Question

The ionisation constant of ammonium hydroxide is $ 1.77 \times 10^{-5} $ at $298\, K.$ Hydrolysis constant of ammonium chloride is

• A. $ 5.65 \times 10^{-10} $
• B. $ 6.50 \times 10^{-12} $
• C. $ 5.65 \times 10^{-13} $
• D. $ 5.65 \times 10^{-12} $

Answer 1

The Correct Option is A

To find the hydrolysis constant of ammonium chloride, we need to understand the relationship between the ionization constant of ammonium hydroxide and the hydrolysis constant of its salt, ammonium chloride. This requires a basic application of the concepts of chemical equilibrium.

Ammonium hydroxide (NH₄OH) partially ionizes in water according to the equilibrium:

NH_4OH_{(aq)} \rightleftharpoons NH_4^+_{(aq)} + OH^-_{(aq)}

Here, the ionization constant (K_{b}) is given as 1.77 \times 10^{-5}.

For ammonium chloride, when it dissolves in water, it forms:

NH_4Cl_{(s)} \rightarrow NH_4^+_{(aq)} + Cl^-_{(aq)}

The hydrolysis of ammonium ions can be represented as:

NH_4^+_{(aq)} + H_2O_{(l)} \rightleftharpoons NH_4OH_{(aq)} + H^+_{(aq)}

The hydrolysis constant (K_{h}) is related to the ionization constant of ammonium hydroxide by the following relation:

K_{w} = K_{b} \times K_{h}

where K_{w} is the ion product of water, typically 10^{-14} at 298\, K.

Rearranging the formula, we find the hydrolysis constant:

K_{h} = \frac{K_{w}}{K_{b}}

Substituting the given values:

K_{h} = \frac{10^{-14}}{1.77 \times 10^{-5}}

Calculating this gives:

K_{h} = 5.65 \times 10^{-10}

Thus, the hydrolysis constant of ammonium chloride is 5.65 \times 10^{-10}, which matches the correct option provided.