Suppose you need to extract an organic substance S from water into ether, where the distribution constant Kdist = [S]org/[S]aq is at first unknown.
a) By testing the solubility of the material in each solvent under saturating conditions at room temperature, you find that it has a solubility in water of 12 g L–1 but it is 36 g L–1 in ether. What is the value of Kdist for this solvent pair? Hint: Assume you start with 48 g.
b. If you use equal volumes of water and ether in your extraction, what percentage of the material will be transferred to the ether phase? Hint: Suppose you have 48 g of substance and use 1 L of each solvent, then use the solubilities in each phase and the Kdist value from part (a).
(c) What is the minimum volume of water required to dissolve 7.2 g of your substance?
d.Suppose now that you have a saturated solution in water. If you then extract this with one sixth the volume of ether as you have water, what percentage of the material will find itself in the ether phase? Hint: The Kdist value will still apply.
We would be interested in knowing how much of this you know how to do. And where do you have the hang ups?
a) To find the value of Kdist, we need to divide the solubility of the substance in ether ([S]org) by its solubility in water ([S]aq). In this case, the solubility of the substance in water is 12 g L–1 and in ether it is 36 g L–1.
Kdist = [S]org / [S]aq
Kdist = 36 g L–1 / 12 g L–1
Kdist = 3
Therefore, the value of Kdist for this solvent pair is 3.
b) To find the percentage of the material transferred to the ether phase, we need to use the solubilities in each phase and the Kdist value.
Let's assume you start with 48 g of substance and use 1 L of each solvent. From part (a), we know that Kdist = 3.
The amount of substance in the water phase = solubility in water * volume of water
= 12 g L–1 * 1 L
= 12 g
The amount of substance in the ether phase = solubility in ether * volume of ether
= 36 g L–1 * 1 L
= 36 g
The percentage of the material transferred to the ether phase = (amount in ether phase / initial amount) * 100
= (36 g / 48 g) * 100
= 75%
Therefore, 75% of the material will be transferred to the ether phase.
c) To find the minimum volume of water required to dissolve 7.2 g of the substance, we can use the solubility in water.
Let's assume the minimum volume of water required is V.
The amount of substance in the water phase = solubility in water * volume of water
7.2 g = 12 g L–1 * V
Solving for V, we get:
V = 7.2 g / 12 g L–1
V = 0.6 L
Therefore, the minimum volume of water required to dissolve 7.2 g of the substance is 0.6 L.
d) To find the percentage of the material in the ether phase when one-sixth the volume of ether is used compared to water, we can still use the Kdist value.
Let's assume you start with a saturated solution of the substance in water.
The amount of substance in the water phase = solubility in water * volume of water
= 12 g L–1 * 1 L
= 12 g
Now, one-sixth of the volume of ether is used compared to water. So, the volume of ether used would be (1/6) * 1 L = 1/6 L.
The amount of substance in the ether phase = solubility in ether * volume of ether
= 36 g L–1 * (1/6) L
= 6 g
The percentage of the material in the ether phase = (amount in ether phase / initial amount) * 100
= (6 g / 12 g) * 100
= 50%
Therefore, 50% of the material will find itself in the ether phase.