Chapter 17, Problem 17.16QP

Calculate the pH of 1.00 L of the buffer 1.00 M CH₃COONa/1.00 M CH₃COOH before and after the addition of (a) 0.080 mol NaOH and (b) 0.12 mol HCl. (Assume that there is no change in volume.)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the given buffer solution before and after the addition of $\text{HCl}\text{and}\text{NaOH}$ have 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 17.16QP

The $\text{pH}$ of buffer solution after addition of $\text{NaOH}$ is $4.82$

Explanation of Solution

The given concentrations of acetic acid and sodium acetate are $\text{1}\text{.00M}$

The number of moles of $\text{NaOH}$ is $\text{0}\text{.080}$

$\begin{array}{l}\text{The}\text{pH}\text{of}\text{acetic}\text{acid}\text{and}\text{sodium}\text{acetate}\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}\text{1}\text{.8}\text{×}{\text{10}}^{\text{-5}}\\ \text{=}\text{-log(1}\text{.8}\text{×}{\text{10}}^{\text{-5}}\text{)}\\ \text{=}\text{4}\text{.74}\\ \text{NaOH}\text{completely}\text{reacts}\text{with}\text{acetic}\text{acid}\text{.}\text{hence,}\text{0}\text{.080}\text{mol}\text{of}{\text{OH}}^{\text{-}}\\ \text{is}\text{added}\text{to}\text{buffer}\\ \text{The}\text{reaction}\text{is}\text{as}\text{follows,}\\ {\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\text{ +}{\text{OH}}^{\text{-}}{}_{\text{(aq)}}\to {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\\ \\ \text{Initial concentration(M): }\text{1}\text{.00 }\text{0}\text{.080}\text{1}\text{.00}\\ \text{Change in concentration (M):-}\text{0}\text{.080}\text{-0}\text{.080}\text{+0}\text{.0010}\\ \text{Equilibrium}\text{concentration (M): 0}\text{.92 }\text{0}\text{1}\text{.08}\\ \\ \text{The}\text{acetic acid equilibrium table is}\text{redfined as}\\ {\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.92 }\text{0}\text{1}\text{.08}\\ \text{Change in concentration (M):}\text{-}\text{x}\text{+x}\text{+x}\\ \text{Equilibrium}\text{concentration (M): 0}\text{.92-x }\text{0}\text{1}\text{.08+x}\\ \\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}{{\text{[CH}}_{\text{3}}\text{COOH]}}\\ \text{1}\text{.8}\text{×}{\text{10}}^{\text{-5}}\text{=}\frac{\text{(x)(1}\text{.08+x)}}{\text{(0}\text{.92-x)}}\\ \text{x}\text{is}\text{very}\text{small}\text{and}\text{we}\text{neglect}\text{it,}\\ \text{1}\text{.8}\text{×}{\text{10}}^{\text{-5}}\text{=}\frac{\text{(x)(1}\text{.08)}}{\text{(0}\text{.92)}}\\ \text{x=}{\text{[H}}^{\text{+}}\text{]}\text{=}\text{1}\text{.5}\text{×}{\text{10}}^{\text{-5}}\\ \text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}\\ \text{=}\text{-log(1}\text{.5}\text{×}{\text{10}}^{\text{-5}}\text{)}\\ \text{pH}\text{=}\text{4}\text{.82}\\ \end{array}$

When sodium hydroxide adds to the buffer solution the acetic acid reacts with it and concentration of acetic acid changes.  From the expression of ${\text{K}}_{\text{a}}$ , the concentration of hydrogen ion can be calculated.  With the help of hydrogen ion concentration, $\text{pH}$ of the buffer solution after addition of base can be found.  The $\text{pH}$ value raised from $\text{4}\text{.74}\text{to}\text{4}\text{.82}$ due to addition of strong base.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the given buffer solution before and after the addition of $\text{HCl}\text{and}\text{NaOH}$ have 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 17.16QP

The $\text{pH}$ of buffer solution after addition of $\text{HCl}$ is $4.64$

Explanation of Solution

The given concentrations of acetic acid and sodium acetate are $\text{1}\text{.00M}$

The number of moles of $\text{HCl}$ is $\text{0}\text{.12}$

$\begin{array}{l}\text{HCl}\text{completely}\text{reacts}\text{with}\text{acetate in buffer}\text{.}\text{Hence,}\text{0}\text{.12}\text{mol}\text{of}{\text{H}}^{+}\\ \text{is}\text{added}\text{to}\text{buffer}\\ \text{The}\text{reaction}\text{is}\text{as}\text{follows,}\\ {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\text{ +}{\text{H}}^{\text{+}}{}_{\text{(aq)}}\to {\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{1}\text{.00 }\text{0}\text{.12}\text{1}\text{.00}\\ \text{Change in concentration (M):}\text{-}\text{0}\text{.12 }\text{-0}\text{.12}\text{+0}\text{.12}\\ \text{Equilibrium}\text{concentration (M): 0}\text{.88 }\text{0}\text{1}\text{.12}\\ \\ \text{The}\text{acetic acid equilibrium table is}\text{redfined as}\\ {\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{1}\text{.12 }\text{0}\text{0}\text{.88}\\ \text{Change in concentration (M):}\text{-}\text{x}\text{+x}\text{+x}\\ \text{Equilibrium}\text{concentration (M): 1}\text{.12-x }\text{0}\text{0}\text{.88+x}\\ \\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}{{\text{[CH}}_{\text{3}}\text{COOH]}}\\ \text{1}\text{.8}\text{×}{\text{10}}^{\text{-5}}\text{=}\frac{\text{(x)(0}\text{.88+x)}}{\text{(1}\text{.12-x)}}\\ \text{x}\text{is}\text{very}\text{small}\text{and}\text{we}\text{neglect}\text{it,}\\ \text{1}\text{.8}\text{×}{\text{10}}^{\text{-5}}\text{=}\frac{\text{(x)(0}\text{.88)}}{\text{1}\text{.12}}\\ \text{x=}{\text{[H}}^{\text{+}}\text{]}\text{=}\text{2}\text{.3}\text{×}{\text{10}}^{\text{-5}}\text{M}\\ \text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}\\ \text{=}\text{-log(2}\text{.3}\text{×}{\text{10}}^{\text{-5}}\text{)}\\ \text{pH}\text{=}\text{4}\text{.64}\\ \end{array}$

When hydrochloric acid adds to the buffer solution the acetate reacts with it and concentration of acetic acid changes.  From the expression of ${\text{K}}_{\text{a}}$, the concentration of hydrogen ion can be calculated.  With the help of hydrogen ion concentration, $\text{pH}$ of the buffer solution after addition of acid can be found.  The $\text{pH}$ value lowered from $\text{4}\text{.74}\text{to}\text{4}\text{.64}$ due to addition of strong acid.