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6.2. Equilibrium - II (icon Equilibrium) — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry6.2. Equilibrium - II (icon Equilibrium)1 Mark
MCQ
A monoprotic acid in $ 1.00 \,M $ solution is $ 0.01\%$  ionised. The dissociation constant of this acid is
  • ✓
    $1 \times {10^{ - 8}}$
  • B
    $1 \times {10^{ - 4}}$
  • C
    $1 \times {10^{ - 6}}$
  • D
    ${10^{ - 5}}$
✓

Answer

Correct option: A.
$1 \times {10^{ - 8}}$
(a)$K = \frac{{{\alpha ^2}C}}{{1 - \alpha }};\,\,\alpha = \frac{{0.01}}{{100}} \approx 1\,\,\,\therefore \,\,K = {\alpha ^2}C = {\left[ {\frac{{0.01}}{{100}}} \right]^2} \times 1$
$ = 1 \times {10^{ - 8}}$.

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