The iodide content of a solution was determined by titration with cerium (IV) sulphate in the presence of HCl, in which I⁻ is converted to ICl. A 250 ml sample of the solution required 20 ml of 0.05 N

Question

The iodide content of a solution was determined by titration with cerium (IV) sulphate in the presence of HCl, in which I⁻ is converted to ICl. A 250 ml sample of the solution required 20 ml of 0.05 N Ce⁴⁺ solution. What is the iodide concentration in the original solution in g/L?

TestHub · Accepted Answer

I⁻ → ICl is I(-1) → I(+1), a 2-electron change, so equivalent mass of I⁻ = 127/2. meq Ce⁴⁺ = 20 x 0.05 = 1 meq = meq I⁻ in 250 ml. Mass of I⁻ = 1 x 63.5 = 63.5 mg in 250 ml = 254 mg/L = 0.254 g/L, option (b).

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