A 20 mL solution of potassium iodate ( KIO ₃ ) is treated with an excess of KI in an acidic medium. The liberated iodine requires 30 mL of 0.1 M sodium thiosulphate solution for complete neutralization. The molarity of the KIO ₃ solution is x 10⁻³ M . The value of x is ________.

Step-by-step solution

The reactions involved are: IO ₃^- + 5 I ^- + 6 H ^+ 3 I ₂ + 3 H ₂ O I ₂ + 2 S ₂ O ₃²⁻ 2 I ^- + S ₄ O ₆²⁻ Millimoles of S ₂ O ₃²⁻ used = 30 0.1 = 3.0 mmol . From the second reaction, 2 moles of thiosulphate react with 1 mole of I ₂ . Millimoles of I ₂ produced = 3.0 2 = 1.5 mmol . From the first reaction, 1 mole of IO ₃^- produces 3 moles of I ₂ . Millimoles of KIO ₃ present = 1.5 3 = 0.5 mmol . Molarity of KIO ₃ = 0.5 mmol 20 mL = 0.025 M = 25 10⁻³ M . Thus, x = 25 . Answer: 25