Question

Mass of $KHC_{2}O_{4}$ (potassium acid oxalate) required to reduce 100 ml of 0.02 M $KMnO_{4}$ in acidic medium (to $Mn^{2+}$) is x g and to neutralise $100$ ml of $0.05$ M Ca$(OH)_{2}$ is y g, then

• A. $x=y$
• B. $2x=y$
• C. $x = 2y$
• D. none is correct

Answer 1

Answer: (B)

The reaction is as follows:
$
5 HC _{2} O _{4}^{-}+2 MnO _{4}^{-} \rightarrow 2 Mn ^{2+}+10 CO _{2}
$
The number of moles of $MnO _{4}^{-}=0.1 \times 0.02=0.002 mol$
This is equal to $0.005 mol$ of $HC _{2} O _{4}^{-}$ion. This corresponds to $xg$.
The number of moles of $Ca ( OH )_{2}$ is $0.1 \times 0.05=0.005 mol$
They corresponds to $0.01 mol$ of hydrogen ions and $0.01 mol$ of $HC _{2} O _{4}{ }^{-}$.
This is equal to $g$.
Hence, $\frac{ x }{ y }=\frac{0.005}{0.01}=\frac{1}{2}$ or $y =2 x$
Hence, the relation is $2 x = y$.