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Using small samples of ether is better in an extraction?

  • #1

Homework Statement



Show that using small volumes of ether solvent in an extraction from aqueous phase is better than using larger volumes. The total volume of organic solvent, amount of material to extract, aqueous volume and distribution constant are the same. You can do this with an equation or words.

Homework Equations



Kdistribution = g/100mL ether layer / g/100mL aqueous layer


The Attempt at a Solution



So say we have 10 small samples of 10g ether and then 2 big samples of 100g ether. Our aqueous phase is constant at 100g lets say.

So, 10/100 = .1 = K
Since there are 10 extraction .1 x 10 = 1

on the other hand, 2 big samples of 100g ether
= 100/100
= 1
2 extractions:
= 1x2
= 2

I'm guessing the distribution constant is better when it's lower so that's why small samples of ether are used. I have a feeling I'm approaching this question completely wrong though.
 

Answers and Replies

  • #2
Borek
Mentor
28,402
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Assume you have 100 mL of ether in total and 100 mL of water solution.

How much of the substance will be left in water after one extraction using 100 mL of the ether?

How much of the substance will be left in water after two extractions each using 50 mL of the ether? Note you have to calculate how much was left after first extraction, and use this number to calculate the result of the second extraction.

Which way gives better yield?
 
  • #3
You're really smart Borek, lol. Assume there is 0.5 g of benzoic acid that needs to be extracted. Then:

.5g/100mL = 0.005g/mL
&
.5g/50mL = 0.01g/mL -> x2 extractions = 0.02g/mL

Now I'm stuck, does this mean that there is 0.02g of benzoic acid per mL of water or ether?
 
  • #4
Borek
Mentor
28,402
2,801
To calculate concentration after extraction you have to use distribution coefficient. You don't have to know its value - if you can express mas of the substance as a function of K, that will be enough. But if that's too difficult for you, assume K=10 and do calculations on numbers.
 
  • #5
k = 50mL / 100 mL = .5
k= 100/100 = 1

Isn't this what I did orginally?
 
  • #6
Borek
Mentor
28,402
2,801
K is not ratio of volumes, K is ratio of concentrations.
 
  • #7
Ok well here's where I'm a little confused. I gave the equation of above as:

Kdistribution = g/100mL ether layer / g/100mL aqueous layer

So am I wrong in saying that all of the sample is in the aqueous layer at first and none in the ether layer? So it would be 0g/100mL in the numerator and watever amount in the denominator?
 
  • #8
Borek
Mentor
28,402
2,801
You add ether to water solution. After the equilibrium is reached benzoic acid is present in both layers. You know two things. First, amount of benzoic acid has not changed - sum of amounts in both layers must be identical to initial amount in the water layer. This is a just a mass conservation. Say initially there is a 1g of benzoic acid in 100 mL of water solution. Second - you know what must be ratio of concentrations (given by the distribution coefficient). That gives two equations in two unknowns (concentrations). Solve - and you will know how much benzoic acid was left in water layer. Thats how you calculate result of a single extraction.
 
  • #9
Ok, I think I understand now:

k=10

10=g/50mL ether/ 1g/100mL water
10= g/50mL ether
.2 = g/mL ether

10=g/100mL ether/ 1g/100mL water
.1=g/mL ether

So there is more benzoic acid left when using 50mL (small amount) than 100mL (large amount) of ether. Please tell me I did it right lol
 
  • #10
Borek
Mentor
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No, unfortunately you are still wrong, I think you are confused by your own notation.

[tex]K=\frac {C_e} {C_w}[/tex]

[tex]C_e V_e + C_w V_w =m[/tex]

Where Ce - concentration in ether, Cw - concentration in water, m - total mass of the acid. We started with 100 mL of water solution that contained 1 g per 100 mL, so

[tex]m = C^i_w V_w = 100 mL \times \frac {1g}{100 mL} = 1g[/tex]

Index i is here to remind you that this was INITIAL concentration, not the one at equilibrium. Now you know enough to solve for equilibrium Ce and Cw.
 
  • #11
LaTeX Code: C_e V_e + C_w V_w =m

LaTeX Code: C_e V_e + C_w V_w = 1

LaTeX Code: (1/100)(100) + (x/100)(100) = 1

1+ 100x/100 = 1

1+ x = 1

x= 0

This would be the initial amount in ether, correct? Arghh I'm still stuck :(
 
  • #12
Borek
Mentor
28,402
2,801
I guess by x you mean something that - in my notation - would be Cie - initial concentration of benzoic acid in ether? If so, that's OK. Initially there is no benzoic acid in ether, that's correct.

However, there was no need to calculate it - it was given. Your unknowns are Ce and Cw.
 

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