Question

Three different solutions of oxidizing agents $K_{2}C{r}_2{O}_7,I_2$, and $KMn{O}_4$ is titrated separately with $0.19g$ of $K_2{S}_2{O}_3$. The molarity of each oxidising agent is $0.1M$ and the reactions are:
i. $C{r}_2{O}_7^{2-}+S_2{O}_3^{2-}\to C{r}^{3+}+S{O}_4^{2-}$
ii. $I_2+S_2{O}_3^{2-}\to I^{\ominus }+S_4{O}_6{}^{2-}$
iii. $Mn{O}_4{}^{\ominus }+S_2{O}_3{}^{2-}\to Mn{O}_2+S{O}_4{}^{2-}$
(molecular weight of $K_2{S}_2{O}_3=190,K_{2}C{r}_2{O}_7=294,KMn{O}_4=158$, and $I_2=254gmo{l}^{-1}$ )
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)

All three are self indicator s i.e., they do not need any indicator for titration
$K_2{S}_2{O}_3=\frac{0.19}{190}=10^{-3}mol=1mmol$
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 $I_2(n=2)$
$1\times 1=0.1\times 2\times V$
$\therefore V_{I_2}=5mL$
iii. $mEq\text{ of }{S}_2{O}_3^{2-}(n=8)=mEq$ of $Mn{O}_4{}^{\ominus }(n=3)$
$1\times 8=0.1\times 3\times V$
$\therefore V_{Mn{O}_4}{}^{\ominus }=\frac{80}{3}mL$
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 $KMn{O}_4=8\times \frac{158}{3}\times 10^{-3}=1.264g$