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

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$3.7 \times 10^{- 14} M$

$3.2 \times 10^{- 7} M$

$3.2 \times 10^{- 2} M$

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

answer is D.

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Detailed Solution

$2 .5 \times \frac{2}{5} = V \times \frac{2}{15}; \to V = 7 .5 mL$

[Salt] $= \frac{2.5 \times \frac{2}{5}}{2.5 + 7.5} = 0.01$

$\underset{0.1 - x}{B^{+}} + H_{2} O \underset{}{⇌} \underset{x}{BOH} + \underset{x}{H^{+}}$

$\begin{array}{l} \frac{x^{2}}{0.1 - x} = \frac{10^{- 14}}{10^{- 2}} \\ = 2.7 \times 10^{- 2} M \end{array}$

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