Chapter 16, Problem 16.115QP

Interpretation Introduction

Interpretation: The $\text{pH}$ of the solution is to be calculated at $25°\text{C}$ after addition of $10.0,20.0,\text{and}30.0\text{ml}$ of base.

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 base.

Answer to Problem 16.115QP

The value of $\text{pH}$ after addition of $10\text{ml}$ of base is $\underset{\_}{4.75}$.

The value of $\text{pH}$ after addition of $20\text{ml}$ of base $\text{pH}$ is $\underset{\_}{8.75}$.

The value of $\text{pH}$ after addition of $30\text{ml}$ of base $\text{pH}$ is $\underset{\_}{12.34}$.

Explanation of Solution

Given

Volume of $\text{NaOH}$, $V_1$ is $10.0\text{ml}$.

Concentration of $\text{NaOH}$, $M_1$ is $0.125\text{M}$.

Concentration of acetic acid, $M_2$ is $0.100\text{M}$.

The addition of acetic acid in a solution, $V_2$ is $25\text{ml}$.

The equation of dissociation of acetic acid is given as,

${\text{CH}}_3\text{COOH}\to {\text{CH}}_3{\text{COOH}}^{-}+{\text{H}}^{+}$

Sodium hydroxide, $\text{NaOH}$ reacts with ${\text{H}}^{+}$ ion to form water and acetate ion is remained, that will act as buffer.

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

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

Where,

• $M_2$ is the molarity of the acid.
• $V_2$ is the volume of the acid.

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

$\begin{array}{c}\text{mmol}\text{of}\text{acid}=0.1\times 25\\ =2.5\text{mmol}\end{array}$

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

$\text{mmol}\text{of}\text{base}=M_1{V}_1$ (2)

Where,

• $M_1$ is the molarity of the base.
• $V_1$ is the volume of the base.

Substitute the value of $M_1\text{and}{V}_1$ in equation (2).

$\begin{array}{c}\text{mmol}\text{of}\text{base,}{\text{A}}^{-}=0.125\times 10\\ =1.25\text{mmol}\end{array}$

Acetic acid left after reaction is calculated by the formula,

$\text{Acid}\text{left}=\text{mmol}\text{of}\text{acid}-\text{mmol}\text{of}\text{base}$ (3)

Substitute the values of $\text{mmol}\text{of}\text{acid,}\text{mmol}\text{of}\text{base}$ in equation (3).

$\begin{array}{c}\text{Acid}\text{left,}\text{HA}=2.5-1.25\\ =1.25\end{array}$

The standard value of $\text{p}{K}_{\text{a}}$ of acid is $4.75$.

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

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\left(\frac{{\text{A}}^{-}}{\text{HA}}\right)$ (4)

Where,

• $\text{p}{K}_{\text{a}}$ is the acidic strength.
• ${\text{A}}^{-}$ is the $\text{mmol}$ of the base.
• $\text{HA}$ is the $\text{mmol}$ of acid left.

Substitute the values of $\text{p}{K}_{\text{a}},{\text{A}}^{-}\text{and}\text{HA}$ in equation (1).

$\begin{array}{c}\text{pH}=4.75+\text{log}\left(\frac{1.25}{1.25}\right)\\ =\underset{\_}{4.75}\end{array}$

Therefore, the $\text{pH}$ value after addition of $10\text{ml}$ base is $\underset{\_}{4.75}$.

Similarly,

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

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

$\text{mmol}\text{of}\text{base}=M_1{V}_1$ (2)

Where,

• $M_1$ is the molarity of the base.
• $V_1$ is the volume of the base.

Substitute the value of $M_1\text{and}{V}_1$ in equation (2).

$\begin{array}{c}\text{mmol}\text{of}\text{base,}{\text{A}}^{-}=0.125\times 20\\ =2.5\text{mmol}\end{array}$

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)

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).

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

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

The value of conjugate concentration is calculated by the formula,

$\left[{\text{BH}}^{+}\right]=\frac{\text{mmol}\text{of}\text{conjugate}\text{salt}}{V_{\text{total}}}$ (6)

Substitute the value of $\text{mmol}\text{of}\text{conjugate}\text{salt}\text{and}{V}_{\text{total}}$ in equation (6).

$\begin{array}{c}\left[{\text{BH}}^{+}\right]=\frac{2.5}{45}\\ =0.05555\text{M}\end{array}$

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

$K_{\text{b}}=\frac{K_{\text{w}}}{K_{\text{a}}}$ (7)

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

The value of $K_{\text{a}}$ is $1.8\times {10}^{-5}$

Substitute the value of $K_{\text{w}}\text{and}{K}_{\text{b}}$ in equation (7).

$\begin{array}{c}{K}_{\text{b}}=\frac{{10}^{-14}}{{10}^{1.8\times {10}^{-5}}}\\ =5.5\times {10}^{-10}\end{array}$

The value of $K_{\text{b}}$ is $5.5\times {10}^{-10}$.

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

$\text{pOH}=-\text{log}\left[x\right]$ (8)

Substitute the value of $x$ in equation (8).

$\begin{array}{c}\text{pOH}=-\text{log}\left(5.55\times {10}^{-6}\right)\\ =5.25\end{array}$

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

$\text{pH}=14-\text{pOH}$ (9)

Substitute the value of $\text{pOH}$ in equation (9).

$\begin{array}{c}\text{pH}=14-5.25\\ =\underset{\_}{8.75}\end{array}$

Therefore, the value of $\text{pH}$ after addition of $20\text{ml}$ of base $\text{pH}$ is $\underset{\_}{8.75}$.

Similarly,

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

The base is in excess at this point.

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

$\text{mmol}\text{of}\text{base}=M_1{V}_1$ (2)

Where,

• $M_1$ is the molarity of the base.
• $V_1$ is the volume of the base.

Substitute the value of $M_1\text{and}{V}_1$ in equation (2).

$\begin{array}{c}\text{mmol}\text{of}\text{base,}{\text{A}}^{-}=0.125\times 30\\ =3.75\text{mmol}\end{array}$

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

$\text{mmol}\text{of}\text{OH}\text{left}=\text{mmol}\text{of}\text{OH}-\text{mmol}\text{of}\text{acid}$ (11)

Substitute the values of $\text{mmol}$ of an acid and base in equation (11).

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

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

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

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

$\left[{\text{OH}}^{-}\right]=\frac{\text{mmol}\text{of}\text{OH}\text{left}}{V_{\text{Total}}}$ (12)

Substitute the value of $\text{mmol}\text{of}\text{OH}\text{left}\text{and}{V}_{\text{Total}}$ in equation (12).

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{1.25}{55}\\ =0.0227\end{array}$

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

$\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$ (8)

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

$\begin{array}{c}\text{pOH}=-\text{log}\left(0.0227\right)\\ =1.644\end{array}$

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

$\text{pH}=14-\text{pOH}$ (9)

Substitute the value of $\text{pOH}$ in equation (9).

$\begin{array}{c}\text{pH}=14-1.644\\ =\underset{\_}{12.34}\end{array}$

Therefore, the value of $\text{pH}$ after addition of $30\text{ml}$ of base $\text{pH}$ is $\underset{\_}{12.34}$.

Conclusion

The value of $\text{pH}$ after addition of $10\text{ml}$ of base is $\underset{\_}{4.75}$.

The value of $\text{pH}$ after addition of $20\text{ml}$ of base $\text{pH}$ is $\underset{\_}{8.75}$.

The value of $\text{pH}$ after addition of $30\text{ml}$ of base $\text{pH}$ is $\underset{\_}{12.34}$.