.0\text{mL}$.

The concentration of trimethylamine is $0.100\text{M}$.

The concentration of hydrochloric acid is $0.125\text{M}$.

The volume of hydrochloric acid added is $10.0\text{mL,}\text{20}\text{.0}\text{mL}$ and $30.0\text{mL}$.

At $10.0mL$ volume of hydrochloric acid:

The concentration of trimethylammonium chloride formed after addition of hydrochloric acid is calculated by multiplying the ratio of hydrochloric acid present in the solution with the concentration of hydrochloric acid solution.

$\begin{array}{c}\text{Concentration}\text{of}\left[{\left({\text{CH}}_3\right)}_3\text{NHCl}\right]=\frac{10.0}{10.0+25.0}\times 0.125\text{M}\\ =0.036\text{M}\end{array}$

The concentration of trimethylamine left in the solution after formation of trimethylammonium chloride is calculated by subtracting the concentration of trimethylammonium chloride from the volume of trimethylamine in the solution.

$\begin{array}{c}\text{Concentration}\text{of}\left[{\left({\text{CH}}_3\right)}_3\text{N}\right]\left(\text{in}\text{solution}\right)=\frac{25.0}{25.0+10.0}\times 0.100\text{M}-0.036\text{M}\\ =0.035\text{M}\end{array}$

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

$\text{14}-\text{pH}=\text{p}{K}_{\text{b}}+\log \frac{\left[{\left({\text{CH}}_3\right)}_3\text{NHCl}\right]}{\left[{\left({\text{CH}}_3\right)}_3\text{N}\right]}$

Where,

• $\text{pH}$ is the hydrogen content of solution.
• $\text{p}{K}_{\text{b}}$ is the dissociation constant of base.
• $\left[{\left({\text{CH}}_3\right)}_3\text{NHCl}\right]$ is the concentration of trimethylammonium chloride solution.
• $\left[{\left({\text{CH}}_3\right)}_3\text{N}\right]$ is the concentration of trimethylamine.

Substitute the values of $\left[{\left({\text{CH}}_3\right)}_3\text{NHCl}\right]$, $\left[{\left({\text{CH}}_3\right)}_3\text{N}\right]$ and $\text{p}{K}_{\text{b}}$ in the above equation.

$\begin{array}{c}\text{14}-\text{pH}=4.19+\log \frac{0.036}{0.035}\\ \text{pH}=\underset{\_}{9.79}\end{array}$

At $20.0mL$ volume of hydrochloric acid:

At $20.0\text{mL}$ volume of hydrochloric acid the whole acid present in the solution gets neutralized and the solution will reach at the equivalence point. The equivalent concentration of trimethylammonium chloride salt in the solution is calculated as,

$\begin{array}{c}\text{Concentration}\text{of}\left[{\left({\text{CH}}_3\right)}_3\text{NHCl}\right]=\frac{20.0}{20.0+25.0}\times 0.125\text{M}\\ =0.056\text{M}\end{array}$

The concentration of hydroxyl ions in the given solution is calculated by using the dissociation constant of trimethylammonium chloride by using the following equation.

${\left({\text{CH}}_3\right)}_3{\text{NH}}^{+}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\to {\left({\text{CH}}_3\right)}_3\text{N}\left(aq\right)+{\text{H}}_3{\text{O}}^{+}\left(aq\right)$

The concentration of trimethylamine and hydrogen ions formed in the given equilibrium reaction is considered to be $x$. The dissociation constant of the trimethylammonium chloride is calculated by using the formula,

$K_{\text{a}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\left({\text{CH}}_3\right)}_3\text{N}\right]}{\left[{\left({\text{CH}}_3\right)}_3{\text{NH}}^{+}\right]}$

Where,

• $K_{\text{a}}$ is the dissociation constant of the acid.
• $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is the hydrogen ion concentration.
• $\left[{\left({\text{CH}}_3\right)}_3\text{N}\right]$ is the concentration of trimethylamine.
• $\left[{\left({\text{CH}}_3\right)}_3{\text{NH}}^{+}\right]$ is the concentration of trimethylammonium ion.

Substitute the values of $K_{\text{a}}$, $\left[{\text{H}}_3{\text{O}}^{+}\right]$, $\left[{\left({\text{CH}}_3\right)}_3\text{N}\right]$ and $\left[{\left({\text{CH}}_3\right)}_3{\text{NH}}^{+}\right]$ in the above equation.

$1.56\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.056-x\right)}$

Since, the value of dissociation constant is very less than the concentration of the solution. Therefore, the value of $x<<<0.056$ due to which $0.056-x\simeq 0.056$.

$\begin{array}{c}1.56\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.056\right)}\\ x=2.96\times {10}^{-6}\end{array}$

Therefore, the concentration of hydrogen ions in the solution is $2.96\times {10}^{-6}$.

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

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

Where,

• $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions in the solution.

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

$\begin{array}{c}\text{pH}=-\log \left(2.96\times {10}^{-6}\right)\\ \text{pH}=\underset{\_}{5.52}\end{array}$

At $30.0mL$ volume of hydrochloric acid:

At $30.0\text{mL}$ volume of hydrochloric acid the volume of hydrochloric acid is present in excess in the solution. Therefore, the $\text{pH}$ of the solution corresponds to the excess acid present in the solution. The concentration of excess acid present in the solution is calculated as,

$\begin{array}{c}\text{Concentration}\text{of}\text{HCl}\left(\text{excess}\right)=\frac{10.0}{30.0+25.0}\times 0.125\text{M}\\ =0.023\text{M}\end{array}$

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

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

Where,

• $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions in the solution.

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

$\begin{array}{c}\text{pH}=-\log \left(0.023\right)\\ \text{pH}=\underset{\_}{1.64}\end{array}$

Conclusion

The $\text{pH}$ of the solution at $10.0\text{mL}$ volume of acid is $\underset{\_}{9.79}$.

The $\text{pH}$ of the solution at $20.0\text{mL}$ volume of acid is $\underset{\_}{5.52}$.

The $\text{pH}$ of the solution at $30.0\text{mL}$ volume of acid is $\underset{\_}{1.64}$