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.
Please help me on this.
1 Answer | Add Yours
`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.
We’ve answered 319,858 questions. We can answer yours, too.Ask a question