Chapter 16, Problem 16.116QP

Interpretation Introduction

Interpretation: The value of $\text{p}{K}_{\text{b}}$ of trimethylamine is given to be $4.19$ at $25°\text{C}$. The $\text{pH}$ of the solution is to be calculated after addition of $10.0,20.0,\text{and}30.0\text{ml}$ of acid.

Concept introduction: The value of $\text{pH}$ is calculated by the use of concentration of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions in the solution. $\text{pH}$ is the numeric value which defines acidity and basicity of the solution.

To determine: The $\text{pH}$ of the solution after addition of $10.0,20.0,\text{and}30.0\text{ml}$ of acid.

Answer to Problem 16.116QP

Answer

The value of $\text{pH}$ after addition of $10\text{ml}$ of acid is $\underset{\_}{9.81}$.

The value of $\text{pH}$ after addition of $20\text{ml}$ of acid is $\underset{\_}{5.53}$.

The value of $\text{pH}$ after addition of $30\text{ml}$ of acid is $\underset{\_}{1.65}$.

Explanation of Solution

Explanation

Given

The value of $\text{p}{K}_{\text{b}}$ of solution is $4.19$.

Volume of trimethylamine, $V_1$ is $25.0\text{ml}$.

Concentration of trimethylamine, $M_1$ is $0.100\text{M}$.

Concentration of $\text{HCl}$, $M_2$ is $0.125\text{M}$.

The addition of $\text{HCl}$ in a solution, $V_2$ is $10\text{ml}$.

The equation for given reaction is,

${\left({\text{CH}}_3\right)}_3\text{N}+\text{HCl}\to {\left({\text{CH}}_3\right)}_3{\text{NH}}^{+}{\text{Cl}}^{-}$

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

$\text{mmol}\text{of}\text{acid}=M_2{V}_2$ (1)

Substitute the value of $M_2\text{and}{V}_2$ in equation (1).

$\begin{array}{c}\text{mmol}\text{of}\text{acid,}{\text{BH}}^{+}=0.125\times 10\\ =1.25\text{mmol}\end{array}$ (2)

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

$\text{mmol}\text{of}\text{base}=M_1{V}_1$ (3)

Substitute the value of $M_2\text{and}{V}_2$ in equation (1).

$\begin{array}{c}\text{mmol}\text{of}\text{base}=0.1\times 25\\ =2.5\text{mmol}\end{array}$ (4)

The value of $\text{mmol}$ of base after reaction is calculated by subtract the equation (2) and (4),

$\begin{array}{c}\text{mmol}\text{of}\text{base}\text{after}\text{reaction,}\text{B}=2.5-1.25\\ =1.25\text{mmol}\end{array}$

The value of $\text{mmol}$ of conjugate that is formed during the reaction is $1.25$. The given reaction forms a buffer solution.

The entire base reacts with acid, so the total volume of solution is calculated by addition of acidic and basic substances.

$V_{\text{Total}}=V_{\text{acid}}+V_{\text{base}}$ (5)

Where,

• $V_{\text{acid}}$ is the volume of acid.
• $V_{\text{base}}$ is the volume of base.

Substitute the values of $V_{\text{acid}}\text{and}{V}_{\text{base}}$ in equation (5).

$\begin{array}{c}{V}_{\text{Total}}=20+25\\ =45\text{ml}\end{array}$

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

$\text{pOH}=\text{pKb}+\text{log}\left({\text{BH}}^{+}/\text{B}\right)$

Substitute the value of $\text{pKb,}{\text{BH}}^{+}\text{and}\text{B}$ in above equation,

$\begin{array}{c}\text{pOH}=\text{4}\text{.19}+\text{log}\left(1.25/1.25\right)\\ =4.19\end{array}$

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

$\text{pH}=14-\text{pOH}$ (5)

Substitute the value of $\text{pOH}$ in equation (5).

$\begin{array}{c}\text{pH}=14-4.19\\ =\underset{\_}{9.81}\end{array}$

Therefore, the value of $\text{pH}$ is $\underset{\_}{9.81}$.

Similarly,

The addition of $\text{HCl}$ in a solution, $V_2$ is $20\text{ml}$.

The equation for given reaction is,

${\left({\text{CH}}_3\right)}_3\text{N}+\text{HCl}\to {\left({\text{CH}}_3\right)}_3{\text{NH}}^{+}{\text{Cl}}^{-}$

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

$\text{mmol}\text{of}\text{acid}=M_2{V}_2$ (1)

Substitute the value of $M_2\text{and}{V}_2$ in equation (1).

$\begin{array}{c}\text{mmol}\text{of}\text{acid}=0.125\times 20\\ =2.5\text{mmol}\end{array}$ (2)

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

$\text{mmol}\text{of}\text{base}=M_1{V}_1$ (3)

Substitute the value of $M_2\text{and}{V}_2$ in equation (1).

$\begin{array}{c}\text{mmol}\text{of}\text{base}=0.1\times 25\\ =2.5\text{mmol}\end{array}$ (4)

At this point value of $\text{mmol}$ of acid becomes equal to the value of $\text{mmol}$ of base.

The total volume of solution is calculated by addition of acidic and basic substances.

$V_{\text{Total}}=V_{\text{acid}}+V_{\text{base}}$ (5)

Substitute the values of $V_{\text{acid}}\text{and}{V}_{\text{base}}$ in equation (5).

$\begin{array}{c}{V}_{\text{Total}}=20+25\\ =45\text{ml}\end{array}$

The value of conjugate formation is equal to $2.5\text{mmol}$.

The value of conjugate concentration is calculated by the formula,

$\begin{array}{c}\left[{\text{BH}}^{+}\right]=\frac{\text{mmol}\text{of}\text{conjugate}\text{salt}}{V_{\text{total}}}\\ =\frac{2.5}{45}\\ =0.05555\text{M}\end{array}$

The equation of reaction of conjugate formed with water is given as,

${\text{BH}}^{+}+{\text{H}}_2\text{O}\to {\text{H}}_3{\text{O}}^{+}+\text{B}$

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

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[\text{B}\right]}{\left[{\text{BH}}^{+}\right]}\\ =\frac{K_{\text{w}}}{K_{\text{b}}}\end{array}$ (6)

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

Substitute the value of $K_{\text{w}}\text{and}{K}_{\text{b}}$ in equation (6).

$\begin{array}{c}{K}_{\text{a}}=\frac{{10}^{-14}}{{10}^{-4.19}}\\ =1.548\times {10}^{-10}\end{array}$

So, $\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]=x=\left[\text{B}\right]\\ \left[{\text{BH}}^{+}\right]=\text{M}-x\\ =0.05555-x\end{array}$

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

$K_a=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[\text{B}\right]}{\left[{\text{BH}}^{+}\right]}$ (7)

Where,

• $K_{\text{a}}$ is equilibrium constant.
• $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is the concentration of an acid.
• $\left[{\text{BH}}^{+}\right]$ is the concentration of conjugate salt.

Substitute the value of $\left[{\text{H}}_3{\text{O}}^{+}\right],\left[\text{B}\right]\text{and}\left[{\text{BH}}^{+}\right]$ in equation (7).

$\begin{array}{c}1.548\times {10}^{-10}=\frac{x\times x}{\left(0.5555-x\right)}\\ x^2=1.548\times {10}^{-10}\times \left(0.5555-x\right)\\ x^2=1.548\times {10}^{-10}\times 0.5555-1.548\times {10}^{-10}x\\ 0=x^2-0.859914\times {10}^{-10}+1.548\times {10}^{-10}x\end{array}$

Simplify the above expression,

$\begin{array}{c}x=\frac{-1.548\times {10}^{-10}\pm \sqrt{\left(-1.548\times {10}^{-10}\right)-4\left(1\times 0.859914\times {10}^{-10}\right)}}{2}\\ x=\frac{-1.548\times {10}^{-10}\pm \sqrt{2.39\times {10}^{-20}+3.43\times {10}^{-10}}}{2}\\ x=\frac{-1.548\times {10}^{-10}+\sqrt{5.82\times {10}^{-30}}}{2}\\ x=2.93\times {10}^{-6}\end{array}$

The value of $\left[{\text{H}}^{+}\right]$ is equal to $x$.

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

$\text{pH}=-\text{log}\left[{\text{H}}^{+}\right]$ (8)

Substitute the value of $\left[{\text{H}}^{+}\right]$ in equation (8).

$\begin{array}{c}\text{pH}=-\text{log}\left(2.93\times {10}^{-6}\right)\\ =\underset{\_}{5.53}\end{array}$

Therefore, the value of $\text{pH}$ is $\underset{\_}{5.53}$.

Similarly,

The addition of $\text{HCl}$ in a solution, $V_2$ is $30\text{ml}$.

The equation for given reaction is,

${\left({\text{CH}}_3\right)}_3\text{N}+\text{HCl}\to {\left({\text{CH}}_3\right)}_3{\text{NH}}^{+}{\text{Cl}}^{-}$

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

$\text{mmol}\text{of}\text{acid}=M_2{V}_2$ (1)

Substitute the value of $M_2\text{and}{V}_2$ in equation (1).

$\begin{array}{c}\text{mmol}\text{of}\text{acid}=0.125\times 30\\ =3.75\text{mmol}\end{array}$ (2)

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

$\text{mmol}\text{of}\text{base}=M_1{V}_1$ (3)

Substitute the value of $M_1\text{and}{V}_1$ in equation (1).

$\begin{array}{c}\text{mmol}\text{of}\text{base}=0.1\times 25\\ =2.5\text{mmol}\end{array}$ (4)

The total volume of solution is calculated by addition of acidic and basic substances.

$V_{\text{Total}}=V_{\text{acid}}+V_{\text{base}}$ (5)

Substitute the values of $V_{\text{acid}}\text{and}{V}_{\text{base}}$ in equation (5).

$\begin{array}{c}{V}_{\text{Total}}=30+25\\ =55\text{ml}\end{array}$

The acid is excess and the value of $\text{mmol}$ of acid left is calculated by the formula,

$\text{mmol}\text{of}\text{remained}\text{acid}=\text{mmol}\text{of}\text{acid}-\text{mmol}\text{of}\text{base}$ (6)

Substitute the values of $\text{mmol}$ of acid and base in equation (6).

$\begin{array}{c}\text{mmol}\text{of}\text{remained}\text{acid}=3.75-2.5\\ =1.25\text{mmol}\end{array}$

The concentration of acid is calculated by the formula,

$\left[\text{acid}\right]=\frac{\text{mmol}\text{of}\text{acid}}{{\text{V}}_{\text{total}}\text{of}\text{acid}}$ (7)

Substitute the value of $\text{mmol}\text{of}\text{acid}\text{and}{\text{V}}_{\text{total}}\text{of}\text{acid}$ in equation (7).

$\begin{array}{c}\left[\text{acid}\right]=\frac{\text{mmol}\text{of}\text{acid}}{{\text{V}}_{\text{total}}\text{of}\text{acid}}\\ =\frac{1.25}{55}\\ =0.0227\text{M}\end{array}$

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

$\text{pH}=-\text{log}\left[{\text{H}}^{+}\right]$ (8)

Substitute the value of $\left[{\text{H}}^{+}\right]$ in equation (8).

$\begin{array}{c}\text{pH}=-\text{log}\left(0.227\text{M}\right)\\ =1.65\end{array}$

Therefore, the value of $\text{pH}$ is $\underset{\_}{1.65}$.

Conclusion

Conclusion

The value of $\text{pH}$ after addition of $10\text{ml}$ of acid is $\underset{\_}{9.81}$.

The value of $\text{pH}$ after addition of $20\text{ml}$ of acid is $\underset{\_}{5.53}$.

The value of $\text{pH}$ after addition of $30\text{ml}$ of acid is $\underset{\_}{1.65}$.