Question

$20mL$ of $F{e}^{2+}$ solution of certain concentration has completely reacted with $20mL$ of $0.01M{K}_{2}C{r}_2{O}_7$ in acidic medium. If
$20mL$ of same $F{e}^{2+}$ solution has reacted completely with $20mL$ of $KMn{O}_4$ solution in acid medium, the molarity of $KMn{O}_4$ 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 $F{e}^{2+}$ and $C{r}_2{O}_7^{2-}$

(i.e. $K_{2}C{r}_2{O}_7$ ) in acidic medium occurs as follows:

$C{r}_2{O}_7^{2-}+14H^{+}+6F{e}^{2+}⟶2C{r}^{3+}+7H_{2}O+6F{e}^{3+}$

The reaction between $F{e}^{2+}$ and $Mn{O}_4^{-}$

(i.e. $KMn{O}_4$ ) in acidic medium occurs as follows :

$Mn{O}_4^{-}+8H^{+}+5F{e}^{2+}⟶M{n}^{2+}+4H_{2}O+5F{e}^{3+}$

$\therefore C{r}_2{O}_7^{2-}$ react with $6F{e}^{2+}{n}_1$ and $Mn{O}_4^{-}$ react with

$5F{e}^{2+}\left(n_2\right)$ in a balanced equation.

Gram equivalent of $F{e}^{2+}$ $=$

= gram equivalent of $K_{2}C{r}_2{O}_7$

= gram equivalent of $KMn{O}_4$

Therefore, on comparing gram equivalent of

$K_{2}C{r}_2{O}_7$ and $KMn{O}_4$, we have

$n_1=6,n_2=5$

Conc. of $K_{2}C{r}_2{O}_7=0.01M\left(M_1\right)$

Volume of $K_{2}C{r}_2{O}_7=20mL\left(V_1\right)$

react with $F{e}^{2+}$ ions $=6mol\left(n_1\right)$

Conc. of $KMn{O}_4=$ to find $($ say $M_2)$

Volume of $KMn{O}_4=20mL\left(V_2\right)$

react with $F{e}^{2+}$ ions $=5mol\left(n_2\right)$

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 $M_2=\frac{0.01\times 6}{5}=0.012M$