Sodium thiosulfate (Na₂S₂O₃,molar mass = 158.2 g/mol)is used in photography.The purity of a Na₂S₂O₃ solution labeled as 0.2043 M is checked by redox titration.It is found that 0.100 L of this solution requires 40.21 mL of 0.246 M iodine solution to reach the equivalence point.What is the percent purity of the sodium thiosulfate solution? I₂(aq)+ 2 S₂O₃²^-(aq)$\to$ 2 I^-(aq)+ S₄O₆²^-(aq)

Question

Quizplus · Accepted Answer

First, let's determine the number of moles of iodine used in the titration:

0.246 mol/L x 0.04021 L = 0.00989866 mol I2

Since the reaction between iodine and sodium thiosulfate is a 1:2 ratio, the number of moles of sodium thiosulfate can be calculated as:

0.00989866 mol I2 x (1 mol Na2S2O3 / 2 mol I2) = 0.00494933 mol Na2S2O3

Now we can calculate the concentration of the sodium thiosulfate solution:

C = n/V

C = 0.00494933 mol / 0.100 L = 0.0494933 mol/L

The percent purity can be calculated by dividing the measured concentration (0.2043 M) by the calculated concentration (0.0494933 M) and multiplying by 100:

Percent purity = (0.0494933 M / 0.2043 M) x 100% = 24.2%

Therefore, the answer is choice letter D.

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

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