Question

2.5 mL of $\frac{2}{5}$M weak monoacidic base $(K_b=1\times {10}^{-12}$ at ${25}^{o}C)$ is titrated with $\frac{2}{15}$ M $HCl$ in water at ${25}^{o}C$. The concentration of $H^{+}$ at equivalence point is:

$[K_w=1\times {10}^{-14}$ at ${25}^{o}C]$

• A. $37\times {10}^{-13}M$
• B. $3.2\times {10}^{-7}M$
• C. $3.2\times {10}^{-2}M$
• D. $2.7\times {10}^{-2}M$

Answer 1

The correct option is D $2.7\times {10}^{-2}M$

Weak monoacidic base, e.g., $BOH$ is neutralized as follows

$BOH+HCl\to BCl+H_{2}O$

At the equivalence point, all $BOH$ get converted into the salt. The concentration of $H^{+}$ (or $pH$ of solution) is due to hydrolysis of the resultant salt ($BCl$, cationic hydrolysis here.)
$\underset{C(1-h)}{B^{+}}+H_{2}O\rightleftharpoons \underset{Ch}{BOH}+\underset{Ch}{H^{+}}$
Volume of $HCl$ used up,

$V_a=\frac{N_b{V}_b}{N_a}$
$V_a=\frac{2.5\times 2\times 15}{2\times 5}$
$V_a=7.5mL$
Concentration of salt,

$\left[BCl\right]=\frac{conc.ofbase}{totalvolume}=\frac{2\times 2.5}{5(7.5+2.5)}=\frac{1}{10}=0.1$
$K_h=\frac{C{h}^2}{1-h}=\frac{K_w}{K_b}=\frac{{10}^{-14}}{{10}^{-12}}={10}^{-2}=\frac{0.1\times h^2}{1-h}$ ---- 1
(h should be estimated whether that can be neglected or not)

On calculating, $h=0.27$ (significant, not negligible)

$\left[H^{+}\right]=Ch=0.1\times 0.27=2.7\times {10}^{-2}M$

Hence, the correct option is $D$