Chloroform can be used to extract caffeine from an aqueous solution. If the distribution?

Question

Chloroform can be used to extract caffeine from an aqueous solution. If the distribution
constant for the equilibrium between caffeine dissolved in water and caffeine dissolved in
chloroform is 0.0714, what volume of chloroform will be needed to extract 85% of the
caffeine from 120.0 mL of an aqueous caffeine solution?
please give more detailes.

Answer 1

Caffeine is soluble to some extent in both water and chloroform, however water and chloroform do not mix. Therefore if we have an aqueous solution of caffeine in physical contact with some liquid cloroform (as in a separation funnel) then some of the caffeine will move from the water and spread into the chloroform. We can represent this kind of reaction by the following equation:

caf(aq) <--> caf(ch)

The caffeine can move back and forth between the water(aq) and the chloroform(ch). We say that the caffeine partitions between the water and the chloroform. Just like any other reaction there is an equillibrium associated with the partioning reaction, and the equillibrium constant (it is called a distribution constant or partition constant in this case) is defined in the same way as for other reactions: concentration of products over concentration of reactants. In this case the distribution constant would be given by:

Kd = [caf(ch)] / [caf(aq)] or rearanging [caf(aq)] * Kd = [caf(ch)].

Suppose we have an initial volume of aqueous caffeine solution V(aq), and we mix it with a volume of chloroform V(ch). After we have mixed them and let them come to equillibrium. There will be a certian concentration of caffeine in the aqueous phase of [caf(aq)], and a certian concentration of caffeine in the chloroform of [caf(ch)].

The moles of caffeine in the water can be calculated by the following:

moles caffeine in water = V(aq)[caf(aq)]

The moles of caffeine in the chloroform can be calculated in a simmilar manner:

moles caffeine in chloroform = V(ch)[caf(ch)]

The total moles of caffeine in the system will be the moles of caffeine in the water plus the moles of caffeine in the chloroform.

total moles caffeine = V(aq)[caf(aq)] + V(ch)[caf(ch)]

Now, the fraction of caffeine left in the water (fraction left = fl) is the moles of caffeine in the water at equillibrium divided by the total moles of caffeine in the system as shown below.

fraction left in water at equillibrium = (moles caffeine in water) / (total moles caffeine) or in other words:

fl = V(aq)[caf(aq)] / (V(aq)[caf(aq)] + V(ch)[caf(ch)])

Using the equillibrium expression we may make the following substitution in the above equation: [caf(aq)]*Kd = [caf(ch)].

substituting this in we get the following:

fl = V(aq)[caf(aq)] / (V(aq)[caf(aq)] + V(ch)[caf(aq)]*Kd)

Now we see that all the [caf(aq)] factors cancel leaving us with:

fl = V(aq) / (V(aq) + V(ch)Kd)

Since we are not actually interested in the fraction left but are rather interested in the fraction extracted (fe) the equation for the fraction extracted will simply be one minus the fraction left or:

fe = 1 - fl = 1 - V(aq) / (V(aq) + V(ch)Kd).

If a percent is desired we simply multiply the fraction by 100% or:

%fe = (1 - V(aq) / (V(aq) + V(ch)Kd))*100%.

For this particular problem you want an 85% extraction, you have 120ml of aqueous volume, and a Kd of 0.0714. Putting these values into the above equation we arrive at:

85% = (1 - 120ml / (120ml + V(ch)0.0714))*100%.

All that is left to do is solve for the volume of chloroform needed (V(ch)). I will leave the rest of the algebra to you, but I will give you a multiple choice so you can check your answer.

Volume of chloroform needed is:
A) 11.9ml
B) 97.5ml
C) 524ml
D) 982ml
E) 2210ml
F) 9520ml
G) 10500ml
H) 53600ml

Answer 2

The distribution constant (or partition ratio), is the equilibrium constant for the distribution of an analyte in two immiscible solvents. For a particular solvent, it is equal to the ratio of its molar concentration in the stationary phase to its molar concentration in the mobile phase, also approximating the ratio of the solubility of the solvent in each phase. The term is often confused with partition coefficient or distribution coefficient, and is often represented by Kd.

Hopefully you can figure out the rest.

Answer 3

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