2.5 mL of $\frac{2}{5}$ M weak monoacidic base ($K_{b} = 1 imes 10^{- 12}\ at\ 25℃$) is titrated with $\frac{2}{15}$ M $HCl$ in water at $25℃$. The concentration of $H^{+}$ at equivalence point is ($K_{w} = 1 imes 10^{- 14}\ at\ 25℃$)

Question

2.5 mL of $\frac{2}{5}$ M weak monoacidic base ($K_{b} = 1 	imes 10^{- 12}\ at\ 25℃$) is titrated with $\frac{2}{15}$ M $HCl$ in water at $25℃$. The concentration of $H^{+}$ at equivalence point is ($K_{w} = 1 	imes 10^{- 14}\ at\ 25℃$)

Dr. Abubacker · Accepted Answer

(d)
Weak monoacidic base e.g., $BOH$ is neutralised as follows
$BOH + HCl ightarrow BCl + H_{2}O$
At equivalence point all $BOH$ gets converted into salt and remember! The concentration of $H^{+}$ (or pH of solution) is due to hydrolysis of resultant salt ($BCl,\ cationic,\ hydrolysis\ here)$
$B^{+}\ \ \ \ \ \ \ \ \ \  + H_{2}O ightleftharpoons BOH\ \ \ \ \ \ \  + \ \ \ \ H^{+}$
$C(1 - h)$ $Ch$ $Ch$
Volume of $HCl$ used up
$V_{a} = \frac{N_{b}V_{b}}{N_{a}} = \frac{2.5 	imes 2 	imes 15}{2 	imes 5} = 7.5\ mL$
Concentration of salt
$\lbrack BClbrack = \frac{conc.of\ base}{total\ volume} = \frac{2 	imes 2.5}{5(7.5 + 2.5)} = \frac{1}{10} = 0.1$
$K_{h} = \frac{Ch^{2}}{1 - h} = \frac{K_{w}}{K_{b}}$
($h$ should be estimated whether that can be neglected or not)
on calculating $h$=0.27 (significant, not negligible)
$\left\lbrack H^{+} ightbrack = Ch = 0.1 	imes 0.27 = 2.7 	imes 10^{- 2}M$

Q. 2.5 mL of $\frac{2}{5}$ M weak monoacidic base ($K_{b} = 1 \times 10^{- 12}\ at\ 25℃$) is titrated with $\frac{2}{15}$ M $HCl$ in water at $25℃$. The concentration of $H^{+}$ at equivalence point is ($K_{w} = 1 \times 10^{- 14}\ at\ 25℃$)

Explanation

Reviewed by Dr. Abubacker

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