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MHT CET 202625 AprilMorning Shift
Exam
MHT CET
Year
2026
Date
25 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

MHT CET 2026 · 25 April · Morning Shift · Chemistry · Ionic Equilibrium

Calculate the dissociation constant of a weak monobasic acid if it...

Calculate the dissociation constant of a weak monobasic acid if it is 0.04% dissociated in a 0.01M solution.

Options

  1. A2.4×10−92.4 \times 10^{-9}
  2. B1.2×10−91.2 \times 10^{-9}
  3. C2.0×10−92.0 \times 10^{-9}
  4. D1.6×10−91.6 \times 10^{-9}

Correct answer

D. 1.6×10−91.6 \times 10^{-9}

Step-by-step solution

Given concentration C=0.01C = 0.01 M =10−2= 10^{-2} M.

Degree of dissociation α=0.04%=0.04100=4×10−4\alpha = 0.04\% = \dfrac{0.04}{100} = 4 \times 10^{-4}.

For a weak monobasic acid, the dissociation constant is given by Ostwald's dilution law:

Ka=Cα21−αK_a = \dfrac{C\alpha^2}{1 - \alpha}

Since α≪1\alpha \ll 1, we can approximate 1−α≈11 - \alpha \approx 1.

Ka=Cα2K_a = C\alpha^2

Substituting the values:

Ka=(10−2)×(4×10−4)2K_a = (10^{-2}) \times (4 \times 10^{-4})^2

Ka=10−2×16×10−8K_a = 10^{-2} \times 16 \times 10^{-8}

Ka=16×10−10=1.6×10−9K_a = 16 \times 10^{-10} = 1.6 \times 10^{-9}

Answer: 1.6×10−91.6 \times 10^{-9}
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