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Arrhenius Acid

In 0.1 M solu...

Question

In 0.1 M solution, weak electrolyte, $C H_{3} C O O H$ is 1.3% dissociated. If 100 molecules of $C H_{3} C O O H$ are placed originally in the solution, how many acid molecules remain nonionized at equilibrium?

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Solution

$C H_{3} C O O H \to C H_{3} C O O^{-} + H^{+}$
No. of acid molecules non ionized $= 100 - \frac{100 \times 1.3}{100}$
$= 98.7 ≃ 99$

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