Chapter 21, Problem 21.8P

(a)

Interpretation Introduction

Interpretation:

The change in $\text{pH}$ has to be calculated when add $\text{5mL of 0}\text{.100 M HCl (aq)}$ solution.

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

Given buffer solution,

$\text{ 0}{\text{.100 M in NH}}_{\text{3}}\text{(aq) and 0}{\text{.100 M in NH}}_{\text{4}}\text{Cl(aq)}$

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

  $\begin{array}{l}{\text{pK}}_{\text{b}}\text{of}\text{ammonia}\text{=4}\text{.75}\\ {\text{pH = pK}}_{\text{a}}\text{ + log}\left(\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]}\right)\\ \text{pH = 4}\text{.75 + log}\left(\frac{\left[0.100\right]}{\left[0.100\right]}\right)\\ \\ \text{= 4}\text{.75 + log}\left(\text{1}\right)\\ \\ {\text{= pK}}_{\text{a}}\\ {\text{pH = pK}}_{\text{a}}\text{ = 9}\text{.25}\text{.}\end{array}$

When add $\text{HCl}$ to the mix, ${\text{H}}^{\text{+}}$ will react with ${\text{NH}}_{\text{3}}$ to form ${\text{NH}}_{\text{4}}^{\text{+}}$ so that $\left[{\text{NH}}_{\text{3}}\right]$ will decrease, while $\left[{\text{NH}}_4^{+}\right]$ will increase.

The changes of concentration can be calculated as $\text{0}\text{.005 L × 0}\text{.1 M = 0}\text{.0005 mol HCl}$ reacting with $\text{0}\text{.1 L × 0}\text{.1 M = 0}{\text{.01 mol NH}}_{\text{3}}\text{.}$

$\text{HCl}$ is the limiting reagent, it will be consumed first so that we have $\text{0}\text{.01-0}\text{.0005 = 0}\text{.0095 mole}\text{NH}{\text{3}}_{}\text{ remaining}\text{.}$

Initial concentration of the solution is $\text{0}\text{.1 L × 0}\text{.1 M = 0}{\text{.01 mol NH}}_{\text{4}}^{\text{+}}$.  That gets increased to $\text{0}\text{.01 + 0}\text{.0005 = 0}{\text{.0105 mol NH}}_{\text{4}}^{\text{+}}$ moles is converted into concentration, $\begin{array}{l}\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]\text{=}\frac{\text{0}\text{.0105 mol}}{\left(\text{0}\text{.1 + 0}\text{.005 L}\right)}\text{=0}\text{.1 M}\\ \left[{\text{NH}}_{\text{3}}\right]\text{=}\frac{\text{0}\text{.0095 mol}}{\left(\text{0}\text{.1 + 0}\text{.005 L}\right)}\text{= 0}\text{.0905 M}\end{array}$

According to Henderson-Hasselbalch equation,

$\begin{array}{l}{\text{pH = pK}}_{\text{a}}\text{ + log}\left(\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]}\right)\\ \text{= 9}\text{.25 + log}\left(\text{0}\text{.0905/0}\text{.1}\right)\\ \text{= 9}\text{.21}\end{array}$ Therefore, the change in $\text{pH = 9}\text{.21 - 9}\text{.25 = -0}\text{.04}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The change in $\text{pH}$ has to be calculated when add $\text{5mL of 0}\text{.100 M NaOH (aq)}$ solution.

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

Given buffer solution,

$\text{ 0}{\text{.100 M in NH}}_{\text{3}}\text{(aq) and 0}{\text{.100 M in NH}}_{\text{4}}\text{Cl(aq)}$

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

When add $\text{HCl}$ to the mix, ${\text{NH}}_{\text{4}}^{\text{+}}$ will react with ${\text{NH}}_{\text{3}}$ to form ${\text{H}}_{\text{2}}\text{O}$ so that $\left[{\text{NH}}_{\text{3}}\right]$ will increase, while $\left[{\text{NH}}_4^{+}\right]$ will decrease.

The changes of concentration can be calculated as $\text{0}\text{.005 L × 0}\text{.1 M = 0}\text{.0005 mol NaOH}$ reacting with $\text{0}\text{.1 L × 0}\text{.1 M = 0}{\text{.01 mol NH}}_4^{+}$

$\text{HCl}$ is the limiting reagent, it will be consumed first so that we have $\text{0}\text{.01-0}\text{.0005 = 0}\text{.0095 mole}\text{NH}{\text{3}}_{}\text{ remaining}\text{.}$

Initial concentration of the solution is $\text{0}\text{.1 L × 0}\text{.1 M = 0}{\text{.01 mol NH}}_{\text{3}}$.  That gets increased to $\text{0}\text{.01 + 0}\text{.0005 = 0}{\text{.0105 mol NH}}_{\text{3}}$ moles is converted into concentration,

$\begin{array}{l}\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]\text{=}\frac{\text{0}\text{.0095 mol}}{\left(\text{0}\text{.1 + 0}\text{.005 L}\right)}\text{=0}\text{.0905 M}\\ \left[{\text{NH}}_{\text{3}}\right]\text{=}\frac{\text{0}\text{.0105 mol}}{\left(\text{0}\text{.1 + 0}\text{.005 L}\right)}\text{= 0}\text{.1 M}\end{array}$

According to Henderson-Hasselbalch equation,

  $\begin{array}{l}{\text{pH = pK}}_{\text{a}}\text{ + log}\left(\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]}\right)\\ \text{= 9}\text{.25 + log}\left(\frac{\text{0}\text{.1}}{0.0905}\right)\\ \text{= 9}\text{.29}\text{.}\end{array}$ Therefore, the change in $\text{pH = 9}\text{.29 - 9}\text{.25 = 0}\text{.04}\text{.}$