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JEE Advance - Chemistry (2008 - Paper 1 Offline - No. 7)

2.5 mL of $$\frac{2}{5}$$M weak monoacidic base (K$$_b$$ = 1 $$\times$$ 10$$^{-12}$$ at 25$$^\circ$$C) is titrated with $$\frac{2}{15}$$M HCl in water at 25$$^\circ$$C. The concentration of H$$^+$$ at equivalence point is (K$$_w$$ = 1 $$\times$$ 10$$^{-14}$$ at 25$$^\circ$$C).
3.7 $$\times$$ 10$$^{-13}$$ M
3.2 $$\times$$ 10$$^{-7}$$ M
3.2 $$\times$$ 10$$^{-2}$$ M
2.7 $$\times$$ 10$$^{-2}$$ M

Explanation

Weak monoacidic base, e.g., BOH is neutralised as follows:

BOH + HCl $$\to$$ BCl + H$$_2$$O

At the equivalence point, all BOH get converted into the salt. The concentration of H$$^+$$ (or pH of solution) is due to hydrolysis of the resultant salt (BCl, cationic hydrolysis here.)

IIT-JEE 2008 Paper 1 Offline Chemistry - Ionic Equilibrium Question 2 English Explanation

Volume of HCl used up,

$${V_a} = {{{N_b}{V_b}} \over {{N_a}}}$$

$${V_a} = {{2.5 \times 2 \times 15} \over {2 \times 5}}$$

$${V_a} = 7.5$$

Concentration of salt,

$$[BCl] = {{conc.\,of\,base} \over {total\,volume}}$$

$$ = {{2 \times 2.5} \over {5(7.5 + 2.5)}} = {1 \over {10}} = 0.1$$

$${K_h} = {{C{h^2}} \over {1 - h}} = {{{K_w}} \over {{K_b}}} = {{{{10}^{ - 14}}} \over {{{10}^{ - 12}}}}$$

$$ = {10^{ - 2}} = {{0.1 \times {h^2}} \over {1 - h}}$$ ..... (i)

(h should be estimated whether that can be neglected or not.)

On calculating, $$h = 0.27$$ (significant, not negligible)

$$[{H^ + }] = Ch = 0.1 \times 0.27 = 2.7 \times {10^{ - 2}}$$ M

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