500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M KMnO4 solution in basic medium. The liberated iodine was titrated with standard 0.1 M Na2S2O3 solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of Na2S2O3 consumed is . (Nearest integer)
Answer
3
Solution
mol of I- = 0.5 × 1.2 = 0.6 mol and mol of MnO4- = 0.5 × 0.2 = 0.1 mol.
In basic medium MnO4- is reduced to MnO2 (Mn: +7 → +4, 3 e⁻):
2MnO4- + 6I- + 4H2O 2MnO2 + 3I2 + 8OH-
0.1 mol MnO4- needs only 0.3 mol I- (0.6 mol is present), so I- is in excess and the I2 liberated = 32 0.1 = 0.15 mol.
In the titration, I2 + 2S2O32- 2I- + S4O62-, so mol of S2O32- = 2 × 0.15 = 0.3 mol.
V = 0.30.1 = 3 L
Answer: 3