Chapter 15, Problem 31QRT

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of buffer solution had to be determined.

Concept Introduction:

A solution is said to be a buffer solution which contains weak acid and its corresponding conjugate base.  The $\text{pH}$ of the buffer solution will remain the same even if a small amount of the acid or base would be added to it.

The solution is said to be a buffer solution when the moles of the weak acid are greater than or equal to the number of moles of its conjugate base.

Answer to Problem 31QRT

The $\text{pH}$ of the buffer solution is $\underset{\_}{5.03}$.

Explanation of Solution

The standard value of $K_{\text{a}}$ for ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}$ is $1.4\times {10}^{-5}$.

The $\text{p}{K}_{\text{a}}$ for ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}$ is calculated by using the formula shown below.

  $\text{p}{K}_{\text{a}}=-\log K_{\text{a}}$        (1)

Substitute the value of $K_{\text{a}}$ in equation (1).

  $\begin{array}{c}\text{p}{K}_{\text{a}}=-\log \left(1.4\times {10}^{-5}\right)\\ =4.85\end{array}$

As per the given data the conjugate acid base pair in the buffer solution is propanoic acid and sodium propanoate.

The concentration of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}$ is $0.20\text{M}$.

The concentration of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COONa}$ is $0.30\text{M}$.

The Henderson-Hasselbalch equation can be represented as:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{Conjugate}\text{base}\right]}{\left[\text{Conjugate}\text{acid}\right]}$        (2)

The concentration of conjugate acid is $0.20\text{M}$.

The concentration of conjugate base is $0.30\text{M}$.

The value of $\text{p}{K}_{\text{a}}$ for ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}$ is $4.85$.

Substitute the value of $\text{p}{K}_{\text{a}}$, concentration of conjugate acid and the concentration of conjugate base in the equation (2).

  $\begin{array}{c}\text{pH}=4.85+\log \frac{0.30\text{M}}{0.20\text{M}}\\ =4.85+\log \left(1.5\right)\\ =4.85+0.1760\\ =5.026\end{array}$

Thus, after rounding to three significant figures, the $\text{pH}$ of the buffer solution is $\underset{\_}{5.03}$.

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of buffer solution when $1.0\text{ mL}$ hydrochloric acid is added to it had to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 31QRT

The $\text{pH}$ of the buffer solution is $\underset{\_}{4.88}$.

Explanation of Solution

The volume of $\text{HCl}$ added is $1.0\text{ mL}$.

The unit conversion of volume of $\text{HCl}$ from $\text{mL}$ to $\text{L}$ is shown below.

  $1.0\text{ mL}\times \frac{1\text{L}}{1000\text{mL}}=0.001\text{L}$

The concentration of $\text{HCl}$ added is $0.10\text{ M}$.  As $\text{HCl}$ is a strong acid therefore it will dissociate completely to give hydrogen ions.

Thus, the concentration of hydrogen ions is $0.10\text{ M}$.

Therefore, new concentration of the solution when $0.001\text{L}$ of $0.10\text{ M}$ of hydrochloric acid is added to the buffer solution is calculated as shown below.

  $\begin{array}{c}\text{New}\text{concentration}=\frac{0.001\text{L}\times 0.10\text{M}}{0.010\text{L}}\\ =0.01\text{M}\end{array}$

When $\text{HCl}$ is added to the initial buffer solution then the total volume of the solution is equal to the sum of the volume of $\text{HCl}$ and the volume of the initial buffer solution.

Therefore, the total volume is calculated as shown below.

  $\begin{array}{c}{V}_{\text{total}}=0.001\text{L}+0.010\text{L}\\ =0.011\text{L}\end{array}$

The chemical reaction between ${\text{H}}^{+}$ and ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ is shown below.

  ${\text{CH}}_{\text{3}}{\text{COO}}^{-}+{\text{H}}^{+}\to {\text{CH}}_{\text{3}}\text{COOH}$

The summarized data for the number of moles of all species after $\text{HCl}$ is added to the buffer solution is shown below.

  $\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COO}}^{-}+{\text{H}}^{+}\to {\text{CH}}_{\text{3}}\text{COOH}\\ \begin{array}{cccc}\text{Initial}& 0.30& 0& 0.20\\ \text{Final}& 0.30-0.01& 0.01& 0.20+0.01\end{array}\end{array}$

Therefore, the new number of moles of ${\text{CH}}_{\text{3}}\text{COOH}$ and ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ is $0.21\text{mol}$ and $0.29\text{mol}$ respectively.

The concentration of ${\text{CH}}_{\text{3}}\text{COOH}$ is calculated by the formula shown below.

  $M=\frac{n}{V}$        (3)

Where,

• $M$ is the molarity.
• $n$ is the number of moles.
• $V$ is the volume.

The value of $n$ for ${\text{CH}}_{\text{3}}\text{COOH}$ is $0.21\text{mol}$.

The value of $V$ for ${\text{CH}}_{\text{3}}\text{COOH}$ is $0.011\text{L}$.

Substitute the values of  $n$ and  $V$ for ${\text{CH}}_{\text{3}}\text{COOH}$ in the equation (3).

  $\begin{array}{c}\left[{\text{CH}}_{\text{3}}\text{COOH}\right]=\frac{\left(0.21\text{ mol}\right)}{\left(0.011\text{L}\right)}\\ =19.09\text{M}\end{array}$

The value of $n$ for ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ is $0.011\text{mol}$.

The value of $V$ for ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ is $0.011\text{L}$.

Substitute the values of $n$ and $V$ for ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ in the equation (3).

  $\begin{array}{c}\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]=\frac{\left(0.29\text{ mol}\right)}{\left(0.011\text{L}\right)}\\ =26.36\text{M}\end{array}$

The standard value of $\text{p}{K}_{\text{a}}$ for ${\text{CH}}_{\text{3}}\text{COOH}$ is $4.74$.

The concentration of conjugate acid ${\text{CH}}_{\text{3}}\text{COOH}$ is $19.09\text{M}$.

The concentration of conjugate base ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ is $26.36\text{M}$.

Substitute the value of $\text{p}{K}_{\text{a}}$, concentration of conjugate acid and concentration of conjugate base in the equation (2).

  $\begin{array}{c}\text{pH}=4.74+\log \frac{26.36\text{M}}{19.09\text{M}}\\ =4.74+\log \left(1.3808\right)\\ =4.74+0.1401\\ =4.88\end{array}$

Thus, the $\text{pH}$ of the buffer solution is $\underset{\_}{4.88}$.

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of buffer solution when $3.0\text{ mL}$ hydrochloric acid is added to it has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 31QRT

The $\text{pH}$ of the buffer solution is $\underset{\_}{4.74}$.

Explanation of Solution

The volume of $\text{HCl}$ added is $3.0\text{ mL}$.

The unit conversion of volume of $\text{HCl}$ from $\text{mL}$ to $\text{L}$ is shown below.

  $3.0\text{ mL}\times \frac{1\text{L}}{1000\text{mL}}=0.003\text{L}$

The concentration of $\text{HCl}$ added is $\text{1}\text{.0 M}$.  As $\text{HCl}$ is a strong acid therefore it will dissociate completely to give hydrogen ions.

Thus, the concentration of hydrogen ions is $\text{1}\text{.0 M}$.

Therefore, new concentration of the solution when $0.003\text{L}$ of $\text{1}\text{.0 M}$ of hydrochloric acid is added to the buffer solution is calculated as shown below.

  $\begin{array}{c}\text{New}\text{concentration}=\frac{0.003\text{L}\times 1.0\text{M}}{0.010\text{L}}\\ =0.3\text{M}\end{array}$

When $\text{HCl}$ is added to the initial buffer solution then the total volume of the solution is equal to the sum of the volume of $\text{HCl}$ and the volume of the initial buffer solution.

Therefore, the total volume is calculated as shown below.

  $\begin{array}{c}{V}_{\text{total}}=0.003\text{L}+0.010\text{L}\\ =0.013\text{L}\end{array}$

The chemical reaction between ${\text{H}}^{+}$ and ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ is shown below.

  ${\text{CH}}_{\text{3}}{\text{COO}}^{-}+{\text{H}}^{+}\to {\text{CH}}_{\text{3}}\text{COOH}$

The summarized data for the number of moles of all species after $\text{HCl}$ is added to the buffer solution is shown below.

  $\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COO}}^{-}+{\text{H}}^{+}\to {\text{CH}}_{\text{3}}\text{COOH}\\ \begin{array}{cccc}\text{Initial}& 0.30& 0& 0.20\\ \text{Final}& 0.30-0.30& 0.30& 0.20+0.30\end{array}\end{array}$

Therefore, the new number of moles of ${\text{CH}}_{\text{3}}\text{COOH}$ and ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ is $0.50\text{mol}$ and $0.0\text{mol}$ respectively.

Since the volume of conjugate acid and base is equal.  Therefore, concentration of conjugate acid and base can be replaced by number of moles in equation (2).

The standard value of $\text{p}{K}_{\text{a}}$ for ${\text{CH}}_{\text{3}}\text{COOH}$ is $4.74$.

The number of moles of conjugate acid ${\text{CH}}_{\text{3}}\text{COOH}$ is $0.50\text{mol}$.

The number of moles of conjugate base ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ is $0.0\text{mol}$.

Substitute the value of $\text{p}{K}_{\text{a}}$, number of moles of conjugate acid and base in the equation (2).

  $\begin{array}{c}\text{pH}=4.74+\log \frac{\text{0}\text{.0}\text{mol}}{0.50\text{mol}}\\ \simeq 4.74+0\\ =4.74\end{array}$

Thus, the $\text{pH}$ of the buffer solution is $\underset{\_}{4.74}$.