Question

$2.5mL$ of $\frac{2}{5}M$ weak monoacidic base $(K_b=1\times {10}^{-12}$ at ${25}^{\circ }C)$ is titrated with $\frac{2}{15}MHCl$ in water at ${25}^{\circ }C$. The concentration of $H^{+}$ at equivalence point is:
[Note : use $K_w=1\times {10}^{-14}$ at ${25}^{\circ }C$].

• A. $3.7\times {10}^{-14}M$
• B. $3.2\times {10}^{-7}M$
• C. $3.2\times {10}^{-2}M$
• D. $2.7\times {10}^{-2}M$

Answer 1

Answer: (D)

$2.7\times {10}^{-2}M$

Given : $V_1=2.5ml$ ; $M_1=0.4M$ ; $K_b=1\times {10}^{-12}$

$\left[HCl\right]=\frac{2}{15}M$

To find the concentration of $H^{+}$

Moles of $BOH$ = 2.5 x $\frac{2}{5}$ x 0.001 = 0.001 moles

Moles of $HCl$ required is = 0.001 mol

V x $\frac{2}{15}$ x 0.001 = 0.001

Volume of $HCl$ required = V = 7.5 ml

Total volume = 7.5 +2.5 = 10 ml

$\left[BCl\right]=\frac{0.001}{10}$ x 1000 = 0.1 M

$\frac{C{h}^2}{1-h}=\frac{K_w}{K_b}\Rightarrow 0.1\times \frac{h^2}{1-h}=0.01\Rightarrow h=0.27$

$\left[H^{+}\right]$ = Ch = 0.027 M