MCQ
If $p{K_b}$ for fluoride ion at ${25\,^o}C$ is $10.83$, the ionisation constant of hydrofluoric acid in water at this temperature is
- A$1.74 \times {10^{ - 3}}$
- B$3.52 \times 10^{-3}$
- ✓$6.75 \times {10^{ - 4}}$
- D$5.38 \times {10^{ - 2}}$
✓
Answer
Correct option: C.
$6.75 \times {10^{ - 4}}$
(c) ${K_a} \times {K_b} = {K_w}$
$\therefore \,{K_a} = \frac{{{K_w}}}{{{K_b}}}$$ = \frac{{{{10}^{ - 14}}}}{{1.48 \times {{10}^{ - 11}}}}$$ = 6.75 \times {10^{ - 4}}$
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