Chapter 6, Problem 6H.2BST

Interpretation Introduction

Interpretation:

The $\text{pH}$ at the stoichiometric point of the titration of 25.0 mL of $\text{0}\text{.020}\text{M}{\text{NH}}_{\text{3}}{}_{\text{(aq)}}$ with $\text{0}\text{.015}\text{M}{\text{HCl}}_{\text{(aq)}}$ has to be calculated.

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base -10 logarithm of the hydrogen or hydronium ion concentration.

  $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

On rearranging, the concentration of hydrogen ion ${\text{[H}}^{+}\text{]}$ can be calculated using pH as follows,

  ${\text{[H}}^{+}\text{]}\text{=}{\text{10}}^{\text{-pH}}$

Answer to Problem 6H.2BST

The $\text{pH}$ at the stoichiometric point of the titration of 25.0 mL of $\text{0}\text{.020}\text{M}{\text{NH}}_{\text{3}}{}_{\text{(aq)}}$ with $\text{0}\text{.015}\text{M}{\text{HCl}}_{\text{(aq)}}$ is 5.54.

Explanation of Solution

The initial amount of ${\text{NH}}_3$ present in the analyte solution is

  $\begin{array}{l}{\text{n(NH}}_3\text{)}\text{=}\text{(0}{\text{.025×10}}^{\text{-3}}\text{L)×(0}\text{.02}{\text{mol×L}}^{\text{-1}}\text{)}\\ \\ \text{=}\text{5}{\text{.0×10}}^{\text{-4}}\text{mol}\\ \\ \text{=}\text{0}{\text{.5×10}}^{\text{-3}}\text{mol}\\ \\ \text{=}\text{0}\text{.5}\text{mmol}\end{array}$

The amount of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ required to react with ${\text{NH}}_{\text{3}}$ is calculated and their reaction stoichiometry is

$\text{1}\text{mol}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{=}\text{1}\text{mol}{\text{NH}}_{\text{3}}$.

  $\begin{array}{l}{\text{n(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\text{=}\text{(initial amount of}{\text{NH}}_{\text{3}}\text{in the analyte)×}\frac{\text{1}\text{mol}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}}{\text{1}\text{mol}{\text{NH}}_{\text{3}}}\\ \\ \text{=}\text{(0}{\text{.5×10}}^{\text{-3}}\text{mol)×}\frac{\text{1}\text{mol}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}}{\text{1}\text{mol}{\text{NH}}_{\text{3}}}\\ \\ \text{=}\text{0}\text{.5}{\text{×10}}^{\text{-3}}\text{mol}\\ \\ \text{=}\text{0}\text{.5}\text{mmol}\end{array}$

Therefore, the ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion required to react with ${\text{NH}}_3$ is 0.5 mmol.

From the stoichiometric point the amount of ${\text{NH}}_{\text{4}}{}^{\text{+}}$ is equal to the amount of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ added.

  ${\text{n(NH}}_4{}^{+}\text{)}\text{=}\text{0}\text{.5}\text{mmol}$

The amount of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions present in the volume of titrant is calculated as.

  $\begin{array}{l}{\text{V}}_{\text{added}}\text{=}\frac{{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{ion required to react with}{\text{NH}}_{\text{3}}}{\text{concentration}\text{}\text{of}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{ion}}\\ \\ \text{=}\frac{\text{0}{\text{.5×10}}^{\text{-3}}\text{mol}}{\text{0}\text{.015}{\text{mol×L}}^{\text{-1}}}\\ \\ \text{=}\text{3}{\text{.33×10}}^{\text{-2}}\text{L}\\ \\ \text{=}\text{33}\text{.3}\text{mL}\end{array}$

Hence, the amount of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions present in the volume of titrant is 33.3 mL.

The total volume of the solution at the stoichiometric point is calculated as,

  $\begin{array}{l}{\text{V}}_{\text{final}}\text{=}{\text{V}}_{\text{initial}}{\text{+V}}_{\text{added}}\\ \\ \text{=}\text{25}\text{.0+33}\text{.3}\text{mL}\\ \\ \text{=}58.3\text{mL}\end{array}$

The concentration of ${\text{ClO}}^{-}$ ions at the stoichiometric point is

  $\begin{array}{l}{\text{[NH}}_4{}^{+}\text{]}\text{=}\frac{\text{amount of }{\text{NH}}_4{}^{+}\text{present}\text{in}\text{the}\text{analyte}\text{solution}}{\text{total volume of the solution}}\\ \\ \text{=}\frac{\text{0}{\text{.5×10}}^{\text{-3}}\text{mol}}{\text{3}{\text{.33×10}}^{\text{-2}}\text{L}}\\ \\ \text{=}\text{0}\text{.015}\text{mol}\cdot {\text{L}}^{\text{-1}}\end{array}$

Therefore, the concentration of ${\text{NH}}_4{}^{+}$ is $\text{0}\text{.015}\text{mol}\cdot {\text{L}}^{\text{-1}}$.

The chemical equilibrium reaction is given below.

  ${\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{NH}}_{\text{3(aq)}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

  ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[NH}}_{\text{3}}\text{]}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[NH}}_{\text{4}}{}^{\text{+}}\text{]}}$

	${\text{NH}}_4{}^{+}$	${\text{NH}}_3$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$
Initial concentration	0.015	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.015-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

  ${\text{K}}_{\text{a}}\text{=}\frac{\text{x×x}}{\text{0}\text{.015-x}}$

The above equation, assume that the x present in 0.015-x is very small than 0.015 then it can be negligible and it follows,

  ${\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.015}}$

Ammonia ${\text{K}}_{\text{a}}$ value is $\text{5}{\text{.6×10}}^{\text{-10}}$ substitute this in above equation.

  $\begin{array}{l}5.6{\text{×10}}^{\text{-10}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.015}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{(}5.6{\text{×10}}^{\text{-10}}\text{)×0}\text{.015}\\ \\ \text{x}\text{=}\sqrt{\text{(}5.6{\text{×10}}^{\text{-10}}\text{)×0}\text{.015}}\\ \\ \text{=}2.90{\text{×10}}^{\text{-6}}\end{array}$

Hence, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $\text{2}{\text{.90×10}}^{\text{-6}}$.

Now, we can calculate the pH of the solution,

  $\begin{array}{l}\text{pH}\text{=}\text{-log(}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\\ \\ \text{=}\text{-log(2}{\text{.90×10}}^{\text{-6}}\text{)}\\ \\ \text{=}\text{5}\text{.54}\end{array}$

Hence, the $\text{pH}$ at the stoichiometric point of the titration of 25.0 mL of $\text{0}\text{.020}\text{M}{\text{NH}}_{\text{3}}{}_{\text{(aq)}}$ with $\text{0}\text{.015}\text{M}{\text{HCl}}_{\text{(aq)}}$ is 5.54.