Question

$2.5\, ml\, 0.4(M)$ weak monoacidic base $\left(k_{b}=1 \times 10^{-12}\right.$. at $25^{\circ} C$ ) is titrated with $\frac{2}{15}(M) H C l$ in water at $25^{\circ} C$. The concentration of $H^{+}$at equivalence point is
$\left(k_{w}=1 \times 10^{-14}\right.$, at $\left.25^{\circ} C \right)$

• A. $3.7 \times 10^{-13}(M)$
• B. $3.2 \times 10^{-7}(M)$
• C. $3.2 \times 10^{-2}(M)$
• D. $2.7 \times 10^{-2}(M)$

Answer 1

Answer: (C)

$N_{1} V_{1}=N_{2} V_{2}($ At equivalence point $)$
$0.4 \times 2.5=\frac{2}{15} \times V_{2}$
$B O H(a q)+H C l(a q) \rightarrow B C l^{-}(a q)+H_{2} O(I)$
$\therefore V_{2}=\frac{0.4 \times 2.5}{\frac{2}{15}}$
$=\frac{0.4 \times 2.5}{2} \times 15=7.5 \,ml$
$\therefore $ Concentration of salt $=\frac{0.4 \times 2.5}{2.5+7.5}$
or, Salt of weak base and strong acid $=0.1$
$\therefore \left[H^{+}\right]=C h=\sqrt{\frac{K_{w} C}{K_{b}}}$
$=\sqrt{\frac{10^{-14} \times 0.1}{10^{-12}}}=\sqrt{10^{-3}}$
$=3.2 \times 10^{-2} M$
as $C h^{2}=\frac{K_{w}}{K_{b}}$
$\therefore h=\sqrt{\frac{K_{w}}{K_{b} C}}$