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

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

Equilibrium

Solution:

$[H_{3}O^{+}]=\sqrt{K_{a} \times C}$
$=(4\times10^{-10}\times2.5\times10^{-1})^{1/2}$
$=(10^{-10})^{1/2}=10^{-5}\,M$
$\therefore PH=5$