Question

Match the Column I(Salt) with Column II(Degree of hydrolysis)

Answer 1

A) ${NH}_{4}Cl\left(s\right)+H_{2}O\left(l\right)\rightleftharpoons {NH}_{4}OH\left(aq\right)+{Cl}^{-}\left(aq\right)$

We may write :

${NH}_4^{+}+H_{2}O\rightleftharpoons BOH+H^{+}$

$C(1-h)$ $Ch$ $Ch$

$\therefore$ $K_h=\frac{Ch.Ch}{C(1-h)}=\frac{C^2{h}^2}{C(1-h)}\approx h^{2}C$

$\therefore$ $h=\sqrt{\frac{K_h}{C}}$

now, $K_h=\frac{C_{BOH}.C_{{OH}^{-}}.C_{H^{+}}}{C_{B^{+}}.C_{{OH}^{-}}}=\frac{K_w}{K_b}$

$\therefore$ $h=\sqrt{\frac{K_w}{K_b.C}}$

So, $A\to 2,3$.

B) $NaCl$ formed from strong acid $HCl$ and strong base $NaOH$. No hydrolysis will take place.

$\therefore B\to 1$.

C) ${CH}_{3}COONa$

${CH}_{3}COONa+H_{2}O\rightleftharpoons {CH}_{3}COOH\left(aq\right)+{Na}^{+}\left(aq\right)+{OH}^{-}\left(aq\right)$

We may write : ${CH}_3{COO}^{-}+H_{2}O\rightleftharpoons {CH}_{3}COOH+{OH}^{-}$

$C(1-h)$ $Ch$ $Ch$

$K_h=\frac{Ch.Ch}{C(1-h)}\approx h^{2}C$

$\therefore$ $h=\sqrt{\frac{K_h}{C}}$ $\therefore$ $C\to 2$

D) ${CH}_3{COONH}_4$

${CH}_3{COO}^{-}+{NH}_4^{+}+H_{2}O\rightleftharpoons {CH}_{3}COOH+{NH}_{4}OH$

$C(1-h)$ $C(1-h)$ $Ch$ $Ch$

$\therefore$ $K_h=\frac{Ch.Ch}{C(1-h).C(1-h)}={\left(\frac{h}{1-h}\right)}^2\approx h^2$ [Since $1-h\approx 1$]

$h=\sqrt{K_h}$

So, $D\to 4$