Chapter 16, Problem 16.36QP

(a)

Interpretation Introduction

Interpretation: The $\text{pH}$ of the given buffer solutions of acetic acid and sodium hydroxide is to be calculated.

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

$\text{pH}=p{K}_{\text{a}}+\log \frac{\left[\text{Conjugate}\text{Base}\right]}{\left[\text{Acid}\right]}$

To determine: The $\text{pH}$ of the given buffer solution of acetic acid and sodium hydroxide.

Answer to Problem 16.36QP

Solution

The $\text{pH}$ of the given buffer solution of acetic acid and sodium hydroxide is $\underset{\_}{4.88}$.

Explanation of Solution

Explanation

Given

The concentration of acetic acid solution is $0.175\text{M}$.

The volume of acetic acid solution in buffer is $100.0\text{mL}$.

The concentration of sodium hydroxide solution is $0.200\text{M}$.

The volume of sodium hydroxide solution in buffer is $50.0\text{mL}$.

The concentration of sodium acetate formed after addition of sodium hydroxide is calculated by multiplying the ratio of sodium hydroxide present in the solution with the concentration of sodium hydroxide solution.

$\begin{array}{c}\text{Concentration}\text{of}{\text{CH}}_3\text{COONa}=\frac{50.0}{50.0+100.0}\times 0.200\text{M}\\ =0.067\text{M}\end{array}$

The concentration of acetic acid left in the solution after formation of sodium acetate is calculated by subtracting the concentration of sodium acetate from the volume of acetic acid in the solution.

$\begin{array}{c}\text{Concentration}\text{of}{\text{CH}}_3\text{COOH}\left(\text{in}\text{solution}\right)=\frac{100.0}{50.0+100.0}\times 0.175\text{M}-0.067\text{M}\\ =0.05\text{M}\end{array}$

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

$\text{pH}=p{K}_{\text{a}}+\log \frac{\left[{\text{CH}}_3\text{COONa}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}$

Where,

• $\text{pH}$ is the hydrogen content of solution.
• $p{K}_{\text{a}}$ is the dissociation constant of acid.
• $\left[{\text{CH}}_3\text{COONa}\right]$ is the concentration of sodium acetate solution.
• $\left[{\text{CH}}_3\text{COOH}\right]$ is the concentration of acetic acid.

Substitute the values of $\left[{\text{CH}}_3\text{COONa}\right]$, $\left[{\text{CH}}_3\text{COOH}\right]$ and $p{K}_{\text{a}}$ in the above equation.

$\begin{array}{c}\text{pH}=4.75+\log \frac{0.067}{0.05}\\ \text{pH}=\underset{\_}{4.88}\end{array}$

(b)

Interpretation Introduction

To determine: The $\text{pH}$ of the solution after dissolving $1.00\text{g}$ sodium hydroxide in the given buffer solution.

Answer to Problem 16.36QP

Solution

The $\text{pH}$ of the buffer solution after addition of $1.00\text{g}$ sodium hydroxide is $\underset{\_}{13.08}$.

Explanation of Solution

Explanation

Given

The concentration of acetic acid present in the buffer solution is $0.05\text{M}$.

The mass of sodium hydroxide added to the solution is $1.00\text{g}$.

The volume of the buffer solution is $150\text{mL}\left(=0.150\text{L}\right)$.

The number of moles of sodium hydroxide in the given solution is calculated by dividing its mass by the molar mass.

$\begin{array}{c}\text{Moles}\text{of}\text{NaOH}=\frac{1.00\text{g}}{39.99\text{g/mol}}\\ =0.025\text{mol}\end{array}$

The molarity of sodium hydroxide in the given solution is calculated by using the formula,

$M=\frac{n}{V}$

Where,

• $M$ is the molarity of the solution.
• $n$ is the number of moles of solute.
• $V$ is the volume of solution in liters.

Substitute the values of $n$ and $V$ in above equation.

$\begin{array}{c}M=\frac{0.025\text{mol}}{0.150\text{L}}\\ =0.17\text{M}\end{array}$

Since, the concentration of sodium is very large in the given solution. Therefore, the sodium hydroxide solution neutralizes the acetic acid present in the buffer solution and the activity of buffer gets exhausted. The concentration of sodium hydroxide contribute to the $\text{pH}$ of the solution is the difference of concentration of sodium hydroxide and acetic acid which is equal to $0.12\text{M}$.

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

$\text{14}-\text{pH}=-\log \left[{\text{OH}}^{-}\right]$

Where,

• $\left[{\text{OH}}^{-}\right]$ is the concentration of hydroxyl ions in the solution.

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

$\begin{array}{c}14-\text{pH}=-\log \left(0.12\right)\\ \text{pH}=\underset{\_}{13.08}\end{array}$

Conclusion

1. a. The $\text{pH}$ of the given buffer solution of acetic acid and sodium hydroxide is $\underset{\_}{4.88}$.
2. b. The $\text{pH}$ of the buffer solution after addition of $1.00\text{g}$ sodium hydroxide is $\underset{\_}{13.08}$