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Question: Calculate the degree of hydrolysis and pH of \(0.02{\text{ M}}\) ammonium cyanide \(\left( {{\text{N...

Calculate the degree of hydrolysis and pH of 0.02 M0.02{\text{ M}} ammonium cyanide (NH4CN)\left( {{\text{N}}{{\text{H}}_{\text{4}}}{\text{CN}}} \right) at 298 K298{\text{ K}}. [K1 of HCN=4.99×109, Kb for NH4OH=1.77×105]\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) 8.2{\text{8}}{\text{.2}}
B) 3.2{\text{3}}{\text{.2}}
C) 9.3{\text{9}}{\text{.3}}
D) 3.9{\text{3}}{\text{.9}}

Explanation

Solution

The negative logarithm of the H+{{\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: Kh=KwKa×Kb{{\text{K}}_{\text{h}}} = \dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{a}}} \times {{\text{K}}_{\text{b}}}}}
pKa=pH+log[HA][A]{\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 (NH4CN)\left( {{\text{N}}{{\text{H}}_{\text{4}}}{\text{CN}}} \right) is a salt of strong base ammonium hydroxide (NH4OH)\left( {{\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}} \right) and weak acid hydrogen cyanide (HCN)\left( {{\text{HCN}}} \right).
Calculate the hydrolysis constant for ammonium cyanide using the equation as follows:
Kh=KwKa×Kb{{\text{K}}_{\text{h}}} = \dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{a}}} \times {{\text{K}}_{\text{b}}}}}
Where Kh{{\text{K}}_{\text{h}}} is the hydrolysis constant,
Kw{{\text{K}}_{\text{w}}} is the ionization constant of water,
Ka{{\text{K}}_{\text{a}}} is the acid dissociation constant,
Kb{{\text{K}}_{\text{b}}} is the base dissociation constant.
Substitute 1014{10^{ - 14}} for the ionization constant of water, 4.99×1094.99 \times {10^{ - 9}} for the acid dissociation constant, 1.77×1051.77 \times {10^{ - 5}} for the base dissociation constant. Thus,
Kh=10144.99×109×1.77×105{{\text{K}}_{\text{h}}} = \dfrac{{{{10}^{ - 14}}}}{{4.99 \times {{10}^{ - 9}} \times 1.77 \times {{10}^{ - 5}}}}
Kh=1.132{{\text{K}}_{\text{h}}} = {\text{1}}{\text{.132}}
Thus, the hydrolysis constant for ammonium cyanide is 1.132{\text{1}}{\text{.132}}.
Calculate the degree of hydrolysis for ammonium cyanide using the equation as follows:
h=Kh1+Kh{\text{h}} = \dfrac{{\sqrt {{{\text{K}}_{\text{h}}}} }}{{1 + \sqrt {{{\text{K}}_{\text{h}}}} }}
Where h{\text{h}} is the degree of hydrolysis.
Thus,
h=1.1321+1.132{\text{h}} = \dfrac{{\sqrt {{\text{1}}{\text{.132}}} }}{{1 + \sqrt {{\text{1}}{\text{.132}}} }}
h=0.51{\text{h}} = {\text{0}}{\text{.51}}
Thus, the degree of hydrolysis for ammonium cyanide is 0.51{\text{0}}{\text{.51}}.
Calculate the pH for ammonium cyanide using the Henderson-Hasselbalch equation as follows:
pKa=pH+log[HA][A]{\text{p}}{{\text{K}}_{\text{a}}} = {\text{pH}} + \log \dfrac{{\left[ {{\text{HA}}} \right]}}{{\left[ {{{\text{A}}^ - }} \right]}}
Thus,
pH=pKalog[HA][A]{\text{pH}} = {\text{p}}{{\text{K}}_{\text{a}}} - \log \dfrac{{\left[ {{\text{HA}}} \right]}}{{\left[ {{{\text{A}}^ - }} \right]}}
pH=log(Ka)log[h][1h]{\text{pH}} = {\text{log}}\left( {{{\text{K}}_{\text{a}}}} \right) - \log \dfrac{{\left[ {\text{h}} \right]}}{{\left[ {{\text{1}} - {\text{h}}} \right]}}
Thus,
pH=log(4.99×109)log(0.51)(10.51){\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)}}
pH=9.3{\text{pH}} = {\text{9}}{\text{.3}}
Thus, the pH for ammonium cyanide is 9.3{\text{9}}{\text{.3}}.
Thus, the degree of hydrolysis for ammonium cyanide is 0.51{\text{0}}{\text{.51}} and the pH for ammonium cyanide is 9.3{\text{9}}{\text{.3}}.

Thus, the correct option is (C) 9.3{\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 pKa{\text{p}}{{\text{K}}_{\text{a}}} or pKb{\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.