Question

One mole of ammonia was completely absorbed in one litre solution each of
I. $1MHCl$
II. $1MC{H}_{3}COOH$
III. $1M{H}_{2}S{O}_4$ at $298$ K
The decreasing order for the $pH$ of the resulting solution is:

(Given: $K_b\left(N{H}_3\right)=4.74$)

• A. $II>III>I$
• B. $I>II>III$
• C. $II>I>III$
• D. $III>II>I$

Answer 1

The correct option is C $II>I>III$

$I$. One litre solution of $1M$ $HCl$ completely neutralises one mole of ${NH}_3$ to form $1M$ ${NH}_{4}Cl$.

$pH=\frac{1}{2}{pK}_w-\frac{1}{2}{pK}_b-\frac{1}{2}\times 1$

$=\frac{1}{2}\times 14-\frac{1}{2}\times 4.74-\frac{1}{2}\times 1$

$=7-2.37-0.5=4.13$

$II$. Here also $1M$ of ${CH}_{3}COO{NH}_4$ will be formed.

$pH=\frac{1}{2}{pK}_w-\frac{1}{2}{pK}_a-\frac{1}{2}{pK}_b$

$=\frac{1}{2}\times 14-\frac{1}{2}\times 4.74-\frac{1}{2}\times 4.74=7$

$III$. $1M$ $H_2{SO}_4$ gives $2M$ of $H^{+}$ ions. Now $1M$ of $H^{+}$ neutralises $1$ mole of ${NH}_3$. Final solution contain $1M$ of $H^{+}$ and $1M$ of ${NH}_{4}Cl$. But $1M$ of $H^{+}$ over when $S$ the hydrolysis of ${NH}_{4}Cl$.

$\therefore$ $pH=-log\left[H^{+}\right]=-{log}_{10}^1=0$

$\therefore$ The decreasing order of $II>I>III$

Answer will be C.