Question

A 3.452 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?

Answer 1

Answer:

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⁻×$\frac{1molI_{3}^-}{2molS_{2}O_{3}^{2-}}$ = 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⁻× $\frac{2molCe^{4+}}{1molI_{3}^-}$ =  4,228x10⁻⁴ moles of Ce(IV).

These moles are:

4,228x10⁻⁴ moles of Ce(IV)×$\frac{140,116g}{1mol}$ = 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)×$\frac{250,0mL}{25,00mL}$ =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!