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Q. The $pH$ of a $0.01 \,M\, HCN$ solution for which $pK_a$ is $4$ is

BITSATBITSAT 2011

Solution:

Given, $pK _{ a }=4$
$\therefore K _{ a }=1 \times 10^{-4}$
$\left[ H ^{+}\right]=\sqrt{ K _{ a } \cdot C }$
$=\sqrt{1 \times 10^{-4} \times 0.01}$
$=\sqrt{10^{-6}}=10^{-3} M$
$pH =-\log \left[ H ^{+}\right]$
$=-\log \left(10^{-3}\right)=3$