Question

When 0.1 mole solid NaOH is added in 1L of 0.1 M $N{H}_3$(aq.) $(Kb=2\times {10}^{-5})$, then select the correct statement(s):

• A. Degree of dissociation of $N{H}_3$ approaches to zero.
• B. Change in pH by adding NaOH would be 1.85.
• C. In solution, $\left[N{a}^{+}\right]=0.1M,\left[N{H}_3\right]=0.1M,\left[O{H}^{-}\right]=0.2M$
• D. On addition of $O{H}^{-},K_b$ of $N{H}_3$ does not change.

Answer 1

Answer: (A), (B), (D)

The correct options are
A

Degree of dissociation of $N{H}_3$ approaches to zero.

B

Change in pH by adding NaOH would be 1.85.

D

On addition of $O{H}^{-},K_b$ of $N{H}_3$ does not change.

Due to common ion effect, degree of dissociation of $N{H}_3$ decreases and starts approaching to zero. pH of 0.1M $N{H}_3$ (aq.):

$\left[O{H}^{-}\right]=\sqrt{K_b.C}=\sqrt{2\times {10}^{-5}\times 0.1}=1.41\times {10}^{-3}pOH=-log\left[O{H}^{-}\right]=2.85$
pH =14-pOH =11.15
pH after adding 0.1 mole solid NaOH:
$\left[O{H}^{-}\right]={10}^{-1}pOH=-log\left[O{H}^{-}\right]=1$
pH = 14 - pOH = 13
Hence change in pH is 13 - 11.15 = 1.85
In solution, $\left[O{H}^{-}\right]=0.1M$
On addition of $O{H}^{-},K_b$ of $N{H}_3$ will not change. Common ion effect does not change the $K_b$ value.