Question

Hydrogen peroxide solution $(20 \,mL$ ) reacts quantitatively with a solution of $KMnO _{4}(20 \, mL )$ acidified with dilute $H _{2} SO _{4}$. The same volume of the $KMnO _{4}$ solution is just decolourised by $10 \, mL$ of $MnSO _{4}$ in neutral medium simultaneously forming a dark brown precipitate of hydrated $MnO _{2}$. The brown precipitate is dissolved in $10 \,mL$ of $0.2 \, M$ sodium oxalate under boiling condition in the presence of dilute $H _{2} SO _{4}$. calculate the molarity of $H _{2} O _{2}$. Round off your answer to nearest integer after multiplying with $100$ .

Answer 1

I $2 KMnO _{4}+3 H _{2} SO _{4}+5 H _{2} O _{2} \rightarrow K _{2} SO _{4}+2 MnSO _{4}+8 H _{2} O +5 O _{2}$
II $2 KMnO _{4}+3 MnSO _{4}+2 H _{2} O \rightarrow 5 MnO _{2}+ K _{2} SO _{4}+2 H _{2} SO _{4}$
III $MnO _{2}+ Na _{2} C _{2} O _{4}+2 H _{2} SO _{4} \rightarrow MnSO _{4}+ Na _{2} SO _{4}+2 CO _{2}+ H _{2} O$
Reaction (III) $\Rightarrow mM$ of $Na _{2} C _{2} O _{4}=10\, mL \times 0.2\, M =2\, mM = mM$ of $MnO _{2}$
Reaction (II) $\Rightarrow mM$ of $KMnO _{4}=2 \times \frac{2}{5}=\frac{4}{5} \,mM$
Reaction (I) $\Rightarrow mM$ of $H _{2} O _{2}=\frac{4}{5} \times \frac{5}{2}=2 \,mM$
$2 \,mM _{2}$ of $H _{2} O _{2}=20 \,mL \times M _{ H _{2} O _{2}}$
$M _{ H _{2} O _{2}}=0.1 \,M$