Chapter 16, Problem 16.43QP

Interpretation Introduction

Interpretation: The $\text{pH}$ of the solution of acetic acid and sodium hydroxide at different volume of base is to be calculated.

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

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

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

Answer to Problem 16.43QP

Solution

The $\text{pH}$ of the solution at $10.0\text{mL}$ volume of base is $\underset{\_}{4.762}$.

The $\text{pH}$ of the solution at $20.0\text{mL}$ volume of base is $\underset{\_}{8.75}$.

The $\text{pH}$ of the solution at $30.0\text{mL}$ volume of base is $\underset{\_}{12.36}$.

Explanation of Solution

Explanation

Given

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

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

The concentration of sodium hydroxide is $0.125\text{M}$.

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

At $10.0mL$ volume of sodium hydroxide:

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.

$\begin{array}{c}\text{Concentration}\text{of}{\text{CH}}_3\text{COONa}=\frac{10.0}{10.0+25.0}\times 0.125\text{M}\\ =0.036\text{M}\end{array}$

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.

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

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

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

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.

$\begin{array}{c}\text{pH}=4.75+\log \frac{0.036}{0.035}\\ \text{pH}=\underset{\_}{4.762}\end{array}$

At $20.0mL$ volume of sodium hydroxide:

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

$\begin{array}{c}\text{Concentration}\text{of}{\text{CH}}_3\text{COONa}=\frac{20.0}{20.0+25.0}\times 0.125\text{M}\\ =0.056\text{M}\end{array}$

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

${\text{CH}}_3{\text{COO}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{CH}}_3\text{COOH}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

The concentration of acetic acid and hydroxyl ions formed in the given equilibrium reaction is considered to be $x$. The dissociation constant of the sodium acetate is calculated by using the formula,

$K_{\text{b}}=\frac{\left[{\text{OH}}^{-}\right]\left[{\text{CH}}_3\text{COOH}\right]}{\left[{\text{CH}}_3{\text{COO}}^{-}\right]}$

Where,

• $K_{\text{b}}$ is the dissociation constant of the acid.
• $\left[{\text{OH}}^{-}\right]$ is the hydroxyl ion concentration.
• $\left[{\text{CH}}_3\text{COOH}\right]$ is the concentration of acetic acid.
• $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is the concentration of acetate ion.

Substitute the values of $K_{\text{b}}$, $\left[{\text{OH}}^{-}\right]$, $\left[{\text{CH}}_3\text{COOH}\right]$ and $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ in the above equation.

$\begin{array}{c}5.55\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.056-x\right)}\\ x=5.6\times {10}^{-6}\end{array}$

Therefore, the concentration of hydroxyl ions in the solution is $5.6\times {10}^{-6}$.

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.

$\begin{array}{c}14-\text{pH}=-\log \left(5.6\times {10}^{-6}\right)\\ \text{pH}=\underset{\_}{8.75}\end{array}$

At $30.0mL$ volume of sodium hydroxide:

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

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

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

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

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.

$\begin{array}{c}14-\text{pH}=-\log \left(0.023\right)\\ \text{pH}=\underset{\_}{12.36}\end{array}$

Conclusion

The $\text{pH}$ of the solution at $10.0\text{mL}$ volume of base is $\underset{\_}{4.762}$.

The $\text{pH}$ of the solution at $20.0\text{mL}$ volume of base is $\underset{\_}{8.75}$.

The $\text{pH}$ of the solution at $30.0\text{mL}$ volume of base is $\underset{\_}{12.36}$