chem question help please

hi, please could I have some help with this question? I thought Kpc= [Hdichloromethane]/[H water] and I would do (x/60)/(2.5/100). H is dissolved in 60cm3 of solution and 100 cm3 water?
Question: https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FChemistry%2FA-level%2FPast-Papers%2FCAIE%2FPaper-4%2FQP%2FNovember%202018%20(v1)%20QP.pdf
question ii on page 5.
Thanks!

Reply 1

Al345

Original post by anonymous56754

Could you upload an image of your working out?

Reply 2

TypicalNerd

Kpc between DCM and water implies they want you to use

Kpc = [H (DCM)]/[H (aq)]

Only 50 cm^3 of the original 100 cm^3 of solution is used, so the total mass of H is (2.50 g)/2 = 1.25 g.

10 cm^3 of DCM is used. But recall this should form a separate layer since water and most halogenated organic solvents are immiscible. As such, you don’t combine the volumes.

Let’s call the amount of H in the DCM layer X g. With this established, the mass of H in the water layer is (1.25 - X) g.

Since Kpc is dimensionless and both solutions contain the same solute, we can cheat and simply use the concentrations in g cm^-3 (or any other unit, provided both concentrations have the same unit) without converting to mol dm^-3. We can therefore say that the concentration of H (aq) is (1.25 - X)/50 and the concentration of H (DCM) is X/10.

We now have

4.75 = (X/10) / ((1.25 - X)/50)

= 5X / (1.25 - X)

5.9375 - 4.75X = 5X

5.9375 = 9.75X => X = 0.61 g (2 dp)