Question

Sodium thiosulfate, Na2S2O3, is used as a fixer in photographic film developing. The amount of Na2S2O3 in a solution can be determined by a titration with iodine, I2, according to the equation: 2Na2S2O3(aq) + I2(aq) --> Na2S4O6 +2NaI(aq).

Calculate the concentration of the Na2S2O3 solution if 48.50 mL of a 0.1780 M I2 solution react exactly with a 100.0 mL sample of the Na2S2O3 solution. Use 4 sig. fig.

Aside, the end of the titration is determined by the color. NaI is a pale yellow and I2 is a deep purple. Just when the purple color persists, all the iodine that can react has reacted.

Please help! I tried the question and got 14.68 which wasn't right.

Answer 1

I can't figure how you came up with that number although I tried several combinations of numbers.
2S2O3^2- + I2 ==> I2 + S4O6^2-

millimols I2 = 48.50 x 0.1780 = 8.633
milimmols S2O23^2- = twice that= 8.633 x 2 = 17.266
M S2O3^- = mmols/ml = 17.266/100 = .17266 M which rounds to 0.1727 M to 4 s.f. for Na2S2O3.

Answer 2

Great! Thanks!