500 mL of 0.2 M MnO4- solution in basic medium when mixed with 500 mL of 1.5 M KI solution, oxidises iodide ions to liberate molecular iodine. This liberated iodine is then titrated with a standard x M thiosulphate solution in presence of starch till the end point. If 300 mL of thiosulphate was consumed, then the value of x is .
Answer
1
Solution
Basic medium: MnO4- + 2H2O + 3e- MnO2 + 4OH-; 2I- I2 + 2e-
n(MnO4-) = 0.5 × 0.2 = 0.1 mol ⇒ electrons accepted = 0.3 mol
n(I-) = 0.5 × 1.5 = 0.75 mol (excess, needs only 0.3 mol)
⇒ n(I2) = 0.32 = 0.15 mol
I2 + 2S2O32- 2I- + S4O62- ⇒ n( S2O32-) = 2 × 0.15 = 0.3 mol
⇒ x = 0.3 mol0.300 L = 1 M
Answer: 1