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

(D) $2.7 \times 10^{-2} \mathrm{M}$

(A) $3.7 \times 10^{-13} \mathrm{M}$

(B) $3.2 \times 10^{-7} \mathrm{M}$

(C) $3.2 \times 10^{-2} \mathrm{M}$

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Equilibrium

2.5 mL of

\frac{2}{5} M

weak monoacidic base

(K_{b} = 1 \times 1 0^{- 12}

2 5^{\circ} C)

is titrated with

\frac{2}{15} M HCl

in water at

2 5^{\circ} C

. The concentration of

H^{+}

at equivalence point is

(K_{w} = 1 \times 1 0^{- 14}

2 5^{\circ} C)

Answer: (D)

At equivalence point, the weak base is completely converted to its conjugate acid (salt).

Volume of HCl used:

V = \frac{2.5 \times \frac{2}{5}}{\frac{2}{15}} = 7.5

mL

Total volume = 2.5 + 7.5 = 10 mL

Concentration of salt:

C = \frac{2.5 \times \frac{2}{5}}{10} = 0.1

M

The salt undergoes hydrolysis:

B^{+} + H_{2} O ⇌ BOH + H^{+}

For hydrolysis:

K_{h} = \frac{K _{w}}{K _{b}} = \frac{1 0 ^{- 14}}{1 0 ^{- 12}} = 1 0^{- 2}

Using

\frac{C h ^{2}}{1 - h} = K_{h}

and solving:

h \approx 0.27

Therefore:

[H^{+}] = C h = 0.1 \times 0.27 = 2.7 \times 1 0^{- 2}

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option