A3.452 g sample containing an unknown amount of a ce(iv) salt is dissolved in 250.0 ml of 1 m h2so4. a 25.00 ml aliquot is analyzed by adding ki and titrating the i3^- that forms with s2o3^2-. the end point was reached following the addition of 13.02 ml of 0.03247 m na2s2o3. calculate the weight percent of ce^4+ in the sample?

The weight percent in the sample is 17,16%

Explanation:

The dissolution of the Ce(IV) salt provides free Ce⁴⁺ that reacts, thus:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃²⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03247M Na₂S₂O₃ = 4,228x10⁻⁴ moles of S₂O₃²⁻.

As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,228x10⁻⁴ moles of S₂O₃⁻× = 2,114x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,114x10⁻⁴ moles of I₃⁻× =  4,228x10⁻⁴ moles of Ce(IV).

These moles are:

4,228x10⁻⁴ moles of Ce(IV)× = 0,05924 g of Ce(IV)

As was taken an aliquot of 25,00mL from the solution of 250,0mL:

0,05924 g of Ce(IV)× =0,5924g of Ce(IV) in the sample

As the sample has 3,452g, the weight percent is:

0,5924g of Ce(IV) / 3,452g × 100 = 17,16 wt%

I hope it helps!

3 1/4 %

explanation:

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