Question

Calculate the degree of hydrolysis and pH of $0.02{\text{ M}}$ ammonium cyanide $\left( {{\text{N}}{{\text{H}}_{\text{4}}}{\text{CN}}} \right)$ at $298{\text{ K}}$. $\left[ {{{\text{K}}_1}{\text{ of HCN}} = 4.99 \times {{10}^{ - 9}},{\text{ }}{{\text{K}}_{\text{b}}}{\text{ for N}}{{\text{H}}_4}{\text{OH}} = 1.77 \times {{10}^{ - 5}}} \right]$.
A) ${\text{8}}{\text{.2}}$
B) ${\text{3}}{\text{.2}}$
C) ${\text{9}}{\text{.3}}$
D) ${\text{3}}{\text{.9}}$

Answer 1

Hint: The negative logarithm of the ${{\text{H}}^ + }$ ion concentration in the solution is known as the pH of the solution. Calculate the pH using the Henderson-Hasselbalch equation.

Formulae Used: ${{\text{K}}_{\text{h}}} = \dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{a}}} \times {{\text{K}}_{\text{b}}}}}$
${\text{p}}{{\text{K}}_{\text{a}}} = {\text{pH}} + \log \dfrac{{\left[ {{\text{HA}}} \right]}}{{\left[ {{{\text{A}}^ - }} \right]}}$

Complete step by step answer:
Ammonium cyanide $\left( {{\text{N}}{{\text{H}}_{\text{4}}}{\text{CN}}} \right)$ is a salt of strong base ammonium hydroxide $\left( {{\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}} \right)$ and weak acid hydrogen cyanide $\left( {{\text{HCN}}} \right)$.
Calculate the hydrolysis constant for ammonium cyanide using the equation as follows:
${{\text{K}}_{\text{h}}} = \dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{a}}} \times {{\text{K}}_{\text{b}}}}}$
Where ${{\text{K}}_{\text{h}}}$ is the hydrolysis constant,
${{\text{K}}_{\text{w}}}$ is the ionization constant of water,
${{\text{K}}_{\text{a}}}$ is the acid dissociation constant,
${{\text{K}}_{\text{b}}}$ is the base dissociation constant.
Substitute ${10^{ - 14}}$ for the ionization constant of water, $4.99 \times {10^{ - 9}}$ for the acid dissociation constant, $1.77 \times {10^{ - 5}}$ for the base dissociation constant. Thus,
${{\text{K}}_{\text{h}}} = \dfrac{{{{10}^{ - 14}}}}{{4.99 \times {{10}^{ - 9}} \times 1.77 \times {{10}^{ - 5}}}}$
${{\text{K}}_{\text{h}}} = {\text{1}}{\text{.132}}$
Thus, the hydrolysis constant for ammonium cyanide is ${\text{1}}{\text{.132}}$.
Calculate the degree of hydrolysis for ammonium cyanide using the equation as follows:
${\text{h}} = \dfrac{{\sqrt {{{\text{K}}_{\text{h}}}} }}{{1 + \sqrt {{{\text{K}}_{\text{h}}}} }}$
Where ${\text{h}}$ is the degree of hydrolysis.
Thus,
${\text{h}} = \dfrac{{\sqrt {{\text{1}}{\text{.132}}} }}{{1 + \sqrt {{\text{1}}{\text{.132}}} }}$
${\text{h}} = {\text{0}}{\text{.51}}$
Thus, the degree of hydrolysis for ammonium cyanide is ${\text{0}}{\text{.51}}$.
Calculate the pH for ammonium cyanide using the Henderson-Hasselbalch equation as follows:
${\text{p}}{{\text{K}}_{\text{a}}} = {\text{pH}} + \log \dfrac{{\left[ {{\text{HA}}} \right]}}{{\left[ {{{\text{A}}^ - }} \right]}}$
Thus,
${\text{pH}} = {\text{p}}{{\text{K}}_{\text{a}}} - \log \dfrac{{\left[ {{\text{HA}}} \right]}}{{\left[ {{{\text{A}}^ - }} \right]}}$
${\text{pH}} = {\text{log}}\left( {{{\text{K}}_{\text{a}}}} \right) - \log \dfrac{{\left[ {\text{h}} \right]}}{{\left[ {{\text{1}} - {\text{h}}} \right]}}$
Thus,
${\text{pH}} = {\text{log}}\left( {4.99 \times {{10}^{ - 9}}} \right) - \log \dfrac{{\left( {{\text{0}}{\text{.51}}} \right)}}{{\left( {1 - {\text{0}}{\text{.51}}} \right)}}$
${\text{pH}} = {\text{9}}{\text{.3}}$
Thus, the pH for ammonium cyanide is ${\text{9}}{\text{.3}}$.
Thus, the degree of hydrolysis for ammonium cyanide is ${\text{0}}{\text{.51}}$ and the pH for ammonium cyanide is ${\text{9}}{\text{.3}}$.

Thus, the correct option is (C) ${\text{9}}{\text{.3}}$.

Note: The fraction or percentage of a salt which is hydrolysed at equilibrium is known as the degree of hydrolysis. The Henderson-Hasselbalch equation describes the relationship between the pH and pOH of a solution and the ${\text{p}}{{\text{K}}_{\text{a}}}$ or ${\text{p}}{{\text{K}}_{\text{b}}}$ and the ratio of the concentration of the chemical species that has been dissociated. If the acid dissociation constant is known we can use the Henderson-Hasselbalch equation.