Chapter 21, Problem 21.7P

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$ change of $\text{500}\text{mL}\text{of}\text{a}\text{0}\text{.10}\text{M}{\text{NH}}_{\text{4}}{\text{Cl}}_{\text{(aq)}}\text{-0}\text{.10}\text{M}{\text{NH}}_{\text{3}}{}_{\text{(aq)}}$ buffer solution when add $1$ gram of potassium hydroxide has to be calculated.

Concept Introduction:

Henderson-Hasselbalch equation is given by

    $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[}\text{Conjugate}\text{base]}}{\text{[Acid]}}$

Explanation of Solution

$\text{pH}$ of buffer solution can be calculated as

    $\begin{array}{l}{\text{The moles of NH}}_{\text{4}}\text{Cl=500}\text{mL×}\frac{\text{0}\text{.1}}{\text{1000}}\text{=0}\text{.05}\text{moles}\hfill \\ \begin{array}{l}{\text{The moles of NH}}_{\text{3}}\text{=500}\text{mL×}\frac{\text{0}\text{.1}}{\text{1000}}\text{=0}\text{.05}\text{moles}\\ \text{pH = 14 - }\left[\text{4}\text{.75 + log}\frac{\text{ 0}\text{.05 }}{\text{0}\text{.05}}\right]\\ \end{array}\hfill \\ \begin{array}{l}\text{pH = 14 - }\left[{\text{pK}}_{\text{b}}\text{ + log1 }\right]\\ \end{array}\hfill \\ \text{pH = 9}\text{.25}\hfill \end{array}$

When $\text{KOH}$ is added

$\text{pOH}$ can be calculated as

    $\begin{array}{l}\text{The moles of KOH =}\frac{\text{1}}{\text{56}\text{.1}}\\ \text{= 0}{\text{.0178 = pK}}_{\text{b}}\text{ + log }\left[\frac{\text{salt - C}}{\text{base + C}}\right]\\ \begin{array}{l}\begin{array}{l}\\ \text{= 4}\text{.74 + log }\left[\frac{\text{0}\text{.05 - 0}\text{.0178 }}{\text{0}\text{.05 + 0}\text{.0178}}\right]\end{array}\hfill \\ \text{pOH = 4}\text{.42}\hfill \\ \text{pH = 14}\text{-}\text{4}\text{.42=9}\text{.58}\hfill \end{array}\end{array}$

Therefore, $\text{pH}$ change $=9.58-9.25$

Hence, the $\text{pH}$Change is $\text{0}\text{.33}\text{.}$

(b)

Interpretation Introduction

Interpretation:

$\text{pH}$ change of $\text{500}\text{mL}\text{of}\text{a}1.00\text{M}{\text{NH}}_{\text{4}}{\text{Cl}}_{\text{(aq)}}\text{-0}\text{.10}\text{M}{\text{NH}}_{\text{3}}{}_{\text{(aq)}}$ buffer solution when add $1$ gram of potassium hydroxide has to be calculated.

Concept Introduction:

Henderson-Hasselbalch equation is given by

    $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[}\text{Conjugate}\text{base]}}{\text{[Acid]}}$

Explanation of Solution

$\text{pH}$ of buffer solution can be calculated as

$\begin{array}{l}{\text{The moles of NH}}_{\text{4}}\text{Cl=500mL×}\frac{1}{\text{1000}}\text{=0}\text{.5}\text{moles}\\ {\text{The moles of NH}}_{\text{3}}\text{=500mL×}\frac{\text{1}}{\text{1000}}\text{=0}\text{.5}\text{moles}\end{array}$

When $\text{KOH}$ is added

$\text{pOH}$ can be calculated as

The moles of $\text{KOH}$$=0.0178$

    $\begin{array}{l}\begin{array}{l}\text{pH = 14 - }\left(\text{4}\text{.74 + log }\left[\frac{\text{0}\text{.5 - 0}\text{.0178}}{\text{0}\text{.5 + 0}\text{.0178}}\right]\right)\\ \end{array}\hfill \\ \begin{array}{l}\text{pH = 9}\text{.29}\\ \end{array}\hfill \\ \text{pH change = 9}\text{.29 - 9}\text{.25}\text{=}\text{0}\text{.04}\text{.}\hfill \end{array}$

Therefore, the $\text{pH}$ Change $\text{= 0}\text{.04}\text{.}$