500 mL of 0.2 M MnO_4^- 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 __________.
A. A
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 1 = 750 meq Since MnO _4^- is the limiting reagent, the equivalents of I _2 produced will be equal to the equivalents of MnO _4^- consumed. Equivalents of I _2 formed = 300 meq The liberated iodine is titrated with thiosulphate ( S _2 O _3^ 2- ): I _2 + 2 S _2 O _3^ 2- 2 I ^- + S _4 O _6^ 2- For thiosulphate, the n-factor is 1 (as 2 moles of S _2 O _3^ 2- lose 2 moles of electrons). From the law of equivalence: Equivalents of S _2 O _3^ 2- = Equivalents of I _2 Molarity Volume (in mL) n-factor = 300 x 300 1 = 300 x = 1 Answer: 1
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