Question

$20mL$ of $0.2MN{H}_{4}OH$ is titrated with $0.2MHCl.$ Find the $pH$ of the solution, when $19mL$ $HCl$ has been added.
Given: $K_b\left(N{H}_{4}OH\right)=1.9\times {10}^{-5}$

• A. 4
• B. 3
• C. 8
• D. 6

Answer 1

The correct option is C 8

mmol of $N{H}_{4}OH=20\times 0.2=4$
mmol of $HCl=19\times 0.2=3.8$

$N{H}_{4}OH\left(aq\right)+HCl\left(aq\right)\to N{H}_{4}Cl\left(aq\right)+H_{2}O\left(l\right)Initial:43.800Final:0.203.83.8$

$\text{Final concentration}=\frac{\text{Moles}}{\text{Total Volume}}$
Total Volume $=20+19=39mL$
$[N{H}_{4}OH{]}_{final}=\frac{0.2}{39}M$
$[N{H}_{4}Cl{]}_{final}=[N{H}_4^{+}{]}_{final}=\frac{3.8}{39}M$
For a buffer solution , the effective range should be : $\frac{1}{10}<\frac{\left[N{H}_{4}Cl\right]}{\left[N{H}_{4}OH\right]}<10$
But here, $\frac{\left[N{H}_{4}Cl\right]}{\left[N{H}_{4}OH\right]}=\frac{3.8/39}{0.2/39}=19>10$
So after titration, $N{H}_{4}Cl$ and $N{H}_{4}OH$ is there but the mixture will not acts as a buffer.

Consider the equilibrium reaction of weak base of $N{H}_{4}OH$ , where $\alpha$ is the degree of dissociation.
$N{H}_{4}OH\left(aq\right)\rightleftharpoons N{H}_4^{+}\left(aq\right)+O{H}^{-}\left(aq\right)Initially:0.23.8Equilibrium:0.2(1-\alpha )3.8+0.2\alpha$

Here , $N{H}_{4}OH$ is a weak base which is trying to dissociate , but in product $N{H}_4^{+}$ concentration is very high which eventually try to react with $O{H}^{-}$ and pulled the overall reaction in backward direction. Therefore degree of hydrolysis $\left(\alpha \right)$ will be very small.
$\therefore 1-\alpha \approx 13.8+0.2\alpha \approx 3.8K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}1.9\times {10}^{-5}=\frac{\left(3.8/39\right)\times \left[O{H}^{-}\right]}{\left(0.2/39\right)}\left[O{H}^{-}\right]=\frac{0.2\times (1.9\times {10}^{-5})}{3.8}\left[O{H}^{-}\right]=1\times {10}^{-6}pOH=-\log (1\times {10}^{-6})pOH=6pH=14-6=8$