Find the pH of a $0.002 N$ acetic acid solution if it is $2.5$ % ionised at a given dilution. (take log 3 = 0.5, log 5 = 0.7)

4.4

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4.3

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5.0

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Solution

The correct option is C $4.3$
Degree of dissociation, $α = \frac{2.5}{100} = 0.025$
concentration of acetic acid, $N = n_{f} \times M$ $\to n_{f} = 1$
$C = 0.002 M$
The equilibrium is,
${C H}_{3} C O O H ⇌ {C H}_{3} C O O^{-} + H^{+}$
$C (1 - α)$ $C α$ $C α$
So, $[H^{+}] = C α = 0.002 \times 0.025 = 5.0 \times 10^{- 5} M$
$p H = - log [H^{+}] = 4.3$

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