Verified Solution • 2025–26 Session moderate

QJEEAdv08Chc581: JEE Advanced 2008 Chemistry - Class 11 JEEAdv08Chc581 Solution: 2.5 mL of M weak monoacidic base (K = 1 10 at 25 C) is titrated …

2.5 mL of 25\frac{2}{5} M weak monoacidic base (K b_b = 1 ×\times 10 −12^{-12} at 25 ∘^\circ C) is titrated with 215\frac{2}{15} M HCl in water at 25 ∘^\circ C. The concentration of H +^+ at equivalence point is (K w_w = 1 ×\times 10 −14^{-14} at 25 ∘^\circ C).

Multiple Choice Options

  • A 3.7 ×\times 10 −13^{-13} M
  • B 3.2 ×\times 10 −7^{-7} M
  • C 3.2 ×\times 10 −2^{-2} M
  • D 2.7 ×\times 10 −2^{-2} M

Step-by-Step Verified Solution

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

BOH + HCl →\to BCl + H 2_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,

Va=NbVbNa{V_a} = {{{N_b}{V_b}} \over {{N_a}}}

Va=2.5×2×152×5{V_a} = {{2.5 \times 2 \times 15} \over {2 \times 5}}

Va=7.5{V_a} = 7.5

Concentration of salt,

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

=2×2.55(7.5+2.5)=110=0.1= {{2 \times 2.5} \over {5(7.5 + 2.5)}} = {1 \over {10}} = 0.1

Kh=Ch21−h=KwKb=10−1410−12{K_h} = {{C{h^2}} \over {1 - h}} = {{{K_w}} \over {{K_b}}} = {{{{10}^{ - 14}}} \over {{{10}^{ - 12}}}}

=10−2=0.1×h21−h= {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=0.27h = 0.27 (significant, not negligible)

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

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