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 __________.

Step-by-step solution

First, we calculate the millimoles of the reactants: Millimoles of MnO ₄^- = 500 mL 0.2 M = 100 mmol Millimoles of I ^- = 500 mL 1.5 M = 750 mmol In a basic medium, MnO ₄^- is reduced to MnO ₂ . The change in oxidation state of Mn is from +7 to +4 , so its n-factor is 3 . The problem explicitly states that iodide ions are oxidized to molecular iodine ( I ₂ ). The change in oxidation state for iodine is from -1 to 0 , so the n-factor for I ^- is 1 . Equivalents of MnO ₄^- = 100 3 = 300 meq Equivalents of I ^- = 750