Calculate the percentage ionization of $0.01 M$ acetic acid in $0.1 M H C l . K_{a}$ of acetic acid is $1.8 \times 10^{- 5}$ .

0.18 %

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0.018 %

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1.8 %

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18 %

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Solution

The correct option is B $0.018 %$
$C H_{3} C O O H 0.01 (1 - x) ⇌ C H_{3} C O O^{-} 0.01 x + H^{+} 0.1$

$K_{a} = \frac{[C H_{3} C O O^{-}] \times [H^{+}]}{[C H_{3} C O O H]} = \frac{(0.01 x) (0.1)}{0.01 (1 - x)} = 1.8 \times 10^{- 5}$

Here only the $H^{+}$ ion from $H C l$ acid is significant and $(1 - x) \approx 1$

$\Rightarrow 0.1 x = 1.8 \times 10^{- 5} (1 - x)$

$\Rightarrow x \approx 1.8 \times 10^{- 4}$

% $x$ or % ionization $= 0.018$ %

Hence, the correct option is $B$

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