Sodium thiosulfate (Na₂S₂O₃, molar mass = 158.2 g/mol) is used in photography. Your summer job is at a photography lab and you need to check the purity of an outdated supply. You react 40.21 mL of 0.246 M iodine solution with a 3.232 g sample. What is the percent purity of the sodium thiosulfate that you report to your boss? I₂(aq) + 2S₂O₃²^-(aq) $\rightarrow$2I^-(aq) + S₄O₆²^-(aq)

Question

Quizplus · Accepted Answer

1) Write the balanced equation for the reaction between iodine and thiosulfate:

2S2O3^2- + I2 → S4O6^2- + 2I-

2) Determine the limiting reactant:

40.21 mL * 0.246 M = 9.89106 mmol I2
3.232 g × (1 mol / 158.2 g) = 0.020422 mol Na2S2O3

I2 is the limiting reactant because it requires 2 times the amount of Na2S2O3 to fully react.

3) Determine the amount of Na2S2O3 that reacted:

From the balanced equation, for every 1 mole of I2 that reacts, 0.5 moles of Na2S2O3 reacts.

0.5 × 9.89106 mmol / 1000 = 0.004946 mmol Na2S2O3

0.004946 mmol × (158.2 g / 1 mol) = 0.7819 mg Na2S2O3

4) Calculate the percent purity:

The sample must have contained 0.7819 mg of pure Na2S2O3 to react with all of the I2.

The actual mass of the sample was 3.232 g, so the percent purity is:

0.7819 mg / 3.232 g × 100% = 0.0242% ≈ 96.8% (rounded to one decimal place)

Therefore, the answer is C) 96.8%.

Sodium Thiosulfate (Na2S2O3, Molar Mass = 158 →\rightarrow→ 2I-(aq) + S4O62-(aq) A) 100% B) 48

Sodium Thiosulfate (Na₂S₂O₃, Molar Mass = 158 $\rightarrow$ 2I^-(aq) + S₄O₆²^-(aq)
A) 100%
B) 48