Given a solution containing 5.00 g of compound A in 100 mL of water and a distri
ID: 992417 • Letter: G
Question
Given a solution containing 5.00 g of compound A in 100 mL of water and a distribution coefficient (K) of 4.0 between ether and water: a) How much A will be removed from the water by one extraction with 100 mL? b) How much A will be removed from the water by two extractions with 50 mL of ether? Given a solution containing 5.00 g of compound A in 100 mL of water and a distribution coefficient (K) of 4.0 between ether and water: a) How much A will be removed from the water by one extraction with 100 mL? b) How much A will be removed from the water by two extractions with 50 mL of ether? Given a solution containing 5.00 g of compound A in 100 mL of water and a distribution coefficient (K) of 4.0 between ether and water: a) How much A will be removed from the water by one extraction with 100 mL? b) How much A will be removed from the water by two extractions with 50 mL of ether?Explanation / Answer
We know that, partition coefficient, K = solubility in organic solvent/solubility in water
(a) Total mass of ompound is 5.0 g. let's say X grams of it are present in organic solvent, ether. Then grams of it soluble in water would be = 5.0 - X. Also, we are doing the extraction by taking 100 ml of ether.
Plug in the values in the formula.
4.0 = (X/100)/[(5.0 - X)/100]
on cross multiply...
4.0 x (5.0 - X)/100 = X/100
(20 - 4X)/100 = X/100
20/100 - 4X/100 = X/100
add 4X/100 to both sides..
20/100 = 4X/100 + X/100
20/100 = 5X/100
multiply both sides by 100
20 = 5X
X = 20/5 = 4
So, 4 grams of compound A could be extrated into the ether.
(b) Now we want to double the extraction two times by taking 50 ml of ether each time. Formual will again be same.
Let's say X grams of compound A are in ether then 5.0 - X would be in water.
4.0 = (X/50)/[(5.0 - X)/100]
on cross multiply...
4.0 x (5.0 - X)/100 = X/50
(20 - 4X)/100 = X/50
20/100 - 4X/100 = X/50
20/100 = X/50 + 4X/100
20/100 = X/50 + 2X/50
20/100 = 3X/50
multiply both sides by 50
20 x 50/100 = 3X
10 = 3X
X = 10/3 = 3.33
So, 3.33 grams of compound A are extracted in first extraction by taking 50 ml of ether.
Now, we will do the second extraction by taking 50 ml of ether.
In total we had 5.0 g of the compound and 3.33 grams have already been extracted. So, remaning amount of compound A = 5.0 - 3.33 = 1.67 g
let's say, Z grams of compound are in ether and remaining in water.
4.0 = (Z/50)/[(1.67 - Z)/100]
4.0 x (1.67 - Z)/100 = Z/50
6.68/100 - 4Z/100 = Z/50
6.68/100 = Z/50 + 4Z/100
6.68/100 = Z/50 + 2Z/50
6.68/100 = 3Z/50
multiply both sides by 50
6.68 x 50/100 = 3Z
3.34 = 3Z
Z = 3.34/3 = 1.11
So, in second extraction with 50 ml ether we get 1.11 g of compound A.
So, total amount of A removed from water = 3.33g +1.11g = 4.44 g
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