$0.01 \mathrm{M}$ acetic acid solution is $1 \%$ ionised, then $\mathrm{pH}$ of this acetic acid solution is :
$0.01 \mathrm{M}$ acetic acid solution is $1 \%$ ionised, then $\mathrm{pH}$ of this acetic acid solution is :
- 3
- 2
- 4
- 1
Solution
Ionisation of $\mathrm{CH}_3 \mathrm{COOH}$ is $1 \%$ in aqueous solution.
So, $\left[\mathrm{H}^{+}\right]=\frac{1}{100} \times$ concentration of $\mathrm{CH}_3 \mathrm{COOH}$
$\begin{aligned}
& =\frac{1}{100} \times 10^{-2}=10^{-4} \\
\mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right]=-\log \left(1 \times 10^{-4}\right) \\
\mathrm{pH} & =4
\end{aligned}$
Asked in: MHT CET Full Test 11
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