The hydrogen ion concentration of $0.2 \mathrm{~N}$ $\mathrm{CH}_{3} \mathrm{COOH}$ which is $40 \%$…
The hydrogen ion concentration of $0.2 \mathrm{~N}$ $\mathrm{CH}_{3} \mathrm{COOH}$ which is $40 \%$ dissociated is
- $0.08 \mathrm{~N}$
- $0.12 \mathrm{~N}$
- $0.80 \mathrm{~N}$
- $1.2 \mathrm{~N}$
Solution
$\left[\mathrm{H}^{+}ight]=\mathrm{C} \alpha=0.2 \times 0.40=0.08 \mathrm{~N}$
$$
\left(\alpha=\frac{40}{100}=0.40ight)
$$
,
Asked in: JEE-TOPICTESTS-CHEMISTRY
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