Question

The distribution ratio of iodine between an organic solvent and water is 7.8. 25.0cm3 of 0.10 mol/L of iodine is shaken with 100cm3 of the organic

solvent until equilibrium is reached. Calculate the volume of 0.05mol/L
Na2S2O3 required to titrate a 25.0cm3 aliquots of the organic phase.​

Answer 1

96.8mL of Na₂S₂O₃ 0.05mol/L

Step-by-step explanation:

The distribution ratio, K, is defined as the ratio of the concentration of a solute in an organic solvent and water, that is:

K = 7.8 = Concentration in organic solvent / Concentration in aqueous phase

7.8 = Concentration in organic solvent / Concentration in aqueous phase

The moles of I₂ that are in the first in the aqueous solution are:

0.0250L ₓ (0.10mol / L) = 0.00250 moles I₂

Replacing in distribution ratio equation:

7.8 = X / 100mL / (0.00250 moles - X) / 25.0mL

Where X are the moles that are transported from the aqueous phase to the organic phase.

7.9 = 25X / 0.250 moles - 100X

1.975 moles - 790X = 25 X

1.975 moles = 815X

0.00242 moles I₂ = X

That means moles of I₂ at the organic phase are 0.00242

The reaction of Na₂S₂O₃ with I₂ is:

2 Na₂S₂O₃ + I₂ → Na₂S₄O₆ + 2NaI

Where 2 moles of Na₂S₂O₃ react per mole of I₂

Moles of Na₂S₂O₃ that react with 0.00242 moles of I₂ are:

0.00242 moles I₂ ₓ (2 moles Na₂S₂O₃ / 1 mole I₂) = 0.00484 moles Na₂S₂O₃

The volume of 0.00484 moles of Na₂S₂O₃ in a 0.05mol/L is:

0.00484 moles Na₂S₂O₃ ₓ (1L / 0.05mol) = 0.0968L Na₂S₂O₃ 0.05mol/L

= 96.8mL of Na₂S₂O₃ 0.05mol/L