MCQ
What will be the value of $\ce{pH}$ of $0.01 \text{ mol}\ \mathrm{ dm}^{-3} \mathrm{CH}_3 \mathrm{COOH}\left(\mathrm{Ka}=1.74 \times 10^{-5}\right) \ ?$
- ✓$3.4$
- B$3.6$
- C$3.9$
- D$3.0$
✓
Answer
Correct option: A.
$3.4$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\text{COOH}+\text{H}_2\text{O}\rightleftharpoons\text{H}_3\text{O}^++\text{CH}_3\text{COOH}^-$
$^\text{Initial conc.} \ \ \ \ \ \ \ \ \ \ \ \ 0.01 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0$
$^\text{At equilibrium} \ \ \ \ \ \ \ \ \ 0.01-\text{x} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{x} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{x}$
$\text{K}_\text{a}=\frac{[\text{H}_3\text{O}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH]}}=\frac{\text{x}^2}{0.01-\text{x}}$
Since $\text{x} << 0.01,$
Therefore, $0.01-\text{x}\approx0.01$
$\frac{\text{x}^2}{0.01}=1.74\times10^{-5}$
$\text{x}^2=1.74\times10^{-7}\text{ or x}=4.2\times10^{-4}$
$\text{pH}=-\log(4.2\times10^{-4})=3.4$
$^\text{Initial conc.} \ \ \ \ \ \ \ \ \ \ \ \ 0.01 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0$
$^\text{At equilibrium} \ \ \ \ \ \ \ \ \ 0.01-\text{x} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{x} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{x}$
$\text{K}_\text{a}=\frac{[\text{H}_3\text{O}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH]}}=\frac{\text{x}^2}{0.01-\text{x}}$
Since $\text{x} << 0.01,$
Therefore, $0.01-\text{x}\approx0.01$
$\frac{\text{x}^2}{0.01}=1.74\times10^{-5}$
$\text{x}^2=1.74\times10^{-7}\text{ or x}=4.2\times10^{-4}$
$\text{pH}=-\log(4.2\times10^{-4})=3.4$
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