Question

In the hydrolysis of a salt of weak acid and weak base, the hydrolysis constant $K_h$ is equal to

• A. $\frac{K_w}{K_b}$
• B. $\frac{K_w}{K_a}$
• C. $\frac{K_w}{K_a\cdot K_b}$
• D. $K_a\cdot K_b$

Answer 1

Answer: (C)

According to hydrolysis of salts of a weak acid and a weak base,
Since, the hydrolysis of such salts will form weak acid and weak base, the solution will be neutral and its $pH$ wiill be $7$
Then, $B^{+}{A}^{-}+H_{2}O\rightleftharpoons \underset{\text{Weak acid}}{\mathrm{HA}}+\underset{\text{Weak base }}{\mathrm{BOH}}$
$K_h=\frac{[HA][BOH]}{[B^{+}][A^{-}]}$
Now, let consider the dissociation of $HA$ and $BOH$
$HA\rightleftharpoons H^{+}+A^{-}$
$BOH\rightleftharpoons B^{+}+O{H}^{-}$
$K_a=\frac{\left[H^{+}\right]\left[A-\right]}{\left[HA\right]}$
$K_b=\frac{\left[B^{+}\right]\left[O{H}^{-}\right]}{\left[HA\right]\left[BO{H}^{-}\right]}$
Also, we know that $K_w=[H^{+}][O{H}^{-}]$
Now, put the value of $K_a$ and $K_b$ in above equation
$K_h=\frac{K_w}{K_a\times K_b}$
$=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]\left[HA\right]\left[BOH\right]}{\left[H^{+}\right]\left[A^{-}\right]\left[B^{+}\right]\left[O{H}^{-}\right]}$
Then $K_h=\frac{K_w}{K_a\times K_b}$