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  4. When 0.2 M solution of acetic acid is neutralized with 0.2 M NaOH in 500 mL of water, the pH of the resulting solution will be: [ p Ka. of acetic acid .=4.74]

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Q. When $0.2 \,M$ solution of acetic acid is neutralized with $0.2\, M\, NaOH$ in $500\, mL$ of water, the pH of the resulting solution will be: $\left[ p K_a\right.$ of acetic acid $\left.=4.74\right]$

Equilibrium

Solution:

$CH _3 COOH + NaOH \rightarrow CH _3 COONa + H _2 O$
$\Rightarrow$ At equivalence point, a salt of weak acid, strong base is formed
$\Rightarrow pH =7+\frac{1}{2}\left( p K_a+\log C\right)$
Here $C=$ concentration of salt $=\frac{0.2 \times 500}{500+500}=0.1 M$
[Check that $500\, mL$ of $CH _3 COOH$ is also required]
$\Rightarrow pH =7+\frac{1}{2}(4.74+\log 0.1)=8.87$