Chapter 21, Problem 21.84P

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution at equivalent point in the titration of $\text{17}\text{.5}\text{mL}\text{of}\text{0}\text{.098}\text{M}$ pyridine with

$\text{0}\text{.117 M}{\text{HI}}_{\text{(aq)}}$ has to be calculated.

Concept Introduction:

In the quantitative analysis the unknown concentration can be calculated using volumetric principle

  $\begin{array}{c}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{=M}}_{\text{2}}{\text{V}}_{\text{2}}\\ \\ \text{M=molarity}\\ \\ \text{V=volume}\end{array}$

Henderson-Hasselbalch equation is given by

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

Explanation of Solution

Given,

$\text{17}\text{.5}\text{mL}\text{of}\text{0}\text{.098}\text{M}$ Pyridine

$\text{0}\text{.117 M}{\text{HI}}_{\text{(aq)}}$

At the equivalent point of acid $\left(\text{HI}\right)$ base $\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right)$ titration,

$n\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right)=n\left(\text{HI}\right)$

The volume of $\left(\text{HI}\right)$ can be calculated as

$\begin{array}{c}\text{17}\text{.5}\text{mL×0}\text{.098}\text{M}\text{=}\text{0}\text{.117}\text{M}\text{×}{\text{V}}_{\text{HI}}\\ \\ {\text{V}}_{\text{HI}}\text{=}\frac{\text{17}\text{.5}\text{mL×0}\text{.098}\text{M}}{\text{0}\text{.117}\text{M}}\\ \text{=}\text{14}\text{.658}\text{=14}\text{.7}\text{mL}\text{.}\end{array}$

$\text{14}\text{.7}\text{mL}$ of acid should be added to attain the equivalent point.

Number of moles of Pyridine hydrochloride salt = Pyridine

$\begin{array}{c}\text{=17}\text{.5}\cancel{\text{mL}}\text{×}\frac{\cancel{\text{L}}}{\text{1000}\cancel{\text{mL}}}\text{×}\frac{\text{0}\text{.098}\text{mol}}{\cancel{\text{L}}}\\ \\ \text{=}\text{1}{\text{.715×10}}^{\text{-3}}\text{mol=1}{\text{.72×10}}^{\text{-3}}\text{mol}\end{array}$

The volume base and acid at equivalent point

$\begin{array}{l}\text{=}\text{(17}\text{.5+14}\text{.7)}\text{mL }\\ \\ \text{=}\text{32}\text{.2}\text{mL}\\ \\ \text{=}\text{0}\text{.0322L}\end{array}$

The concentration of ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}$= $\begin{array}{l}\frac{\text{1}{\text{.72×10}}^{\text{-3}}\text{mol}}{\text{0}\text{.0322}\text{L}}\\ \text{=0}\text{.0534}\text{M}\end{array}$

ICE table can be shown in below

$\begin{array}{l}{\text{K}}_{\text{a}}=\frac{K_w}{K_b}=\frac{1.0\times {10}^{-14}}{1.7\times {10}^{-9}}\\ =5.9\times {10}^{-6}\end{array}$

	${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}$	${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$
Initial	0.0533	0	0
Change	-x	+x	+x
Equilibrium	0.0533	x	x

The ionization constant of pyridine is ${\text{K}}_{\text{b}}\text{=1}{\text{.7×10}}^{\text{-9}}$

Acid ionization constant of ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}$

  $\begin{array}{l}{\text{K}}_{\text{a}}=\frac{K_w}{K_b}=\frac{1.0\times {10}^{-14}}{1.7\times {10}^{-9}}\\ =5.9\times {10}^{-6}\end{array}$

Acid ionization constant,

  $\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}\right]}\\ \\ \text{5}\text{.9}{\text{×10}}^{\text{-6}}\text{=}\frac{\text{(x)(x)}}{\text{(0}\text{.0533-x)}}\\ \text{5}\text{.9}{\text{×10}}^{\text{-6}}\text{=}\frac{{\text{(x}}^{\text{2}}\text{)}}{\text{(0}\text{.0533-x)}}\end{array}$

It can assume that

  $\begin{array}{c}\text{(0}\text{.0533-x)}\text{M}\text{=}\text{0}\text{.0533}\text{M}\\ \text{5}\text{.9}\times {\text{10}}^{\text{-6}}=\frac{x^2}{\text{0}\text{.0533}}\\ x^2=(\text{5}\text{.9}\times {\text{10}}^{\text{-6}})(\text{0}\text{.0533})\\ =31.4\times {10}^{-8}\end{array}$

  $\begin{array}{l}\text{x}\text{=}\sqrt{\text{31}{\text{.4×10}}^{\text{-8}}}\\ \text{=}\text{5}{\text{.61×10}}^{\text{-4}}\text{M}\end{array}$

Based on the equilibrium table

  $\begin{array}{c}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]=x}\\ \text{=5}{\text{.61×10}}^{\text{-4}}\text{M}\end{array}$

  $\begin{array}{c}\text{pH=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{pH=}\text{-log[5}{\text{.61×10}}^{\text{-4}}\text{]}\\ \text{=}3.25\end{array}$

Hence, the $\text{pH}$ of the solution at equivalent point is $\mathrm{3.25.}$