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Factors Affecting Alpha

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Question

Find the percentage ionisation of $0.2 M$ acetic acid solution, whose dissociation constant is $1.8 \times 10^{- 6}$ .

0.198

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0.290

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0.950

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n o n e o f t h e s e

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Solution

The correct option is B $0.950$
$C H_{3} C O O H ⇌ C H_{3} C O O^{-} + H^{+}$
$i n i t i a l c o n c . 0.2 00$
$A t e q u i b r m 0.2 (1 - α) 0.2 α 0.2 α$

$K_{a} = \frac{[C H_{3} C O O^{-}] [H^{+}]}{C H_{3} C O O H}$

$1.8 \times 10^{5} = \frac{(0.2 α)^{2}}{0.2 - 0.2 α} α = 0.95 \times 10^{- 2}$
$p e r c e n t a g e I o n i s a t i o n = 0.95 \times 10^{- 2}$
$p e r c e n t a g e I o n i s a t i o n = 0.95 %$

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