Question

A chemist dissolves 155mg of iodine in 1L of dististilled water. this solution is equilibrated at 20\deg C with 50ml of carbon teterachloride. titration of the organic solvent requires 22.50ml of 0.005M sodium thiosulphate solution. calculate the distribution coefficient of iodine between the two solvents

Answer 1

(i) For organic fase:

$$I_2 + 2Na_2S_2O_3 = Na_2S_4O_6 + 2NaI$$

$$n(I_2) : n(Na_2S_2O_6) = 1 : 2$$

$$n(I_2) = {n(Na_2S_2O_3) \over 2}$$

$$n(Na_2S_2O_3) = c*V = 0.005 * 0.02250 = 1.125*10^{-4} mol$$

$$n(I_2) = {1.125*10^{-4} \over 2} = 5.625*10^{-5} mol$$ $$c(I_2) = {n(I_2) \over V} = {5.625*10^{-5} \over 0.050} = 1.125*10^{-3} M$$

(ii) For inorganic fase:

$$c(I_2) = {n(I_2) \over V}$$

$$n(I_2) = m/M = 0.155/253.808 = 6.107*10^{-4} mol$$

$$c(I_2) = 6.107*10^{-4}/1 = 6.107*10^{-4} M$$

(iii) Total:

$$k = {c_{org} \over c_{inorg}} = {1.125*10^{-3} \over 6.107*10^{-4}} = 1.842$$

Answer: 1.842