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x mole KIO 3 is treated with excess of KI, liberated I 2 which was dissolved in freshly prepared starch solution which was titrated with by 60 mL 0.1 N Na 2 S 2 O 3 until the end point. What is x?

Options

  1. A1 0 - 3 mol
  2. B1 0 - 5 mol
  3. C5 mol
  4. D6 mol

Correct answer

A. 1 0 - 3 mol

Step-by-step solution

I - ⟶ 1 2 I 2 + e ] × 5 (HR) IO 3 - 1 + 6 H + + 5 e → 1 2 I 2 + 3 H 2 O (RHR) 5 I - + I O 3 - + 6 H + ⟶ 3 I 2 I 2 + 2 N a 2 S 2 O 3 ⟶ 2 N a I + N a 2 S 4 O 6 6 0 mL of O . 1 N Na 2 S 2 O 3 = 6 Meqs = 6 mM Na 2 S 2 O 3 ≡ 3 mm of I 2 ≡ 1 mm of KIO 3 = x ∴ x = 1 0 - 3 moles

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