Question

The ionization constant of ammonium hydroxide is $1.77 \times {{10}^{-5}}$ at 298K.
Hydrolysis constant of ammonium chloride is:
A). 5.65 x ${{10}^{-12}}$
B). 5.65 x${{10}^{-10}}$
C). 6.50 x ${{10}^{-12}}$
D). 5.65 x ${{10}^{-13}}$

Answer 1

Hint: The ionization reaction of ammonium hydroxide is given by the following equation
$\mathrm{NH}_{4} \mathrm{OH} \stackrel{\mathrm{K}_{\mathrm{b}}}{\rightleftharpoons} \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-}$
Similarly, the hydrolysis reaction for ammonium hydroxide is given by the following equation $\mathrm{NH}_{4}^{+}+\mathrm{H_2O}^{-} \stackrel{\mathrm{K}_{\mathrm{b}}}{\rightleftharpoons} \mathrm{NH}_{4}\mathrm{OH}+\mathrm{H}^{+}$

Complete step by step answer:
A base ionization constant is defined as the equilibrium constant for the ionization of a base. In our case the base is ammonia.
A hydrolysis constant is defined as an equilibrium constant for a hydrolysis reaction
For the ionization reaction of ammonium hydroxide is given by the following equation
$\mathrm{NH}_{4}^{+}+\mathrm{H_2O}^{-} \stackrel{\mathrm{K}_{\mathrm{b}}}{\rightleftharpoons} \mathrm{NH}_{4}\mathrm{OH}+\mathrm{H}^{+}$
The ionization constant $k_b$ from the above-mentioned ionization reaction of ammonium hydroxide can be written as
${{K}_{b}}=\dfrac{[NH{{4}^{+}}][O{{H}^{-}}]}{[N{{H}_{4}}OH]}$ (1)

Similarly, for the hydrolysis reaction for ammonium hydroxide is given by the following equation $\mathrm{NH}_{4}^{+}+\mathrm{H_2O}^{-} \stackrel{\mathrm{K}_{\mathrm{b}}}{\rightleftharpoons} \mathrm{NH}_{4}\mathrm{OH}+\mathrm{H}^{+}$

The hydrolysis constant $k_h$ from the above-mentioned hydrolysis reaction of ammonium hydroxide can be written as:
${{K}_{h}}=\dfrac{[N{{H}_{4}}OH]}{[NH{{4}^{+}}]}\times \dfrac{[O{{H}^{-}}]}{[O{{H}^{-}}]}$
Upon rearranging the equation for hydrolysis constant, we get:
${{K}_{h}}=\dfrac{[N{{H}_{4}}OH]}{[NH{{4}^{+}}][O{{H}^{-}}]}\times \dfrac{[{{H}^{+}}][O{{H}^{-}}]}{{}}$
${{K}_{w}}$= $[{{H}^{+}}][O{{H}^{-}}]$
Using equation (1) and using the two equations above gives

The relation between hydrolysis constant and ionization constant given by ${{k}_{h}}=\dfrac{{{k}_{w}}}{{{k}_{b}}}$
$k_w$ for water at 298 K is $10^{-14}$
Putting the value of $k_w$ and ionization constant mentioned above and, in the question, respectively gives the value of hydrolysis constant as
${{k}_{h}}=5.65\times {{10}^{-10}}$
So, the correct answer is “Option B”.

Note: The value of $k_w$ (also known as ionic product) used in the question is only valid for 298K. The value of $k_w$ will vary with temperature. The $k_w$ of water increases with increase in temperature and decreases with decrease in temperature of water.