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Question: The ionization constant of ammonium hydroxide is \(1.77 \times {{10}^{-5}}\) at 298K. Hydrolysis c...

The ionization constant of ammonium hydroxide is 1.77×10−51.77 \times {{10}^{-5}} at 298K.
Hydrolysis constant of ammonium chloride is:
A). 5.65 x 10−12{{10}^{-12}}
B). 5.65 x10−10{{10}^{-10}}
C). 6.50 x 10−12{{10}^{-12}}
D). 5.65 x 10−13{{10}^{-13}}

Explanation

Solution

The ionization reaction of ammonium hydroxide is given by the following equation
NH4OH⇌KbNH4++OH−\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 NH4++H2O−⇌KbNH4OH+H+\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
NH4++H2O−⇌KbNH4OH+H+\mathrm{NH}_{4}^{+}+\mathrm{H_2O}^{-} \stackrel{\mathrm{K}_{\mathrm{b}}}{\rightleftharpoons} \mathrm{NH}_{4}\mathrm{OH}+\mathrm{H}^{+}
The ionization constant kbk_b from the above-mentioned ionization reaction of ammonium hydroxide can be written as
Kb=[NH4+][OH−][NH4OH]{{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 NH4++H2O−⇌KbNH4OH+H+\mathrm{NH}_{4}^{+}+\mathrm{H_2O}^{-} \stackrel{\mathrm{K}_{\mathrm{b}}}{\rightleftharpoons} \mathrm{NH}_{4}\mathrm{OH}+\mathrm{H}^{+}

The hydrolysis constant khk_h from the above-mentioned hydrolysis reaction of ammonium hydroxide can be written as:
Kh=[NH4OH][NH4+]×[OH−][OH−]{{K}_{h}}=\dfrac{[N{{H}_{4}}OH]}{[NH{{4}^{+}}]}\times \dfrac{[O{{H}^{-}}]}{[O{{H}^{-}}]}
Upon rearranging the equation for hydrolysis constant, we get:
Kh=[NH4OH][NH4+][OH−]×[H+][OH−]{{K}_{h}}=\dfrac{[N{{H}_{4}}OH]}{[NH{{4}^{+}}][O{{H}^{-}}]}\times \dfrac{[{{H}^{+}}][O{{H}^{-}}]}{{}}
Kw{{K}_{w}}= [H+][OH−][{{H}^{+}}][O{{H}^{-}}]
Using equation (1) and using the two equations above gives

The relation between hydrolysis constant and ionization constant given by kh=kwkb{{k}_{h}}=\dfrac{{{k}_{w}}}{{{k}_{b}}}
kwk_w for water at 298 K is 10−1410^{-14}
Putting the value of kwk_w and ionization constant mentioned above and, in the question, respectively gives the value of hydrolysis constant as
kh=5.65×10−10{{k}_{h}}=5.65\times {{10}^{-10}}
So, the correct answer is “Option B”.

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