Question

Three different solutions of oxidizing agents $K _2 Cr _2 O _7, I _2$, and $KMnO _4$ is titrated separately with $0.19 \,g$ of $K _2 S _2 O _3$. The molarity of each oxidising agent is $0.1 \,M$ and the reactions are:
i. $Cr _2 O _7^{2-}+ S _2 O _3^{2-} \rightarrow Cr ^{3+}+ SO _4^{2-}$
ii. $I _2+ S _2 O _3^{2-} \rightarrow I ^{\ominus}+ S _4 O _6{ }^{2-}$
iii. $MnO _4{ }^{\ominus}+ S _2 O _3{ }^{2-} \rightarrow MnO _2+ SO _4{ }^{2-}$
(molecular weight of $K _2 S _2 O _3=190, K _2 Cr _2 O _7=294, KMnO _4=158$, and $I _2=254 \,g\, mol ^{-1}$ )
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 $K _2 Cr _2 O _7$ used is maximum
• D. Weight of $KmnO _4$ 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
$K _2 S _2 O _3=\frac{0.19}{190}=10^{-3} mol =1\, mmol$
i. $mEq$ of $S _2 O _3^{2-}(n=8)= mEq$ of $Cr _2 O _7^{2-}(n=6)$
$1 \times 8=0.1 \times 6 \times V$
$\therefore V_{ Cr _2 O _7^{2-}}=\frac{80}{6} mL$
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}=5 \,mL $
iii. $mEq \text { of } S _2 O _3^{2-}(n=8)= mEq $ of $ MnO _4{ }^{\ominus}(n=3) $
$ 1 \times 8=0.1 \times 3 \times V $
$\therefore V_{ MnO _4}{ }^{\ominus}=\frac{80}{3} mL$
Also,
Weight of $K _2 Cr _2 O _7=8 \times \frac{294}{6} \times 10^{-3}=0.392 \,g$
Weight of $I _2=1 \times \frac{254}{2} \times 10^{-3}=0.127\, g$
Weight of $KMnO _4=8 \times \frac{158}{3} \times 10^{-3}=1.264 \,g$