Chapter 6, Problem 6H.9E

Interpretation Introduction

Interpretation:

The $\text{pH}$ at the stoichiometric point of the titration of 25.0 mL of $\text{0}\text{.120}\text{M}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COOH}}_{\text{(aq)}}$ with $\text{0}\text{.0230}\text{M}{\text{NaOH}}_{\text{(aq)}}$ 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.9E

The $\text{pH}$ at the stoichiometric point of the titration of 25.0 mL of $\text{0}\text{.120}\text{M}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COOH}}_{\text{(aq)}}$ with $\text{0}\text{.0230}\text{M}{\text{NaOH}}_{\text{(aq)}}$ is 8.54.

Explanation of Solution

The initial amount of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ present in the analyte solution is

  $\begin{array}{c}{\text{n(C}}_6{\text{H}}_5\text{COOH)}\text{=}\text{(0}{\text{.025×10}}^{\text{-3}}\text{L)×(0}\text{.120}\text{mol}\cdot {\text{L}}^{\text{-1}}\text{)}\\ \text{=}\text{0}{\text{.003×10}}^{\text{-3}}\text{mol}\\ \text{=}\text{0}\text{.003}\text{mmol}\end{array}$

The amount of ${\text{OH}}^{-}$ required to react with ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ is calculated and their reaction stoichiometry is

$\text{1}\text{mol}{\text{OH}}^{\text{-}}\text{=}\text{1}\text{mol}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$.

  $\begin{array}{c}{\text{n(OH}}^{\text{-}}\text{)}\text{=}\text{(initial amount of}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\text{in the analyte)×}\frac{\text{1}\text{mol}{\text{OH}}^{\text{-}}}{\text{1}\text{mol}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}}\\ \text{=}\text{(0}{\text{.003×10}}^{\text{-3}}\text{mol)×}\frac{\text{1}\text{mol}{\text{OH}}^{\text{-}}}{\text{1}\text{mol}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}}\\ =\text{0}\text{.003}{\text{×10}}^{\text{-3}}\text{mol}\\ =\text{0}\text{.003}\text{mmol}\end{array}$

Therefore, the ${\text{OH}}^{-}$ ion required to react with ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ is 0.003mmol.

From the stoichiometric point the amount of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{-}$ is equal to the amount of ${\text{OH}}^{-}$ added.

  ${\text{n(C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\text{)}\text{=}\text{0}\text{.003}\text{mmol}$

The amount of ${\text{OH}}^{-}$ ionspresent in the volume of titrant is calculated as.

  $\begin{array}{l}{\text{V}}_{\text{added}}\text{=}\frac{{\text{OH}}^{\text{-}}\text{ion required to react with}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}}{\text{concentration}\text{}\text{of}{\text{OH}}^{\text{-}}\text{ion}}\\ \\ \text{=}\frac{\text{0}{\text{.003×10}}^{\text{-3}}\text{mol}}{\text{0}\text{.0230}\text{mol}\cdot {\text{L}}^{\text{-1}}}\\ \\ \text{=}1.30{\text{×10}}^{\text{-2}}\text{L}\\ \\ \text{=}13.0\text{mL}\end{array}$

Hence, the amount of ${\text{OH}}^{-}$ ionspresent in the volume of titrant is 13.0mL.

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

  $\begin{array}{l}{\text{V}}_{\text{final}}\text{=}{\text{V}}_{\text{initial}}{\text{+V}}_{\text{added}}\\ \\ \text{=}\text{25}\text{.0+13}\text{.0}\text{mL}\\ \\ \text{=}38\text{mL}\end{array}$

The concentration of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{-}$ ions at the stoichiometric point is

  $\begin{array}{l}{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\text{]}\text{=}\frac{\text{amount of }{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\text{present}\text{in}\text{the}\text{analyte}\text{solution}}{\text{total volume of the solution}}\\ \\ \text{=}\frac{\text{0}{\text{.003×10}}^{\text{-3}}\text{mol}}{\text{0}{\text{.038×10}}^{\text{-3}}\text{L}}\\ \\ \text{=}\text{0}\text{.079}\text{mol}\cdot {\text{L}}^{\text{-1}}\end{array}$

Therefore, the concentration of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{-}$ is $\text{0}\text{.079}\text{mol}\cdot {\text{L}}^{\text{-1}}$.

The chemical equilibrium reactionis given below.

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COOH}}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH]}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\text{]}}$

	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{-}$	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.079	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.079-x	x	x

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.079-x}}$

The above equation, assume that the x present in 0.079-x is very small than 0.079 then it can be negligible and it follows,

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{x}}^2}{\text{0}\text{.079}}$

Benzoic acid ${\text{K}}_{\text{a}}$ value is $\text{6}{\text{.5×10}}^{\text{-6}}$ from this we can calculate the ${\text{K}}_{\text{b}}$ value, is given below.

  $\begin{array}{l}{\text{K}}_{\text{a}}{\text{×K}}_{\text{b}}{\text{=K}}_{\text{w}}\\ \\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ \text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{6}{\text{.5×10}}^{\text{-5}}}\\ \\ \text{=}1.54{\text{×10}}^{\text{-10}}\end{array}$

Therefore, the ${\text{K}}_{\text{b}}$ value of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ is $1.54\times {10}^{-10}$ substitute this in below equation.

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.079}}\\ \\ \text{1}{\text{.54×10}}^{\text{-10}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.079}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{(1}{\text{.54×10}}^{\text{-10}}\text{)×0}\text{.079}\\ \\ \text{x}\text{=}\sqrt{\text{(1}{\text{.54×10}}^{\text{-10}}\text{)×0}\text{.079}}\\ \\ \text{=}3.49{\text{×10}}^{\text{-6}}\end{array}$

Hence, the concentration of ${\text{OH}}^{\text{-}}$ is $\text{3}{\text{.49×10}}^{\text{-6}}$.

Now, we can calculate the pH of the solution,

  $\begin{array}{l}\text{pOH}\text{=-log(}3.49{\text{×10}}^{\text{-6}}\text{)}\\ \\ \text{=}5.46\\ \\ \text{pH+pOH}\text{=}\text{14}\\ \\ \text{pH}\text{=}\text{14-5}\text{.46}\\ \\ \text{=}8.54\end{array}$

Hence, the $\text{pH}$ at the stoichiometric point of the titration of 25.0 mL of $\text{0}\text{.120}\text{M}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COOH}}_{\text{(aq)}}$ with $\text{0}\text{.0230}\text{M}{\text{NaOH}}_{\text{(aq)}}$ is 8.54.