A 1.150 g sample containing an unknown amount of arsenic trichloride, with the rest being inerts, was dissolved into a NaHCO3 and HCl aqueous solution. To this solution was added 1.920 g of KI and 50.00 mL of a 0.00885 M KIO3 solution. The excess 13 - was titrated with 50.00 mL of a 0.02000 M Na2S2O3 solution.

What was the mass percent of arsenic trichloride in the original sample?

13.04%

Explanation:

KIO₃, IO₃⁻, reacts with KI, I⁻, thus:

IO₃⁻ + 8I⁻ + 6H⁺ → 3I₃⁻ + 3H₂O

Producing triiodide ion.

KIO₃ is limiting reactant, moles are:

0.05000L ₓ (0.00885mol / L) = 4.425x10⁻⁴ moles.

Moles of I₃⁻ produced are:

4.425x10⁻⁴ moles KIO₃ ₓ (3 moles I₃⁻ / 1 mole IO₃⁻) = 1.3275x10⁻³ moles I₃⁻

This iodine reacts with Na₂S₂O₃ and AsCl₃, As³⁺, thus:

I₃⁻ + 2S₂O₃²⁻ → 3I⁻ + S₄O₆²⁻

I₃⁻ + As³⁺ → 3I⁻ + As⁵⁺

Moles of I₃⁻ that react with Na₂S₂O₃ are:

0.0500L ₓ (0.02000mol / L) = 1.000x10⁻³ moles Na₂S₂O₃

1.000x10⁻³ moles Na₂S₂O₃ ₓ (1 mole I₂ / 2 moles Na₂S₂O₃) = 5.000x10⁻⁴ moles I₃⁻.

That means moles of I₃⁻ that react with As³⁺ are:

1.3275x10⁻³moles - 5.000x10⁻⁴moles = 8.275x10⁻⁴ moles I₃⁻. As 1 mole of I₃⁻ reacts per mole of As³⁺, moles of As³⁺ are 8.275x10⁻⁴ moles.

Molar mass of AsCl₃ is 181.28g/mol. 8.275x10⁻⁴ moles weight:

8.275x10⁻⁴ moles ₓ (181.28g / mol) = 0.1500g of AsCl₃

As the weight of the sample is 1.150g, mass percent of AsCl₃ is:

0.1500g / 1.150g ₓ 100 =

13.04%