Skip to content
dronacharya.ai
← JEE Main 2026 · 6 Apr · Shift 2
ChemistryInteger+4 / -1Question 73 of 75

JEE Main 2026 · 6 Apr · Shift 2, Chemistry question 73

500 mL of 0.2 M MnOX4X−\ce{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 xx M thiosulphate solution in presence of starch till the end point. If 300 mL of thiosulphate was consumed, then the value of xx is ________.

Answer and worked solution

Official answer: 1

Permanganate in basic medium. MnOX4X−→MnOX2\ce{MnO4- -> MnO2} takes 3 electrons. 0.5×0.2=0.10.5\times0.2=0.1 mol gives 0.3 mol of electrons, so 0.15 mol IX2\ce{I2}. (Iodide, 0.75 mol, is in excess.)

Thiosulphate. IX2+2 SX2OX3X2−→2 IX−+SX4OX6X2−\ce{I2 + 2S2O3^2- -> 2I- + S4O6^2-} uses 0.30 mol:
x=0.300.300=1 M.x=\frac{0.30}{0.300}=1\text{ M}.

Answer: 1

Topics:Redox titrationsBalancing redox reactions
← Previous questionNext question →