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, $B^{+} A^{-} +H_{2}O \rightleftharpoons \underset{\text{Weak acid}}{{HA}}+\underset{\text{Weak base }}{{BOH}}$
$K_{h}=\frac{[HA] [BOH]} {[B^{+}] [A^{-}]}$
Now, let consider the dissociation of $HA$ and $BOH$
$HA \rightleftharpoons H^{+}+A^{-}$
$BOH \rightleftharpoons B^{+}+OH^{-}$
$ K_{a}=\frac{\left[H^{+}\right]\left[A-\right]}{\left[HA\right]}$
$ K_{b}=\frac{\left[B^{+}\right]\left[OH^{-}\right]}{\left[HA\right]\left[BOH^{-}\right]}$
Also, we know that $K_{w}=[H^{+}] [OH^{-}]$
Now, put the value of $K_{a}$ and $K_{b}$ in above equation
$ K_{h}=\frac{K_{w}}{K_{a}\times K_{b}}$
$=\frac{\left[H^{+}\right]\left[OH^{-}\right]\left[HA\right]\left[BOH\right]}{\left[H^{+}\right]\left[A^{-}\right]\left[B^{+}\right]\left[OH^{-}\right]}$
Then $K_{h}=\frac{K_{w}}{K_{a}\times K_{b}}$