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  4. pH of 0.0002 M formic acid [Ka=2× 10-4] approximately is

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Q. $pH$ of $0.0002 \,M$ formic acid $ [K_{a}=2\times 10^{-4}] $ approximately is

J & K CETJ & K CET 2012Equilibrium

Solution:

$\left[H^{+}\right]_{H C O O H}=\sqrt{K_{a} . C}$
$=\sqrt{2 \times 10^{-4} \times 0.0002}$
$p H=-\log \left[H^{+}\right]=-\log \left(2 \times 10^{-4}\right)$
$=3.69 \approx 3.7$