Sodium thiosulfate (Na₂S₂O₃, molar mass = 158.2 g/mol) is used in the development of photographic film. Your summer job is at a photography lab and you need to check the purity of an outdated supply.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

Sodium thiosulfate (Na₂S₂O₃, molar mass = 158.2 g/mol) is used in the development of photographic film. Your summer job is at a photography lab and you need to check the purity of an outdated supply.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)

Quizplus · Accepted Answer

First, we need to find the limiting reactant in the reaction between iodine and sodium thiosulfate. Using stoichiometry, we can calculate that 40.21 mL of 0.246 M iodine solution is equivalent to 5.0 x 10^-3 moles of iodine.

Next, we need to use the balanced chemical equation to determine how many moles of sodium thiosulfate are required to react with 5.0 x 10^-3 moles of iodine. From the balanced equation, we see that 1 mole of iodine reacts with 2 moles of sodium thiosulfate. Therefore, we need 2.5 x 10^-3 moles of sodium thiosulfate to react completely with the iodine.

Using the molar mass of sodium thiosulfate, we can convert the mass of the sample (3.232 g) to moles. We find that the sample contains 2.04 x 10^-2 moles of sodium thiosulfate.

To calculate the percent purity, we can use the following formula:

percent purity = (moles of pure compound / moles of impure compound) x 100%

In this case, the moles of pure compound is the 2.5 x 10^-3 moles of sodium thiosulfate we calculated earlier. The moles of impure compound is the 2.04 x 10^-2 moles of sodium thiosulfate in the sample.

Therefore, the percent purity is:

percent purity = (2.5 x 10^-3 / 2.04 x 10^-2) x 100% = 12.3%

However, we need to remember that the sample was outdated, which means some of the sodium thiosulfate may have decomposed over time. Therefore, the actual percent purity is likely higher than this calculated value. The closest answer choice to our result is C, which is 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