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 23.50 mL of a 0.1710 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.

Answer 1

Answer:

The concentration of the Na₂S₂O₃ solution is 0,08037 M

Explanation:

The reaction of the titration is:

2 Na₂S₂O₃(aq) + I₂(aq) → Na₂S₄O₆ +2 NaI(aq).

The moles of I₂ that react are:

0,02350L×$\frac{0,1710mol}{L}$ = 4,0185x10⁻³ moles of I₂

By the reaction of the titration, 1 mol of I₂ react with 2 moles of Na₂S₂O₃, the moles of Na₂S₂O₃ that react are:

4,0185x10⁻³ moles of I₂ ×$\frac{2molNa_{2}S_{2}O_{3}}{1molI_{2}}$= 8,037x10⁻³ moles of Na₂S₂O₃

As the volume of the sample is 100,0mL≡0,1000L. The concentration of Na₂S₂O₃ is:

$\frac{8,037x10^{-3}mol}{0,1000L}$ = 0,08037 M

I hope it helps!