Question

When $0.1$ mole of solid NaOH is added in $1$L of $0.1$M $N{H}_3$ (aq) then which statement is going to be wrong?
$(K_b=2\times {10}^{-5},log2=0.3)$

• A. degree of dissociation of $N{H}_3$ approaches to zero.
• B. Change in pH would be $1.85$
• C. Concentration of $\left[N{a}^{+}\right]=0.1M,\left[N{H}_3\right],=0.1M,\left[O{H}^{-}\right]=0.2M$
• D. on addition of $O{H}^{-},K_b$ of $N{H}_3$ does not changes

Answer 1

The correct option is A degree of dissociation of $N{H}_3$ approaches to zero.

$K_b=2\times {10}^{-5}\left(N{H}_3\right)$

$N{H}_3\to N{H}_4+O{H}^{-}$

$C{\alpha }^2=Ka$

$\alpha =\sqrt{\frac{Ka}{c}}=\sqrt{\frac{2\times {10}^{-5}}{0.1}}=\sqrt{2}\times {10}^{-3}$

$\left[O{H}^{-}\right]=0.1\times \sqrt{2}\times {10}^{-3}=\sqrt{2}\times {10}^{-4}$

$NaOH\to N{a}^{+}+O{H}^{-}$

$\left[O{H}^{-}\right]=0.1M$

$P^H=14-log0.1$

$=10$

$P^H$ due to $N{H}_3=14-{\log }^2\times {10}^{-4}$

$=14(\frac{1}{2}{\log }^2+\log {10}^{-4})$

$=14-(-4+0.15)=10.15$

$\left(A\right)$ degree of dissociation approaches to zero

$\alpha \to 0$

$\left(B\right)$ change in $P^H$ be $13-10.15=2.85$

$\left(C\right)\left(N{a}^{+}\right)=0.1M$ $N{H}_3\simeq 0.1M$

$\left[O{H}^{-}\right]=0.1+\sqrt{2}\times {10}^{-4}\simeq 0.1$