Question

The toxic compound \[2,4 - \] dinitrophenol has ${{\text{K}}_{\text{a}}} = {10^{ - 4}}{\text{M}}$. In an experiment, a buffer solution of \[2,4 - \] dinitrophenol was prepared with the ${\text{pH}}$ adjusted to $5$. Calculate the ratio of the concentrations of the dissociated ion to the undissociated acid.
A. $0.01$
B. $0.1$
C. $10$
D. $100$

Answer 1

Hint:${\text{pH}}$ is a unit of measure which describes the degree of acidity or alkalinity of a solution. Buffer solution is a solution which resists changes in pH when a small amount of acid or base is added.
Henderson-Hasselbalch equation is useful for estimation of ${\text{pH}}$ of a buffer solution.

Complete step by step answer:
Buffer solution is typically a mixture of a weak acid and a salt of its conjugate base or weak base and a salt of its conjugate acid.
Consider a Bronsted acid-base reaction given below:
${\text{HA}} + {{\text{H}}_2}{\text{O}} \rightleftharpoons {{\text{H}}_3}{{\text{O}}^ + } + {{\text{A}}^ - }$
${{\text{K}}_{\text{a}}}$ is known as acid dissociation constant. It is a quantitative measurement of the strength of an acid in solution. In this chemical equation, ${{\text{K}}_{\text{a}}}$ can be expressed as:
${{\text{K}}_{\text{a}}} = \dfrac{{\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}} \to (1)$
${{\text{pK}}_{\text{a}}}$ is the negative base-$10$ logarithm of ${{\text{K}}_{\text{a}}}$ of a solution.
${{\text{pK}}_{\text{a}}} = - \log {{\text{K}}_{\text{a}}} \to (2)$
${\text{pH}} = - \log \left( {{{\text{H}}^ + }} \right) \to (3)$
Taking logarithm on both sides of $(1)$, we get
${\text{ - log}}{{\text{K}}_{\text{a}}} = - \log \dfrac{{\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}$
Or ${\text{ - log}}{{\text{K}}_{\text{a}}} = - \log \left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] + \left( { - \log \dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}} \right)$
From equation $(2),(3)$,
${\text{p}}{{\text{K}}_{\text{a}}} = {\text{pH}} + \left( { - \log \dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}} \right)$
We can rearrange the equation by
${\text{pH}} = {\text{p}}{{\text{K}}_{\text{a}}} + \left( {\log \dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}} \right) \to (4)$
$\left[ {{{\text{A}}^ - }} \right]$ is the concentration of conjugate base. It can also be referred as the concentration of salt.
Equation $(4)$ is the equation used here.
It is given that ${{\text{K}}_{\text{a}}} = {10^{ - 4}}{\text{M}}$ and ${\text{pH = 5}}$
From ${{\text{K}}_{\text{a}}}$ value, we can find ${\text{p}}{{\text{K}}_{\text{a}}}$.
${\text{p}}{{\text{K}}_{\text{a}}} = - \log \left( {{{10}^{ - 4}}} \right) = 4$
Substituting these values in $(4)$, we get
$5 = {\text{4}} + \left( {\log \dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}} \right) \Leftrightarrow 1 = \log \dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}$
$\therefore \dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}} = 10$
Thus the ratio of concentrations of dissociated ion to the undissociated acid is $10$.

Hence, the correct option is C.

Note:
Sometimes, the concentration of acid and its conjugate base at equilibrium will not be the same. At this time, this equation cannot be used. This equation also neglects the dissociation of acid and the hydrolysis of base.