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Ionic Equilibrum question

2007 · Q101
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Ionic Equilibrum question

2007 · Q101

NEETChemistryIonic EquilibrumMCQ+4 / −1
A weak acid, HA, has a Ka of 1.00 ×\times× 10−-−5. If 0.100 mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closest to
  1. A
    1.00%
  2. B
    99.9%
  3. C
    0.100%
  4. D
    99.0%
View written solutionFree

Correct answer: A

For weak acid degree of dissociation,

α\alpha α = KaC\sqrt {{{{K_a}} \over C}} CKa​​​

= 1×10−50.1=10−2\sqrt {{{1 \times {{10}^{ - 5}}} \over {0.1}}} = {10^{ - 2}}0.11×10−5​​=10−2 = 1.00 %

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