# 40cm^3 of 0.06 mol dm^-3 aqueous ethanoic acid solution has been shaked with a 10 cm^3 of solvent called L which is not soluable in water.and then came in to the equilibrium state in 300K.30 cm^3...

40cm^3 of 0.06 mol dm^-3 aqueous ethanoic acid solution has been shaked with a 10 cm^3 of solvent called L which is not soluable in water.and then came in to the equilibrium state in 300K.30 cm^3 of 0.15 mol dm^-3 NaOH has been used to react with 10cm^3 of water layer in above system.Then the remaining amount of the above system has been boiled in 315K and made it to a new equilibrium state.29.5 cm^3 of 0.15 mol dm^-3 NaOH has been used to completely react with 10 cm^3 of water layer in the new equilibrium state.

Find the Distribution constant of ethanoic acid in water and L solvent in each temperature.

jeew-m | Certified Educator

`CH_3COOH_(aq) harr CH_3COOH(L)`

`K_D = [CH_3COOH_(L)]/[CH_3COOH_(aq)]`

Amount of `CH_3COOH` initially mixed `= 0.06/1000xx40 = 0.0024mol`

Let us say xmol from this 0.0024mol has gone to the organic liquid. So in the water remaining `CH_3COOH` is (0.0024-x).

So at the equilibrium we have (0.0024-x) moles in 40ml water and x moles in 10ml of organic solution.

Then when take 10ml of water layer and titrated it with NaOH.

`NaOH+CH_3COOH rarr CH_3COONa+H_2O`

Amount of NaOH consumed `= 0.15/1000xx30 = 0.0045`

This means there were 0.0045 moles in 10ml of water. But initially we had only 0.0024moles in the system. So this cant happen. Something wrong in the question.