A chemical "X" is included in a water in a pond.

Question

1)5dm3 of water from the pond is taken and is mixed with 200cm3 CHCl3.Then CHCl3 layer is taken out separately and left to vaporize. Then the remaining mass of X is measured.There was 10mg.

2)The remaining aqueous zone after (1) is again mixed with another 200cm3 of CHCl3 and again the mass of remaining "X" is measured. There was 5mg.

1-Find kD 
2- Find the percentage of X will remain after four times of the above described is done?
3-Calculate the X in the pond in ppm
4-If ,[X(CHCl3)]/[X(CCl4)]=2,and if 200cm3 of CCl4 is used in (1),except for CHCl3, find the remaining X after one time the above described is done?

My thoughts:

I don't have much to say cause I don't quite really know how to answer this question properly.

If [ X(CHCl3)]/[X(H2O)]=2 ,I know that X dissolves in CHCl3 two times as it dissolves in water.

1-I took the molar mass of X as M(mg) ( which I think won't be needed in calculations as we are considering ratios) and the mass of X in a 1dm3 of water as m(mg) as it simply gives us the concentration of X in water.

So after (1), KD=[X(CHCl3)]/[X(H2O)]=[(10/200)*1000]/[ (m-10)]
kD=[50/(m-10) ]--->(#)

So X(CHCl3) : X(H2O)
     50     :  (m-10)

So from (2) kD=[5]/[ (m-10-5)] =[5]/[(m-15)-->(@)

So from (#) and (@)

[(5/(m-15)] = [50/(m-10)]
Then I get m=(140/9)

So from (@),kD=[(5)/( (140/9) - 15 )]=9

I think I didn't make any mistakes.

But I really don't think this is the method.I feel like there should me a more quicker and elegant method to solve this...

2-For this part I don't think we should do calculations 4 times as I did in the above part.
There MUST be a cleverer method.I hope someone will explain that.

3- Do we need to find "M" to convert moldm-3 to ppm(mgdm-3)?

Shenaya · Answer

Or for the first part we should consider m as the mass of in 5dm3 and continue the calculation,cause we are taking two parts of water each 1dm3,from 5dm3 of water?

DrBob222 · Answer

I assume when you say 10 mg remained that it was 10 mg in the aqueous layer.
1. f = [1+ (Kd*Vo/Va)]^-n
f is the fraction X recovered.
Kd is distribution coefficient with organic layer/aqueous layer usually written as Ko/a - Kc.
Vo is volume of organic layer.
Va is volume of aqueous layer.
n = number of extractions.
Use the numbers (in mg) from 1 and 2 for this to calculate Kd.
Vo = 200 cc
Va = 5000 cc
Kd = ?
f = 0.5 (first time left 10 mg in aq layer; second time left 5 mg in aq layer so that is 0.5 for fraction.
n = 1. Solve for Kd

2. Straight forward.

3. First you need to go through knowing Kd = [(org mg/200 cc)/(aq mg/5000)]. You know you are left with 10 mg in the aq layer, Knowing Kd you can solve for how much is extracted in the CHCl3 (organic) layer. Add amount in CHCL3 layer to amout in aq layer to find total amount you started with. Then ppm = mg/dm^3

4. Kd = 2 = [(mg org layer/200)/(mg aq layer/5000)]
and solve for mg in aq layer. You will have 2 = [(total mg X/200)/(z/5000)] and solve for z and you let z = total mg X (you have that from above) -x where x is mg left left in the aq layer.

Hope this helps.