JEE Advance - Chemistry (2001 - No. 4)
Hydrogen peroxide solution (20 ml) reacts quantitatively with a solution of KMnO4 solution is just decolourised by 10 ml of MnSO4 in neutral medium simultaneously forming a dark brown precipitate of hydrated MnO2. The brown precipitated is dissolved in 10 ml of 0.2 M sodium oxalate under boiling condition in the presence of dilute H2SO4. Write the balanced equations involved in the reactions and calculate the molarity of H2O2.
0.05 M
0.1 M
0.15 M
0.2 M
0.25 M
Explanation
Let the molarity of $\mathrm{KMnO}_4$ be $M_1$. Moles of $\mathrm{KMnO}_4$ in $20 \mathrm{ml}$
$$ =\frac{20 \times M_1}{1000} $$
$n$-factor of $\mathrm{KMnO}_4$ when it reacts with $\mathrm{MnSO}_4$ is 3.
$\therefore$ Equivalent of $\mathrm{KMnO}_4$ reacting with $\mathrm{MnSO}_4=\frac{20 \times M_1}{1000} \times 3$
$\therefore$ Equivalent of $\mathrm{MnSO}_4$ reacting with $\mathrm{KMnO}_4=\frac{20 \times M_1}{1000} \times 3$
Since $\mathrm{MnSO}_4$ has $n$-factor 2, the mole of $\mathrm{MnSO}_4$ reacting
$$ =\frac{20 \times M_1}{1000} \times \frac{3}{2} $$
Total mole of $\mathrm{MnO}_2$ produced $=$ mole of $\mathrm{KMnO}_4+$ mole of $$ \mathrm{MnSO}_4 $$
Equivalent of $\mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4$ reacting with $\mathrm{MnO}_2=\frac{10 \times 0.2}{1000} \times 2$
$\therefore$ Equivalent of $\mathrm{MnO}_2=\frac{10 \times 0.2}{1000} \times 2$
Mole of $\mathrm{MnO}_2$ reacting with sodium oxalate $=\frac{10 \times 0.2 \times 2}{1000 \times 2}$
[as $n$-factor for $\mathrm{MnO}_2$ is 2].
Therefore, $\frac{20 \times M_1}{1000}+\left[\frac{20 \times M_1}{1000} \times \frac{3}{2}\right]=\frac{10 \times 0.2}{1000} ; M_1=0.04$
Equivalent of $\mathrm{KMnO}_4$ reacting with $\mathrm{H}_2 \mathrm{O}_2=\frac{20 \times 0.04}{1000} \times 5$ $$ =0.004 $$
If molarity of $\mathrm{H}_2 \mathrm{O}_2$ is $M_2$, then $=\frac{20 \times M_2 \times 2}{1000}=0.004$
$\therefore M_2=0.1 \mathrm{M}$
$$ =\frac{20 \times M_1}{1000} $$
$n$-factor of $\mathrm{KMnO}_4$ when it reacts with $\mathrm{MnSO}_4$ is 3.
$\therefore$ Equivalent of $\mathrm{KMnO}_4$ reacting with $\mathrm{MnSO}_4=\frac{20 \times M_1}{1000} \times 3$
$\therefore$ Equivalent of $\mathrm{MnSO}_4$ reacting with $\mathrm{KMnO}_4=\frac{20 \times M_1}{1000} \times 3$
Since $\mathrm{MnSO}_4$ has $n$-factor 2, the mole of $\mathrm{MnSO}_4$ reacting
$$ =\frac{20 \times M_1}{1000} \times \frac{3}{2} $$
Total mole of $\mathrm{MnO}_2$ produced $=$ mole of $\mathrm{KMnO}_4+$ mole of $$ \mathrm{MnSO}_4 $$
Equivalent of $\mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4$ reacting with $\mathrm{MnO}_2=\frac{10 \times 0.2}{1000} \times 2$
$\therefore$ Equivalent of $\mathrm{MnO}_2=\frac{10 \times 0.2}{1000} \times 2$
Mole of $\mathrm{MnO}_2$ reacting with sodium oxalate $=\frac{10 \times 0.2 \times 2}{1000 \times 2}$
[as $n$-factor for $\mathrm{MnO}_2$ is 2].
Therefore, $\frac{20 \times M_1}{1000}+\left[\frac{20 \times M_1}{1000} \times \frac{3}{2}\right]=\frac{10 \times 0.2}{1000} ; M_1=0.04$
Equivalent of $\mathrm{KMnO}_4$ reacting with $\mathrm{H}_2 \mathrm{O}_2=\frac{20 \times 0.04}{1000} \times 5$ $$ =0.004 $$
If molarity of $\mathrm{H}_2 \mathrm{O}_2$ is $M_2$, then $=\frac{20 \times M_2 \times 2}{1000}=0.004$
$\therefore M_2=0.1 \mathrm{M}$
The reactions involved are:
$2 \mathrm{KMnO}_4+5 \mathrm{H}_2\mathrm{O}_2+3 \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{K}_2\mathrm{SO}_4+2 \mathrm{MnSO}_4+8 \mathrm{H}_2\mathrm{O}+5 \mathrm{O}_2$,
$2 \mathrm{KMnO}_4+3 \mathrm{MnSO}_4+2 \mathrm{H}_2\mathrm{O} \rightarrow 5 \mathrm{MnO}_2+2 \mathrm{H}_2\mathrm{SO}_4+\mathrm{K}_2\mathrm{SO}_4$,
$\mathrm{MnO}_2+\mathrm{Na}_2\mathrm{C}_2\mathrm{O}_4+2 \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{MnSO}_4+2 \mathrm{CO}_2+\mathrm{Na}_2\mathrm{SO}_4+2 \mathrm{H}_2\mathrm{O}$.
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