Chapter 15, Problem 29QRT

(a)

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ change when $0.10\text{ M}$ of $\text{NaOH}$ is mixed with $0.10\text{ M}$ of ${\text{CH}}_{\text{3}}\text{COOH}$ and $0.10\text{ M}$ of ${\text{CH}}_{\text{3}}\text{COONa}$ has to be calculated.

Concept Introduction:

The Henderson-Hasselbalch equation can be used to determine the $\text{pH}$ of the buffer which is formed by the known concentrations of conjugate base and conjugate acid.  This equation can also be applied to calculate the ratio of conjugate base to conjugate acid concentration.

The Henderson-Hasselbalch equation can be represented as:

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

Answer to Problem 29QRT

The change in $\text{pH}$ value is $\underset{\_}{0.1}$.

Explanation of Solution

The value of $\text{pH}$ of the buffer solution before $\text{NaOH}$ is added to the buffer solution is calculated using the Henderson-Hasselbalch equation shown below.

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{Conjugate base}\right]}{\left[\text{Conjugate}\text{acid}\right]}$        (1)

The concentration of conjugate acid is $0.1\text{M}$.

The concentration of conjugate base is $0.1\text{M}$.

Substitute the value of concentration of conjugate acid and the concentration of conjugate base in the equation (1).

  $\begin{array}{c}\text{pH}=4.74+\log \frac{0.1\text{M}}{0.1\text{M}}\\ =4.74\end{array}$

The number of moles of $\text{NaOH}$ is calculated by the formula shown below.

  $n=M\times V$        (2)

Where,

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

The value of $M$ for $\text{NaOH}$ is $1.0\text{ M}$.

The value of $V$ for $\text{NaOH}$ is $1.0\text{mL}$.

Substitute the values of  $M$ and  $V$ for $\text{NaOH}$ in the equation (2).

  $\begin{array}{c}{n}_{\text{NaOH}}=\left(1.0\text{ M}\right)\left(1.0\text{ mL}\times \frac{1\text{ L}}{1000\text{mL}}\right)\\ =0.001\text{mol}\end{array}$

Therefore, the number of moles of $\text{NaOH}$ is $0.001\text{ mol}$.

The value of $M$ for ${\text{CH}}_{\text{3}}\text{COOH}$ is $0.1\text{ M}$.

The value of $V$ for ${\text{CH}}_{\text{3}}\text{COOH}$ is $0.1\text{ L}$.

Substitute the values of  $M$ and $V$ for ${\text{CH}}_{\text{3}}\text{COOH}$ in the equation (2).

  $\begin{array}{c}{n}_{{\text{CH}}_{\text{3}}\text{COOH}}=\left(0.1\text{ M}\right)\left(0.1\text{ L}\right)\\ =0.01\text{mol}\end{array}$

The concentration of ${\text{CH}}_{\text{3}}\text{COOH}$ and ${\text{CH}}_{\text{3}}\text{COONa}$ are same.  Thus, the mole of ${\text{CH}}_{\text{3}}\text{COONa}$ is equal to the mole of ${\text{CH}}_{\text{3}}\text{COOH}$.

Therefore, the number of mole of ${\text{CH}}_{\text{3}}\text{COONa}$ is $0.01\text{mol}$.

The total volume is calculated is as shown below:

  $\begin{array}{c}{V}_{\text{total}}\text{ =}\left(1.0\text{ L}\times \frac{1000\text{mL}}{1.0\text{ L}}\right)+\text{0}\text{.1 mL}\\ =1000.1\text{ mL}\end{array}$

The chemical reaction between ${\text{CH}}_{\text{3}}\text{COOH}$ with $\text{NaOH}$ is shown below.

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

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

  $\begin{array}{l}{\text{CH}}_{\text{3}}\text{COOH}+\text{NaOH}\to {\text{CH}}_{\text{3}}\text{COONa}+{\text{H}}_{\text{2}}\text{O}\\ \begin{array}{cccc}\text{Initial}& 0.01& 0& 0.01\\ \text{Final}& 0.01-0.001& 0.001-0.001& 0.01+0.001\end{array}\end{array}$

Therefore, the new number of moles of ${\text{CH}}_{\text{3}}\text{COOH}$ and ${\text{CH}}_{\text{3}}\text{COONa}$ is $0.009\text{mol}$ and $0.011\text{mol}$ 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  $0.009\text{mol}$.

The value of  $V$ for ${\text{CH}}_{\text{3}}\text{COOH}$ is  $1000.1\text{ mL}$.

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

  $\begin{array}{c}\left[{\text{CH}}_{\text{3}}\text{COOH}\right]=\frac{\left(0.009\text{ mol}\right)}{\left(\text{1000}\text{.1 mL}\times \frac{1\text{ L}}{1000\text{mL}}\right)}\\ =0.0089\text{M}\end{array}$

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

The value of  $V$ for ${\text{CH}}_{\text{3}}\text{COONa}$ is $1000.1\text{ mL}$.

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

  $\begin{array}{c}\left[{\text{CH}}_{\text{3}}\text{COONa}\right]=\frac{\left(0.011\text{ mol}\right)}{\left(\text{1000}\text{.1 mL}\times \frac{1\text{ L}}{1000\text{mL}}\right)}\\ =0.0109\text{M}\end{array}$

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 $0.0089\text{M}$.

The concentration of conjugate base ${\text{CH}}_{\text{3}}\text{COONa}$ is $0.0109\text{M}$.

Substitute the value of $\text{p}{K}_{\text{a}}$, concentration of conjugate acid and concentration of conjugate base in the equation (1).

  $\begin{array}{c}\text{pH}=4.74+\log \frac{0.0109\text{M}}{0.0089\text{M}}\\ =4.74+\log \left(1.2247\right)\\ =4.74+0.088\\ =4.828\end{array}$

Therefore, the $\text{pH}$ of the buffer solution is $4.828$.

The change in $\text{pH}$ that is, $\Delta \text{pH}$ is calculate as shown below.

  $\begin{array}{c}\Delta \text{pH}=4.828-4.74\\ =0.088\end{array}$

Thus, after rounding to one significant figure, the $\text{pH}$ change of the buffer solution is $\underset{\_}{0.1}$.

(b)

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ change when $0.10\text{ M}$ of $\text{NaOH}$ is mixed with $0.010\text{ M}$ of ${\text{CH}}_{\text{3}}\text{COOH}$ and $0.010\text{ M}$ of ${\text{CH}}_{\text{3}}\text{COONa}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 29QRT

The change in $\text{pH}$ value is $\underset{\_}{3.8}$.

Explanation of Solution

The value of $\text{pH}$ of the buffer solution before $\text{NaOH}$ is added to the buffer solution is calculated using equation (1).

The concentration of conjugate acid is $0.1\text{M}$.

The concentration of conjugate base is $0.1\text{M}$.

Substitute the value of concentration of conjugate acid and the concentration of conjugate base in the equation (1).

  $\begin{array}{c}\text{pH}=4.74+\log \frac{0.1\text{M}}{0.1\text{M}}\\ =4.74\end{array}$

The value of $M$ for $\text{NaOH}$ is $1.0\text{ M}$.

The value of $V$ for $\text{NaOH}$ is $1.0\text{mL}$.

Substitute the values of $M$ and $V$ for $\text{NaOH}$ in the equation (2).

  $\begin{array}{c}{n}_{\text{NaOH}}=\left(1.0\text{ M}\right)\left(1.0\text{ mL}\times \frac{1\text{ L}}{1000\text{mL}}\right)\\ =0.001\text{mol}\end{array}$

The value of $M$ for ${\text{CH}}_{\text{3}}\text{COOH}$ is $0.01\text{ M}$.

The value of $V$ for ${\text{CH}}_{\text{3}}\text{COOH}$ is $0.1\text{ L}$.

Substitute the values of $M$ and $V$ for ${\text{CH}}_{\text{3}}\text{COOH}$ in the equation (2).

  $\begin{array}{c}{n}_{{\text{CH}}_{\text{3}}\text{COOH}}=\left(0.01\text{ M}\right)\left(0.1\text{ L}\right)\\ =0.001\text{mol}\end{array}$

The concentration of ${\text{CH}}_{\text{3}}\text{COOH}$ and ${\text{CH}}_{\text{3}}\text{COONa}$ are equal. Thus, the number of moles of ${\text{CH}}_{\text{3}}\text{COONa}$ is equal to that of ${\text{CH}}_{\text{3}}\text{COOH}$.

Therefore, the number of moles of ${\text{CH}}_{\text{3}}\text{COONa}$ is $0.001\text{ M}$.

The total volume is calculated is as shown below:

  $\begin{array}{c}{V}_{\text{solution}}=\left(1.0\text{ L}\times \frac{1000\text{mL}}{1.0\text{ L}}\right)+\text{0}\text{.1 mL}\\ =1000.1\text{ mL}\end{array}$

The chemical reaction between ${\text{CH}}_{\text{3}}\text{COOH}$ and $\text{NaOH}$ is shown below.

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

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

  $\begin{array}{l}{\text{CH}}_{\text{3}}\text{COOH}+\text{NaOH}\to {\text{CH}}_{\text{3}}\text{COONa}+{\text{H}}_{\text{2}}\text{O}\\ \begin{array}{cccc}\text{Initial}& 0.001& 0& 0.001\\ \text{Final}& 0.001-0.001& 0.001-0.001& 0.001+0.001\end{array}\end{array}$

Therefore, the new number of moles of ${\text{CH}}_{\text{3}}\text{COOH}$ and ${\text{CH}}_{\text{3}}\text{COONa}$ is $0\text{mol}$ and $0.002\text{mol}$ respectively.

The value of $n$ for ${\text{CH}}_{\text{3}}\text{COONa}$ is $0.002\text{mol}$.

The value of $V$ for ${\text{CH}}_{\text{3}}\text{COONa}$ is $1000.1\text{ mL}$.

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

  $\begin{array}{c}\left[{\text{CH}}_{\text{3}}\text{COONa}\right]=\frac{\left(0.002\text{ mol}\right)}{\left(\text{1000}\text{.1 mL}\times \frac{1\text{ L}}{1000\text{mL}}\right)}\\ =0.02\text{M}\end{array}$

The equation for the ionization of ${\text{CH}}_{\text{3}}\text{COONa}$ is shown below.

  ${\text{CH}}_{\text{3}}\text{COONa}\to {\text{CH}}_{\text{3}}{\text{COO}}^{-}+{\text{Na}}^{+}$

The equation to calculate $K_{\text{b}}$ value of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ is shown below.

  $K_{\text{b}}=\frac{K_{\text{w}}}{K_{\text{a}}}$        (4)

Where,

• $K_{\text{a}}$ is the acid dissociation constant.
• $K_{\text{b}}$ is the base dissociation constant.
• $K_{\text{w}}$ is the ionic product constant of water.

The value of $K_{\text{a}}$ for ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ is $1.8\times {10}^{-5}$.

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

Substitute the value $K_{\text{a}}$ and $K_{\text{w}}$ in the equation (4).

  $\begin{array}{c}{K}_{\text{b}}=\frac{{10}^{-14}}{1.8\times {10}^{-5}}\\ =5.55\times {10}^{-10}\end{array}$

The equation for the hydrolysis of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ is as follows:

  ${\text{CH}}_{\text{3}}{\text{COO}}^{-}+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{CH}}_{\text{3}}\text{COOH}+{\text{OH}}^{-}$

In the above equation, one mole of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ is treated with one mole of ${\text{H}}_{\text{2}}\text{O}$ will be formed one mole of ${\text{CH}}_{\text{3}}\text{COOH}$ and one mole of ${\text{OH}}^{-}$.  Thus, concentration of ${\text{CH}}_{\text{3}}\text{COOH}$ and ${\text{OH}}^{-}$ are same.

The ionization constantexpression for the above equation is as follows:

  $K_{\text{b}}=\frac{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]}$        (5)

Since concentration ${\text{CH}}_{\text{3}}\text{COOH}$ and ${\text{OH}}^{-}$ are same.  Rearrange equation (5) to calculate $\left[{\text{OH}}^{-}\right]$.

  $\left[{\text{OH}}^{-}\right]=\sqrt{\left(K_{\text{b}}\right)\left(\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]\right)}$        (6)

The value for the concentration of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ is 0.02 M.

The value of $K_{\text{b}}$ is $5.55\times {10}^{-10}$.

Substitute the value of $K_{\text{b}}$ and $\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]$ in the equation (6).

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]=\sqrt{\left(5.55\times {10}^{-10}\right)\left(0.02\text{ M}\right)}\\ =3.33\times {10}^{-6}\text{ M}\end{array}$

The expression to calculate the concentration of $\left[{\text{H}}^{+}\right]$ is shown below.

  $\left[{\text{OH}}^{-}\right]\left[{\text{H}}^{+}\right]={10}^{-14}$        (7)

The value of $\left[{\text{OH}}^{-}\right]$ is $3.33\times {10}^{-6}\text{ M}$

Substitute value of $\left[{\text{OH}}^{-}\right]$ in the equation (7).

  $\begin{array}{c}\left[{\text{H}}^{+}\right]\left(3.33\times {10}^{-6}\right)={10}^{-14}\\ \left[{\text{H}}^{+}\right]=\frac{{10}^{-14}}{3.33\times {10}^{-6}}\\ =3.0\times {10}^{-9}\text{ M}\end{array}$

Thus, the value of $\left[{\text{H}}^{+}\right]$ is $3.0\times {10}^{-9}\text{ M}$.

The relation between $\text{pH}$ and $\left[{\text{H}}^{+}\right]$ is shown below.

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$        (8)

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

  $\begin{array}{c}\text{pH}=-\log 3.0\times {10}^{-9}\text{ M}\\ =8.52\end{array}$

Therefore, the $\text{pH}$ of the buffer solution is $8.52$.

The change in $\text{pH}$ that is, $\Delta \text{pH}$ is calculate as shown below.

  $\begin{array}{c}\Delta \text{pH}=8.52-4.74\\ =3.78\end{array}$

Thus, after rounding to one significant figure, the $\text{pH}$ change of the buffer solution is $\underset{\_}{3.8}$.

(c)

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ change when $0.10\text{ M}$ of $\text{NaOH}$ is mixed with $0.0010\text{ M}$ of ${\text{CH}}_{\text{3}}\text{COOH}$ and $0.0010\text{ M}$ of ${\text{CH}}_{\text{3}}\text{COONa}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 29QRT

The change in $\text{pH}$ value is $\underset{\_}{7.21}$.

Explanation of Solution

The value of $\text{pH}$ of the buffer solution before $\text{NaOH}$ is added to the buffer solution is calculated using equation (1).

The concentration of conjugate acid is $0.1\text{M}$.

The concentration of conjugate base is $0.1\text{M}$.

Substitute the concentration of conjugate acid and the concentration of conjugate base in the equation (1).

  $\begin{array}{c}\text{pH}=4.74+\log \frac{0.1\text{M}}{0.1\text{M}}\\ =4.74\end{array}$

The value of $M$ for $\text{NaOH}$ is $1.0\text{ M}$.

The value of $V$ for $\text{NaOH}$ is $1.0\text{mL}$.

Substitute the values of  $M$ and  $V$ for $\text{NaOH}$ in the equation (2).

  $\begin{array}{c}{n}_{\text{NaOH}}=\left(1.0\text{ M}\right)\left(1.0\text{ mL}\times \frac{1\text{ L}}{1000\text{mL}}\right)\\ =0.001\text{mol}\end{array}$

The value of $M$ for ${\text{CH}}_{\text{3}}\text{COOH}$ is $0.0010\text{ M}$.

The value of $V$ for ${\text{CH}}_{\text{3}}\text{COOH}$ is $0.1\text{ L}$.

Substitute the values of  $M$ and  $V$ for ${\text{CH}}_{\text{3}}\text{COOH}$ in the equation (2).

  $\begin{array}{c}{n}_{{\text{CH}}_{\text{3}}\text{COOH}}=\left(0.0010\text{ M}\right)\left(0.1\text{ L}\right)\\ =0.0001\text{mol}\end{array}$

The concentration of ${\text{CH}}_{\text{3}}\text{COOH}$ and ${\text{CH}}_{\text{3}}\text{COONa}$ are same. Thus, the mole of ${\text{CH}}_{\text{3}}\text{COONa}$ is equal to the mole of ${\text{CH}}_{\text{3}}\text{COOH}$.

Therefore, the number of moles of ${\text{CH}}_{\text{3}}\text{COONa}$ is $0.0001\text{ M}$.

The total volume is calculated as shown below.

  $\begin{array}{c}{V}_{\text{solution}}=\left(1.0\text{ L}\times \frac{1000\text{mL}}{1.0\text{ L}}\right)+\text{0}\text{.1 mL}\\ =1000.1\text{ mL}\end{array}$

The chemical reaction between ${\text{CH}}_{\text{3}}\text{COOH}$ with $\text{NaOH}$ is shown below.

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

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

  $\begin{array}{l}{\text{CH}}_{\text{3}}\text{COOH}+\text{NaOH}\to {\text{CH}}_{\text{3}}\text{COONa}+{\text{H}}_{\text{2}}\text{O}\\ \begin{array}{cccc}\text{Initial}& 0.0001& 0& 0.0001\\ \text{Final}& 0.0001-0.0001& 0.01-0.0001& 0.0001+0.0001\end{array}\end{array}$

There are greater number of moles of $\text{NaOH}$ as compared to ${\text{CH}}_{\text{3}}\text{COONa}$ and ${\text{CH}}_{\text{3}}\text{COOH}$. Thus, total number of moles of ${\text{CH}}_{\text{3}}\text{COOH}$ will be consumed in the reaction while $\text{NaOH}$ will remains unreacted.

Therefore, the new number of moles of ${\text{CH}}_{\text{3}}\text{COOH}$, ${\text{CH}}_{\text{3}}\text{COONa}$ and $\text{NaOH}$ is $0.0\text{mol}$, $0.0002\text{mol}$ and $0.009\text{mol}$ respectively.

The value of $n$ for $\text{NaOH}$ is $0.009\text{mol}$.

The value of $V$ for $\text{NaOH}$ is $1000.1\text{ mL}$.

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

  $\begin{array}{c}\left[\text{NaOH}\right]=\frac{\left(0.009\text{ mol}\right)}{\left(\text{1000}\text{.1 mL}\times \frac{1\text{ L}}{1000\text{mL}}\right)}\\ =0.009\text{M}\end{array}$

As $\text{NaOH}$ is the strong base so, it will be dissociated completely into aqueous solution.

Thus, the concentration of $\left[{\text{OH}}^{-}\right]$ is $0.009\text{M}$.

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the equation (7).

  $\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{{10}^{-14}}{0.009\text{M}}\\ =1.11\times {10}^{-12}\text{ M}\end{array}$

The value of $\left[{\text{H}}^{+}\right]$ is $1.11\times {10}^{-12}\text{ M}$.

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

  $\begin{array}{c}\text{pH}=-\log \left(1.11\times {10}^{-12}\right)\\ =11.95\end{array}$

Therefore, the $\text{pH}$ of the buffer solution is $11.95$.

The change in $\text{pH}$ that is, $\Delta \text{pH}$ is calculate as shown below.

  $\begin{array}{c}\Delta \text{pH}=11.95-4.74\\ =\underset{\_}{7.21}\end{array}$

Thus, the $\text{pH}$ change of the buffer solution is $\underset{\_}{7.21}$.