Question

$10 \,mL$ of $NaHC _2 O _4$ solution is neutralized by $10 \,mL$ of $0.1\, M \,NaOH$ solution. $10\, mL$ of same $NaHC _2 O _4$ solution is oxidized by $10 \,mL$ of $KMnO _4$ solution in acidic medium. Hence, molarity of $KMnO _4$ is

• A. $0.1 \,M$
• B. $0.2 \,M$
• C. $0.04 \,M$
• D. $0.02\, M$

Answer 1

Answer: (C)

$NaHC _2 O _4 \rightleftharpoons Na ^{+}+ H ^{+}+ C _2 O _4^{2-}$
$H ^{+} \equiv OH ^{-}$
and $C _2 O _4^{2-} \equiv MnO _4^{-}$
$H ^{+}+ OH ^{-} \rightarrow H _2 O$
$C _2 O _4^{2-} \rightarrow 2 CO _2+2 e ^{-}$
$MnO _4^{-}+8 H ^{+}+5 e ^{-} \rightarrow Mn ^{2+}$
$10 \,mL$ of $NaHC _2 O _4 \equiv 10\, mL$ of $0.1 \,M\,NaOH$
Thus, $10 \times M\left( NaHC _2 O _4\right)=10 \times 0.1$
$M=0.1 \,M =0.1 \,N$ ( since $H ^{+}$is neutralized)
$NaHC _2 O _4=0.1 \,N$ as an acid
$=0.2 \,N$ as $C _2 O _4^{2-}$ oxidised by $MnO _4^{-}$
Let $MnO _4^{-}= xM =5 x N$
$10 \,mL$ of $0.2\, N\, NaHC _2 O _4 \equiv 10\, mL$ of $5 \times NMnO _4$
$10 \times 0.2=10 \times 5 x$
$\therefore x =\frac{0.2}{5}=0.04\, M$