Chapter 16, Problem 16.110QP

(a)

Interpretation Introduction

To determine: An equation for the reaction that occurs when sodium acetate is added to $\text{HCl}\left(aq\right)$.

Interpretation: An equation for the reaction that occurs when sodium acetate is added to $\text{HCl}\left(aq\right)$ is to be stated.

Concept introduction: $\text{HCl}$ is a strong acid and it will dissociate completely into ${\text{H}}^{+}$ and ${\text{Cl}}^{-}$ whereas ${\text{CH}}_{\text{3}}\text{COONa}$ is a salt form of weak acid $\left({\text{CH}}_{\text{3}}\text{COOH}\right)$ which is a strong base. The products formed are acetic acid and sodium chloride.

Answer to Problem 16.110QP

An equation for the reaction that occurs when sodium acetate is added to $\text{HCl}\left(aq\right)$ is,

${\text{CH}}_{\text{3}}\text{COONa}+\text{HCl}\to {\text{CH}}_{\text{3}}\text{COOH}+\text{NaCl}$

Explanation of Solution

Sodium acetate is a salt of acetic acid which is considered as a weak acid.

$\text{HCl}$ being a strong acid displace it from its salt leads to the formation of sodium chloride and acetic acid. The equation is shown below,

${\text{CH}}_{\text{3}}\text{COONa}+\text{HCl}\to {\text{CH}}_{\text{3}}\text{COOH}+\text{NaCl}$

The product formed is a mixture of common salt and vinegar (acetic acid).

(b)

Interpretation Introduction

To determine: The $\text{pH}$ of the resulting buffer if $10.0\text{g}$ of ${\text{CH}}_{\text{3}}\text{COONa}$ is added to $275\text{mL}$ of $0.225\text{M}\text{HCl}$.

Interpretation: The $\text{pH}$ of the resulting buffer if $10.0\text{g}$ of ${\text{CH}}_{\text{3}}\text{COONa}$ is added to $275\text{mL}$ of $0.225\text{M}\text{HCl}$ is to be calculated.

Concept introduction: The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation,

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$

Answer to Problem 16.110QP

The $\text{pH}$ of the buffer is $\underset{\_}{4.75}$.

Explanation of Solution

Given

The given weight of ${\text{CH}}_{\text{3}}\text{COONa}$ is $10.0\text{g}$.

The volume of $\text{HCl}$ is $275\text{mL}$.

The molarity of $\text{HCl}$ is $0.225\text{M}$.

The molar mass of ${\text{CH}}_{\text{3}}\text{COONa}$ is $82\text{g}/\text{mol}$.

The number of moles is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the values of given mass and molar mass of ${\text{CH}}_{\text{3}}\text{COONa}$.

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{10\text{g}}{82\text{g}/\text{mol}}\\ =0.122\end{array}$

The molarity of the solution is calculated by the formula,

$\begin{array}{c}\text{Molarity}=\frac{\text{Amount}\text{of}\text{solute}\left(\text{in}\text{moles}\right)}{\text{Volume}}\\ \text{Moles}=\text{Molarity}\times \text{Volume}\end{array}$

Substitute the values of molarity and volume.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{HCl}=\text{Molarity}\times \text{Volume}\\ =\text{0}\text{.225}\text{M}\times \text{275}\times {\text{10}}^{-3}\text{L}\\ =0.062\text{mol}\end{array}$

Consider the reaction between sodium acetate and $\text{HCl}$,

${\text{CH}}_{\text{3}}\text{COONa}+\text{HCl}\to {\text{CH}}_{\text{3}}\text{COOH}+\text{NaCl}$

$\begin{array}{l}\text{moles}\text{of}{\text{CH}}_{\text{3}}\text{COONa}\text{reacted}=\text{moles}\text{of}\text{HCl}\text{added}=0.062\\ \text{moles}\text{of}{\text{CH}}_{\text{3}}\text{COOH}\text{formed}=\text{moles}\text{of}\text{HCl}\text{added}=0.062\end{array}$

So,

$\text{Moles}\text{of}{\text{CH}}_{\text{3}}\text{COOH}=0.062$

$\begin{array}{c}\text{Moles}\text{of}{\text{CH}}_{\text{3}}\text{COONa}=\text{Initial}\text{moles}-\text{reacted}\text{moles}\\ =\text{0}\text{.122}-\text{0}\text{.062}\\ =\text{0}\text{.06}\end{array}$

The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation,

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$

Substitute the values in the above equation.

$\begin{array}{l}\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)\\ \text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[{\text{CH}}_{\text{3}}\text{COONa}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\\ \text{pH}=4.76+\log \frac{\left[0.060\right]}{\left[0.062\right]}\\ \text{pH}=\underset{\_}{4.75}\end{array}$

So, the $\text{pH}$ of the buffer is $\underset{\_}{4.75}$.

(c)

Interpretation Introduction

To determine: The $\text{pH}$ of the solution in part (b) after the addition of $1.26\text{g}$ of solid $\text{NaOH}$.

Interpretation: The $\text{pH}$ of the solution in part (b) after the addition of $1.26\text{g}$ of solid $\text{NaOH}$ is to be calculated.

Concept introduction: The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation,

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$

Answer to Problem 16.110QP

The $\text{pH}$ of the solution in part (b) after the addition of $1.26\text{g}$ of solid $\text{NaOH}$ is $\underset{\_}{5.23}$.

Explanation of Solution

The given weight of $\text{NaOH}$ is $1.26\text{g}$.

The molecular mass of $\text{NaOH}$ is $4\text{0}\text{g}/\text{mol}$.

The number of moles is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the values of given mass and molar mass of ${\text{CH}}_{\text{3}}\text{COONa}$.

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{1.26\text{g}}{40\text{g}/\text{mol}}\\ =0.0315\text{mol}\end{array}$

The reaction between ${\text{CH}}_{\text{3}}\text{COOH}$ and $\text{NaOH}$ is,

${\text{CH}}_{\text{3}}\text{COOH}+\text{NaOH}\to {\text{CH}}_{\text{3}}\text{COONa}+{\text{H}}_{\text{2}}\text{O}$

$\begin{array}{c}\text{Moles}\text{of}{\text{CH}}_{\text{3}}\text{COOH}\text{reacted}=\text{Moles}\text{of}\text{NaOH}\text{added}=0.0315\\ \text{Moles}\text{of}{\text{CH}}_{\text{3}}\text{COONa}\text{formed}=\text{Moles}\text{of}\text{NaOH}\text{added}=0.0315\end{array}$

Finally,

$\begin{array}{c}\text{moles}\text{of}{\text{CH}}_{\text{3}}\text{COOH}=\text{initial}\text{moles}-\text{reacted}\text{moles}\\ =\text{0}\text{.062-0}\text{.0315}\\ =\text{0}\text{.0305}\end{array}$

$\begin{array}{c}\text{moles}\text{of}{\text{CH}}_{\text{3}}\text{COONa}=\text{initial}\text{moles}+\text{moles}\text{formed}\\ =\text{0}\text{.060}+\text{0}\text{.0315}\\ =\text{0}\text{.0915}\end{array}$

The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation,

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$

Substitute the values in the above equation.

$\begin{array}{c}\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)\\ \text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[{\text{CH}}_{\text{3}}\text{COONa}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\\ =4.76+\log \frac{\left[0.0915\right]}{\left[0.0305\right]}\\ =5.23\end{array}$

Conclusion

1. a. ${\text{CH}}_{\text{3}}\text{COONa}+\text{HCl}\to {\text{CH}}_{\text{3}}\text{COOH}+\text{NaCl}$
2. b. The $\text{pH}$ of the buffer is $4.75$.
3. c. The $\text{pH}$ of the solution in part (b) after the addition of $1.26\text{g}$ of solid $\text{NaOH}$ is $5.23$.