Question

$20 \,mL$ of $Fe ^{2+}$ solution of certain concentration has completely reacted with $20\, mL$ of $0.01 \,M K _{2} Cr _{2} O _{7}$ in acidic medium. If
$20\, mL$ of same $Fe ^{2+}$ solution has reacted completely with $20 \,mL$ of $KMnO _{4}$ solution in acid medium, the molarity of $KMnO _{4}$ solution is

• A. 0.01 M
• B. 0.12 M
• C. 0.10 M
• D. 0.012 M

Answer 1

Answer: (D)

The reaction between $Fe ^{2+}$ and $Cr _{2} O _{7}^{2-}$

(i.e. $K _{2} Cr _{2} O _{7}$ ) in acidic medium occurs as follows:

$Cr _{2} O _{7}^{2-}+14 H ^{+}+6\, Fe ^{2+} \longrightarrow 2 \,Cr ^{3+}+7 \,H _{2} O +6 \,Fe ^{3+}$

The reaction between $Fe ^{2+}$ and $MnO _{4}^{-}$

(i.e. $KMnO _{4}$ ) in acidic medium occurs as follows :

$MnO _{4}^{-}+8 H ^{+}+5 Fe ^{2+} \longrightarrow Mn ^{2+}+4\, H _{2} O +5 \,Fe ^{3+}$

$\therefore Cr _{2} O _{7}^{2-}$ react with $6 Fe ^{2+} n_{1}$ and $MnO _{4}^{-}$ react with

$5 Fe ^{2+}\left(n_{2}\right)$ in a balanced equation.

Gram equivalent of $Fe ^{2+}$ $=$

= gram equivalent of $K _{2} Cr _{2} O _{7}$

= gram equivalent of $KMnO _{4}$

Therefore, on comparing gram equivalent of

$K _{2} Cr _{2} O _{7}$ and $KMnO _{4}$, we have

$n_{1}=6, n_{2}=5$

Conc. of $K _{2} Cr _{2} O _{7}=0.01 \,M \left(M_{1}\right)$

Volume of $K _{2} Cr _{2} O _{7}=20\, mL \left(V_{1}\right)$

react with $Fe ^{2+}$ ions $=6\, mol \left(n_{1}\right)$

Conc. of $KMnO _{4}=$ to find $\left(\right.$ say $\left.M_{2}\right)$

Volume of $KMnO _{4}=20 \,mL \left(V_{2}\right)$

react with $Fe ^{2+}$ ions $=5 \,mol \left(n_{2}\right)$

Thus, $M_{1} \times V_{1} \times n_{1}=M_{2} \times V_{2} \times n_{2}$

or, $0.01 \times 20 \times 6=M_{2} \times 20 \times 5$

or $M_{2}=\frac{0.01 \times 6}{5}=0.012\, M$