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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Ionic Equilibrium

NTA Abhyas 2020 · Chemistry · Ionic Equilibrium

The degree of dissociation of a weak monoprotic acid of...

The degree of dissociation of a weak monoprotic acid of concentration 1.2×10–3M1.2 \times {10}^{– 3} \text{M} having Ka=1.0×10–4{K}_{a} = 1.0 \times {10}^{– 4} is

Options

  1. A1
  2. B10
  3. C15
  4. D25

Correct answer

D. 25

Step-by-step solution

By α=KaC=10−41.2×10−3=112\alpha = \sqrt{\dfrac{{K}_{a}}{C}} = \sqrt{\dfrac{1{0}^{- 4}}{1.2 \times 1{0}^{- 3}}} = \sqrt{\dfrac{1}{12}}
%\alpha = 28.86% , as value of α\alpha is greater than 5% so we can not take assumptions and will have to use the exact formula.
So α\alpha will be given by:
α=−10−4+10−8+4×10−3×1.2×10−42×1.2×10−3\alpha = \dfrac{- 1{0}^{- 4} + \sqrt{1{0}^{- 8} + 4 \times 1{0}^{- 3} \times 1.2 \times 1{0}^{- 4}}}{2 \times 1.2 \times 1{0}^{- 3}}
So the correct value of α=25\alpha = 25%
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