Find the pH of a $0.002 N$ acetic acid solution if it is $2.3$ % ionised at a given dilution.

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Solution

Degree of dissociation, $α = \frac{2.3}{100} = 0.023$
concentration of acetic acid, $C = 0.002 M$

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, $[H^{+}] = C α = 0.002 \times 0.023 = 4.6 \times 10^{- 5} M$

$p H = - log [H^{+}] = 4.3372$

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Find the pH of a 0.002N acetic acid solution if it is 2.3% ionised at a given dilution.

Degree of dissociation, α=2.3100=0.023concentration of acetic acid, C=0.002MThe equilibrium is,CH3COOH⇌CH3COO−+H+C(1−α) Cα CαSo, [H+]=Cα=0.002×0.023=4.6×10−5MpH=−log[H+]=4.3372

Find the pH of a $0.002 N$ acetic acid solution if it is $2.3$ % ionised at a given dilution.

Degree of dissociation, $α = \frac{2.3}{100} = 0.023$
concentration of acetic acid, $C = 0.002 M$

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, $[H^{+}] = C α = 0.002 \times 0.023 = 4.6 \times 10^{- 5} M$

$p H = - log [H^{+}] = 4.3372$