Chapter 17, Problem 17.13QP

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the given buffer solution before and after the addition of $\text{NaOH}$ has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
•  $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base. In general, addition of acid or base does not affect the $\text{pH}$ in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
• Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: the $\text{pH}$ of buffer solution acetic acid and sodium acetate on addition of $\text{NaOH}$

Answer to Problem 17.13QP

The $\text{pH}$ of buffer solution before addition of $\text{NaOH}$ is $\text{3}\text{.261}$

The $\text{pH}$ of buffer solution after addition of $\text{NaOH}$ is $\underset{\_}{3.329}$

Explanation of Solution

The given concentrations of ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ and ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}$ are $\text{1}\text{.00M}$

The number of moles of $\text{NaOH}$ is $\text{0}\text{.070}$

$\begin{array}{l}\text{The}\text{pH}\text{of}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{and}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}\text{buffer}\text{system}\text{is}\text{equal}\text{to}{\text{pK}}_{\text{a}}\\ \text{pH}\text{=}{\text{pK}}_{\text{a}}\text{=}{\text{-logK}}_{\text{a}}\\ \text{The}{\text{K}}_{\text{a}}\text{value}\text{for}\text{acetic acid}\text{is}4.4\text{×}{\text{10}}^{\text{-4}}\\ \text{=}\text{-log(}4.4\text{×}{\text{10}}^{\text{-4}}\text{)}\\ \text{=}3.357\end{array}$

According to the ${\text{p}}_{\text{ka}}\text{and}{\text{p}}_{\text{kb}}$ formula is,

  $\begin{array}{l}{\text{p}}_{\text{ka}}\text{+}{\text{p}}_{\text{kb}}=14\\ {\text{p}}_{\text{kb}}=14-{\text{p}}_{\text{ka}}\\ {\text{p}}_{\text{kb}}=14-3.357\\ =10.643\end{array}$

$\begin{array}{l}\text{The}\text{pH}\text{of}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{and}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}\text{buffer}\text{system}\text{is}\text{equal}\text{to}{\text{pK}}_{\text{a}}\frac{1}{2}\\ \text{Before }\text{adding }\\ \text{pOH}\text{=}\text{pKb}\text{+}\text{log}\frac{[\text{Salt}]}{[\text{Base}]}\\ \text{pOH}\text{=}10.643+\text{og}\frac{[\text{Salt}]}{[\text{Base}]}\end{array}$

Given data’s

  $\begin{array}{l}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}=1.00\text{ is [Salt]}\\ {\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}=0.80\text{M}\text{[Base]}\\ \text{pOH}\text{=}\text{10}\text{.643}\text{+}\text{log}\frac{1.00}{0.80}\\ \text{pOH}\text{=}\text{10}\text{.643}\text{+}0.0969\\ \text{pOH}\text{=}10.739\\ \therefore \text{pH=}\text{14-pOH}\\ \text{pH=}14-10.739\\ \text{pH}\text{=}\text{3}\text{.261}\end{array}$

The addition of $0.070\text{mol}$ sodium hydroxide, the $\text{NaOH}$ react with ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}$ and concentration of ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}$ decrease and concentration of base will increase.  

  $\begin{array}{l}[{\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl]}=1.00-0.070=0.930\text{M}\\ {\text{[CH}}_{\text{3}}{\text{NH}}_{\text{2}}]=0.80+0.070=0.870\text{M}\\ \text{pOH}\text{=}\text{10}\text{.643}\text{+}\text{log}\frac{0.930}{0.80}\\ \text{pOH}\text{=}\text{10}\text{.643}\text{+}\text{log}(1.0689)\\ \text{pOH}\text{=}\text{10}\text{.643}\text{+}0.028\\ \text{pOH}\text{=}10.671\\ \therefore \text{pH=}\text{14-pOH}\\ \text{pH=}14-10.671\\ \text{pH}\text{=}\text{3}\text{.329}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the given buffer solution before and after the addition of $\text{HCl}$ has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
•  $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base. In general, addition of acid or base does not affect the $\text{pH}$ in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
• Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: the $\text{pH}$ of buffer solution acetic acid and sodium acetate on addition of $\text{HCl}$.

Answer to Problem 17.13QP

The $\text{pH}$ of buffer solution before addition of $\text{HCl}$ is $\text{3}\text{.261}$

The $\text{pH}$ of buffer solution after addition of $\text{HCl}$ is $\underset{\_}{3.151}$

Explanation of Solution

$\begin{array}{l}\text{The}\text{pH}\text{of}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{and}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}\text{buffer}\text{system}\text{is}\text{equal}\text{to}{\text{pK}}_{\text{a}}\\ \text{pH}\text{=}{\text{pK}}_{\text{a}}\text{=}{\text{-logK}}_{\text{a}}\\ \text{The}{\text{K}}_{\text{a}}\text{value}\text{for}\text{acetic acid}\text{is}4.4\text{×}{\text{10}}^{\text{-4}}\\ \text{=}\text{-log(}4.4\text{×}{\text{10}}^{\text{-4}}\text{)}\\ \text{=}3.357\end{array}$

According to the ${\text{p}}_{\text{ka}}\text{and}{\text{p}}_{\text{kb}}$ formula is,

  $\begin{array}{l}{\text{p}}_{\text{ka}}\text{+}{\text{p}}_{\text{kb}}=14\\ {\text{p}}_{\text{kb}}=14-{\text{p}}_{\text{ka}}\\ {\text{p}}_{\text{kb}}=14-3.357\\ =10.643\end{array}$

$\begin{array}{l}\text{The}\text{pH}\text{of}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{and}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}\text{buffer}\text{system}\text{is}\text{equal}\text{to}{\text{pK}}_{\text{a}}\frac{1}{2}\\ \text{Before }\text{adding }\\ \text{pOH}\text{=}\text{pKb}\text{+}\text{log}\frac{[\text{Salt}]}{[\text{Base}]}\\ \text{pOH}\text{=}10.643+\text{og}\frac{[\text{Salt}]}{[\text{Base}]}\end{array}$

Given data’s

  $\begin{array}{l}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}=1.00\text{ is [Salt]}\\ {\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}=0.80\text{M}\text{[Base]}\\ \text{pOH}\text{=}\text{10}\text{.643}\text{+}\text{log}\frac{1.00}{0.80}\\ \text{pOH}\text{=}\text{10}\text{.643}\text{+}0.0969\\ \text{pOH}\text{=}10.739\\ \therefore \text{pH=}\text{14-pOH}\\ \text{pH=}14-10.739\\ \text{pH}\text{=}\text{3}\text{.261}\end{array}$

The addition of $0.11\text{mol}$$\text{HCl}$, the hydrogen chloride $\text{HCl}$ react with ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}$ and concentration of ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}$ increase and concentration of ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}$ will increase.  

  $\begin{array}{l}[{\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl]}=1.00+0.11=1.110\text{M}\\ {\text{[CH}}_{\text{3}}{\text{NH}}_{\text{2}}]=0.80-0.11=0.69\text{M}\\ \text{pOH}\text{=}\text{10}\text{.643}\text{+}\text{log}\frac{1.110}{0.69}\\ \text{pOH}\text{=}\text{10}\text{.643}\text{+}\text{log}(1.60869)\\ \text{pOH}\text{=}\text{10}\text{.643}\text{+}0.206\\ \text{pOH}\text{=}10.849\\ \therefore \text{pH=}\text{14-pOH}\\ \text{pH=}14-10.849\\ \text{pH}\text{=}\text{3}\text{.151}\end{array}$