Question

When 0.1 mole solid NaOH is added in 1 lt of 0.1 M $N{H}_3\left(aq\right)$ then which statement is wrong? $(K_b=2\times {10}^{-5},\log 2=0.3)$

• 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 changes.

Answer 1

The correct option is C In solution, $\left[N{a}^{+}\right]=0.1M,\left[N{H}_3\right]=0.1M,\left[O{H}^{-}\right]=0.2M.$

Ionisation of $N{H}_3$ follow as
$N{H}_3\stackrel{H_{2}O}{\to }N{H}_{4}OH\to N{H}_4^{+}+O{H}^{-}$
Initially 1
After dissociation $1-C\alpha C\alpha C\alpha$
$K_b=\frac{C^2{\alpha }^2}{C-C\alpha }$
{$N{H}_3$ weak base, hence dissociation very less. So, we can write $C-C\alpha \equiv C$}
$\therefore \alpha =\sqrt{\frac{K_b}{C}}=\sqrt{\frac{2\times {10}^{-5}}{0.1}}=\sqrt{2}\times {10}^{-2}$
(1) $\because \alpha$ is very less $(\sqrt{2}\times {10}^{-2})$ for $N{H}_3$. Hence, degree of dissociation for $N{H}_3$ is nearly zero
(2) Initial pH of weak base $N{H}_3$
$\{p{K}_b=-\log K_b=-\log (2\times {10}^{-5})=4.7\}pOH=\frac{1}{2}[P{K}_b-\log C]\Rightarrow pOH=\frac{1}{2}[4.7-\log (0.1\left)\right]\Rightarrow pOH=2.85\Rightarrow pH=14-2.85=11.15$
If we add NaOH in $N{H}_3$ solution, NaOH is a strong base so dissociation 100 and if we compare to NaOH with $N{H}_3^{\prime }$ the dissociation of $N{H}_3$ is very less. So, we leave it.
$pH$ of 0.1 M NaOH solution
$\left[O{H}^{-}\right]={10}^{-1}M\Rightarrow pOH=-\log \left[O{H}^{-}\right]=1\Rightarrow pH=14-1=13$
Change in $pH=13-11.15=1.85$
(3) In solution
$$\begin{array}{ccc}NaOH\to & N{a}^{+}& +O{H}^{-}\\ 0.1M& 0.1M& 0.1M\\ N{H}_{4}OH\to & N{H}_4^{+}& +O{H}^{-}\end{array}$$ 1- C$\alpha$ C$\alpha$ C$\alpha$ $=0.1\times \sqrt{2}\times {10}^{-2}$
$\Rightarrow \left[O{H}^{-}\right]=Ca=\sqrt{2}\times {10}^{-3}$
Hence, $\Rightarrow \left[N{a}^{+}\right]=0.1M,\left[N{H}_3\right]=0.1-\sqrt{2}\times {10}^{-3}\equiv 0.1M\Rightarrow \left[O{H}^{-}\right]=0.1+\sqrt{2}\times {10}^{-3}\equiv 0.1M$
Hence, concentration of $\left[O{H}^{-}\right]=0.2M$ not possible
(4) Addition of $NaOH{K}_b$ of $N{H}_3$ doesn't change.