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Highly selective Backlog Qs for JEE MainChemistryIonic Equilibrium

2.5 ml of 0.4( 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

  1. A3.7 10⁻¹³ (M)
  2. B3.2 10⁻⁷( M )
  3. C3.2 10⁻²( M )
  4. D2.7 10⁻²( M )

Correct answer

C. 3.2 10⁻²( M )

Step-by-step solution

N ₁ ~V ₁= N ₂ ~V ₂ (At equivalence point) 0.4 2.5= 2 15 V ₂ BOH ( aq )+ HCl ( aq ) B Cl ⁻( aq )+ H ₂ O (I) V ₂= 0.4 2.5 2 15 = 0.4 2.5 2 15=7.5 ml Concentration of salt = 0.4 2.5 2.5+7.5 or, Salt of weak base and strong acid = 0.1 [ H ⁺ ]= Ch = K _ w C K _ b = 10⁻¹⁴ 0.1 10⁻¹² = 10⁻³ =3.2 10⁻² M as C h²= K_ w K_ b h= K_ w K_ b C

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