Question

40 ml M 20 KMnO4 Sol + 50 ml KI solution in presence of h2so4 and the mixture diluted to 300 ml. 25 ml mixture needs 5 ml of s M/50 K2 Cr2 07 Soln for titration Molarity of KI solution used is ?

Answer 1

0.344 M

Answer 2

1. First Reaction: $KMnO_4$ reacts with $KI$ in acidic medium. The balanced equation is: $2MnO_4^- + 10I^- + 16H^+ \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O$ The stoichiometry shows that 1 mole of $KMnO_4$ reacts with 5 moles of $KI$.

2. Calculate moles of $KMnO_4$: Molarity of $KMnO_4 = \frac{M}{20} = 0.05$ M. Volume of $KMnO_4 = 40$ ml $= 0.040$ L. Moles of $KMnO_4 = \text{Molarity} \times \text{Volume} = 0.05 \text{ M} \times 0.040 \text{ L} = 0.002$ moles.

3. Calculate moles of $KI$ reacted: Moles of $KI$ reacted $= 5 \times$ Moles of $KMnO_4 = 5 \times 0.002 \text{ moles} = 0.010$ moles.

4. Excess $KI$: The subsequent titration implies that $KI$ was in excess. Let $M_{KI}$ be the molarity of the initial $KI$ solution. Volume of $KI$ solution $= 50$ ml $= 0.050$ L. Initial moles of $KI = M_{KI} \times 0.050$ L. Moles of $KI$ remaining (excess) $= (M_{KI} \times 0.050) - 0.010$ moles.

5. Excess $KI$ in the titrated portion: The mixture is diluted to 300 ml. A 25 ml portion is taken for titration. Moles of excess $KI$ in 25 ml $= [(M_{KI} \times 0.050) - 0.010] \times \frac{25 \text{ ml}}{300 \text{ ml}}$.

6. Second Reaction: The excess $KI$ is titrated by $K_2Cr_2O_7$ in acidic medium. The balanced equation is: $Cr_2O_7^{2-} + 6I^- + 14H^+ \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O$ The stoichiometry shows that 1 mole of $K_2Cr_2O_7$ reacts with 6 moles of $KI$.

7. Calculate moles of $K_2Cr_2O_7$ used: Molarity of $K_2Cr_2O_7 = \frac{M}{50} = 0.02$ M. Volume of $K_2Cr_2O_7 = 5$ ml $= 0.005$ L. Moles of $K_2Cr_2O_7 = 0.02 \text{ M} \times 0.005 \text{ L} = 0.0001$ moles.

8. Calculate moles of excess $KI$ that reacted: Moles of excess $KI$ reacted with $K_2Cr_2O_7 = 6 \times$ Moles of $K_2Cr_2O_7 = 6 \times 0.0001 \text{ moles} = 0.0006$ moles. This is the amount of excess $KI$ present in the 25 ml mixture.

9. Equate and solve for $M_{KI}$: From step 5 and step 8: $[(M_{KI} \times 0.050) - 0.010] \times \frac{25}{300} = 0.0006$ $(M_{KI} \times 0.050) - 0.010 = 0.0006 \times \frac{300}{25}$ $(M_{KI} \times 0.050) - 0.010 = 0.0006 \times 12$ $(M_{KI} \times 0.050) - 0.010 = 0.0072$ $M_{KI} \times 0.050 = 0.010 + 0.0072$ $M_{KI} \times 0.050 = 0.0172$ $M_{KI} = \frac{0.0172}{0.050} = 0.344$ M.