1. Calculate the distribution coefficient of a solution of 15.1g caffeine dissolved in 25mL of water. The solution is extracted once with 50mL of chloroform. The solubility of caffeine in water is 1.5g/100mL at 25C. The solubility of caffeine in chloroform is 14.0g/100mL at 25C
Ignore first two phrases, they are irrelevant to the question.
2. A 50mL aliquots of a 0.3000M of compound A is extracted with 50.0mL of diethyl ether. Titration of the extracted aqueous solution required 23.50mL of 0.5000M NaOH. Calculate the distribution coefficient of compound A.
Not sure if this is right but this is what I did:
Kd = (0.5000M * 23.5mL)/(0.3000M * 50mL) = 0.783
Is this right?
No, for two reasons - one is your fault, the other is fault of whoever wrote the question. You are not told how the substance A reacts with NaOH (1:1? 1:2? something else?) so technically it is impossible to answer the problem. And even assuming - as you did, probably not even knowing you were assuming anything - they react 1:1, you still did wrong. Try to systematically calculate concentrations in both phases remembering, that the total amount of substance doesn't change.
3. A 7.3g sample of propanoic acid is dissolved in 300mL of water. How much 0.1N NaOH is required to neutralize a 50mL aliquot of propanoic acid?
I used C1V1 = C2V2
Right and wrong. It will give the correct answer, but do you know why?
So the concentration of propanoic acid is:
moles = 7.3g/80.06g/mol = 0.0912moles
Check the molar mass of the propanoic acid.
And yes, the final volume looks rather large - but you are in the right ballpark.
#1 So is it Kd = (14.0/100)/(1.5/100) = 9.3?
- why do you ignore the first two phases? Is that to throw you off or something? Does that mean that Kd doesn't change no matter the volumes/weights of caffeine or chloroform?
#2 So I don't assume they react 1:1? How do you know this?
You find the concentrations of both and then Kd = concentration of the organic phase/concentration of the aqueous phase?
So this is what I did: I found the moles of compound A = 0.015 (before extraction) and moles of NaOH = 0.01175 (after extraction with ether)
So 0.015 - 0.01175 = 0.00325 mol (ether phase)
Kd = conc'n of ether phase/conc'n of aqueous phase
= (0.00325 mol/0.05L)/(0.01175 mol/0.05L) = 0.277
#3 My bad! The molar mass of propanoic acid is 74g/mol so moles = 0.0986
So the final volume is 164 mL (0.164L) - is this right?
- why is C1V1 = C2V2 right and wrong?