JEE Advanced2008ChemistryIonic EquilibriumActual
2.5 ~mL of 2 5 M weak monoacidic base (K_b=1 10⁻¹² . at .25^ C ) is titrated with 2 15 M HCl in water at 25^ C . The concentration of H ⁺ at equivalence point is (K_w=1 10⁻¹⁴ . at .25^ C )
Options
- A3.7 10⁻¹³ M
- B3.2 10⁻⁷ M
- C3.2 10⁻² M
- D2.7 10⁻² M
Correct answer
D. 2.7 10⁻² M
Step-by-step solution
Weak monoacidic base, e.g. BOH is neutralised. BOH + HCl BCl + H ₂ O At equivalence point all BOH gets converted into salt and remember! the concentration of H ⁺ (or pH of solution) is due to hydrolysis of resultant salt ( BCl , cationic hydrolysis here) C (1-h) B ⁺ + H ₂ O Ch OH + Ch H ⁺ Volume of HCl used up, V_a= N_b V_b N_a = 2.5 2 15 2 5 =7.5 ~mL Concentration of salt, aligned [ BCl ] & = Concentration of base Total volume = 2 25 5(7.5+2.5) = 1 10 =0.1 K_h & = C h^2 1-h = K_w K_b aligned ( h should be estimate