Question

$p{K}_b$ of $N{H}_3$ is $4.74$ and $p{K}_b$ of $A^{-},B^{-}$ and $C^{-}$ are $4,5$ and $6$ respectively. Aqueous solution of $0.01M$ has $pH$ in the increasing order.

• A. $N{H}_{4}A<N{H}_{4}B<N{H}_{4}C$
• B. $N{H}_{4}C<N{H}_{4}B<N{H}_{4}A$
• C. $N{H}_{4}C<N{H}_{4}A<N{H}_{4}B$
• D. All have equal $pH$

Answer 1

Answer: (B)

$N{H}_{4}C<N{H}_{4}B<N{H}_{4}A$

$\because SaltsN{H}_{4}A,N{H}_{4}B,N{H}_{4}C$ are of weak acid and weak base.

$\therefore pH=7+\frac{p{K}_a}{2}-\frac{p{K}_b}{2}$

Thus, greater the value of $p{K}_a$ of $HA$, greater the $pH$.

$p{K}_b(A^{-},B^{-},C^{-})|\underset{4}{A^{-}}<\underset{5}{B^{-}}<\underset{6}{C^{-}}$

$\therefore p{K}_a(HA,HB,HC)|\underset{8}{HC}<\underset{9}{HB}<\underset{10}{HA}$

Thus, order of $pH$ values is

$(N{H}_{4}C<N{H}_{4}B<N{H}_{4}A)$