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  4. The pH of 0.10 N acetic acid having Ka=1.8 × 10-5 is

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Q. The $pH$ of $0.10 \,N$ acetic acid having $K_{a}=1.8 \times 10^{-5}$ is

Equilibrium

Solution:

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$\therefore \left[ H ^{+}\right]=C \alpha$
or $C \cdot \sqrt{\frac{K_{a}}{C}}=\sqrt{K_{a} C} $
$\left[\because \alpha=\sqrt{\frac{K_{a}}{c}}\right]$
$\Rightarrow\sqrt{1.8 \times 10^{-5} \times 0.1}$
$=1.34 \times 10^{-3} M$
$\Rightarrow pH =-\log \left[ H ^{+}\right]$
$=-\log \left(1.34 \times 10^{-3}\right)=2.9$