Question

Three different solutions of oxidising agents, i.e. $K_{2}C{r}_2{O}_7,I_2$ and $KMn{O}_4$ is titrated separately with $0.19$ g of $K_2{S}_2{O}_3$. The molarity of each oxidising agent is $0.1M$ and the reactions are as follows:
i. $C{r}_2{O}_7^{2-}+S_2{O}_3^{2-}\to C{r}^{3+}+S{O}_4^{2-}$
ii. $I_2+S_2{O}_3^{2-}\to I^{⊝}+S_4{O}_6^{2-}$
iii. $Mn{O}_4^{⊝}+S_2{O}_3^{2-}\to Mn{O}_2+S{O}_4^{2-}$
(Molecular weight of $K_2{S}_2{O}_3=190,K_{2}C{r}_2{O}_7=294,KMn{O}_4=158,$ and $I_2=254g$ $mo{l}^{-1}$
Which of the following statements is/are correct?

• A. All three oxidising agents can act as self-indicators.
• B. Volume of $I_2$ used is minimum.
• C. Volume of $K_{2}C{r}_2{O}_7$ used is maximum.
• D. Weight of $KMn{O}_4$ used in the titration is maximum.

Answer 1

Answer: (A), (B), (D)

The correct options are
A All three oxidising agents can act as self-indicators.
B Weight of $KMn{O}_4$ used in the titration is maximum.
D Volume of $I_2$ used is minimum.

All the three ($K_{2}C{r}_2{O}_7,I_2$ and $KMn{O}_4$) are self indicators.
Therefore, they do not need any indicator for titration.
Moles of $K_2{S}_2{O}_3=\frac{0.19}{190}={10}^{-3}mol=1$ mmol.
i. mEq. of $S_2{O}_3^{2-}(n=8)$ = mEq. of $C{r}_2{O}_7^{2-}(n=6)$
$1\times 8=0.1\times 6\times V$
$\therefore V_{C{r}_2{O}_7^{2-}}=\frac{80}{6}mL$
ii. mEq of $S_2{O}_3^{2-}(n=1)$=mEq of $I_2(n=2)$
$I\times 1=0.1\times 2\times V$
$\therefore V_{I_2}=5mL$
iii. mEq of $S_2{O}_3^{2-}(n=8)$=mEq of $Mn{O}_4^{⊝}(n=3)$
$1\times 8=0.1\times 3\times V$
$\therefore V_{Mn{O}_4^{⊝}}=\frac{80}{3}mL$
Also,
Weight of $K_{2}C{r}_2{O}_7=8\times \frac{294}{6}\times {10}^{-3}=0.392g$
Weight of $I_2=1\times \frac{254}{2}\times {10}^{-3}=0.127g$
Weight of $KMn{O}_4=8\times \frac{158}{3}\times {10}^{-3}=1.264g$
Only option C is wrong.
Hence, options A, B & D are correct.