Question

A mixture containing $0.05$ mol of $K_{2}C{r}_2{O}_7$ and $0.02$ mol of $KMn{O}_4$ was treated with excess of $KI$ in acidic medium. The liberated iodine required $1L$ of $N{a}_2{S}_2{O}_3$ solution for titration. Determine the concentration of $N{a}_2{S}_2{O}_3$ solution.

• A. $0.40N$
• B. $0.30N$
• C. $0.25N$
• D. $0.20N$

Answer 1

The correct option is A $0.40N$

$C{r}_2{O}_7^{2-}+Mn{O}_4^{-}+I^{-}+H^{+}\to I_2+C{r}^{3+}+M{n}^{2+}+H_{2}O$
From the reaction we can find n-factors of required species
n-factor of $C{r}_2{O}_7^{2-}=6$
n-factor of $Mn{O}_4^{-}=5$
n-factor of $I_2=2$
By applying law of equivalence
$eq.ofMn{O}_4^{-}+eq.ofC{r}_2{O}_7^{2-}=eq.of{I}_2$
$(0.02\times 5)+(0.05\times 6)=eq.of{I}_2$
Also it is given that liberated $I_2$ reacts with $N{a}_2{S}_2{O}_3$
$\therefore$ eq.of $N{a}_2{S}_2{O}_3=$ eq. of $I_2$
$=$ eq. of $Mn{O}_4^{-}+$ eq. of $C{r}_2{O}_7^{2-}$
$\therefore N\times 1=0.02\times 5+0.05\times 6$
$\therefore N=0.4N$

Hence $\left(a\right)$ is the correct option.