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  4. If Ka of HCN =4 × 10-10, then the pH of 2.5 × 10-1 molar HCN (a q) is

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Q. If $K_{a}$ of $HCN =4 \times 10^{-10}$, then the $pH$ of $2.5 \times 10^{-1}$ molar $HCN (a q)$ is

ManipalManipal 2012Equilibrium

Solution:

$\left[ H _{3} O ^{+}\right] =\sqrt{K_{a} \times C}$
$=\sqrt{4 \times 10^{-10} \times 2.5 \times 10^{-1}}$
$=\sqrt{10^{-10}}=10^{-5} M$
$\because pH = -\log \left[ H _{3} O ^{+}\right]$
$\therefore pH =-\log \left(10^{-5}\right)=5$