Question

Hydrolysis constant $\left(K_h\right)$ of ammonium acetate is :
(Given that $K_a=$ $1.8\times$ ${10}^{-5}$, $K_b=$ $2\times$ ${10}^{-5}$, $K_w={10}^{-14}$)

• A. $28\times {10}^{-5}$
• B. $2.8\times {10}^{-5}$
• C. $0.28\times {10}^{-10}$
• D. $1.8\times {10}^{-5}$

Answer 1

The correct option is B $2.8\times {10}^{-5}$

The expression for the relation between hydrolysis constant and the dissociation constants is given below:
$K_h=\frac{K_w}{K_a\times K_b}$

Given that $K_a=$ $1.8\times$ ${10}^{-5}$, $K_b=$ $2\times$ ${10}^{-5}$, $K_w={10}^{-14}$.

Substituting values in the above expression, we get,
$K_h=\frac{{10}^{-14}}{1.8\times {10}^{-5}\times 2\times {10}^{-5}}=2.8\times {10}^{-5}$.

Option B is correct.