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By:

Prof. Dr. Ashadi

Dr. M. Masykuri, M.Si.

Chemical Education Study Program

FKIP Sebelas Maret University

Analitycal Chemistry III

Solvent Extraction

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Chemical Separations

• What is a chemical separation?• Examples:

• Filtration• Precipitations• Crystallizations• Distillation• HPLC• GC• Solvent Extraction• Zone Melting• Electrophoresis• Mass Spectroscopy

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Chemical Separations

• What is the object of the separation.• Collection of a pure product• Isolation for subsequent analysis for either quantification or identification

– Analysis » How Much?» What is it?

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Chemical Separations

• Major Industries

– Petroleum Distillation

– Distilled Spirits

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Chemical Separations

• Petroleum is a mixture of hydrocarbons.

– The larger the molecular weight the less volatile.

– So we must separate into various molecular weight fractions (different boiling points)

– The results are still complex mixtures

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Chemical Separations

•

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Distillation

• As heat is added to the system the lower volatility compounds will boil away and can be collected.– In the spirits industry the low boilers are call foreshots (~75% EtOH)– The high boilers are called feints– Congeners - Chemical compounds produced during fermentation and maturation. Congeners include esters, acids, aldehydes and higher alcohols. Strictly speaking they are impurities, but they give whisk(e)y its flavour. Their presence in the final spirit must be carefully judged; too many would make it undrinkable.

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Extraction

• Liquid – Solid

• Liquid - Liquid

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Liquid Solid Extraction (Soxhlet)whale.wheelock.edu/ bwcontaminants/analysis.html

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Liquid Solid Extractor (Bunn)

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Solvent Extraction (Liquid-Liquid)

• We want to partition a solute between two immiscible phases.– Oil and vinegar– Ethyl ether and water– Hexane water– Octanol and water.

• The two solvent should have low miscibility• Have different densities – avoid emulsion formation.

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http://oceanexplorer.noaa.gov/explorations/02sab/background/products/media/chroma1.html

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Solvent Extraction

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Replace concentration with moles over volume and let q equal the fraction in the aqueous phase.m will be the moles of solute in the entire system

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Define a new term for the ratio of the volumes of the phases

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We can do a little algebra and find an expression for q

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If it does not end up in the aqueous phase it must be in the organic phase.

• p is the term for the fraction in the organic• p + q = 1

• Giving

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Sample Problem

• You have 100.0 mL of an aqueous solution that is 100.0 mM in compound C.   This solution is extracted with 50.0 mL of diethyl ether and the aqueous phase is  assayed and it is found that the concentration of compound C that remains is 20.0 mM.  What is the equilibrium constant for this extraction system.

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Solution

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We can do multiple extraction from the aqueous phase.

• We end up with the following expression for what is left in the aqueous phase.

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Example

• How many extractions would be required to remove 99.99% of aspirin from an aqueous solution with an equal volume of n-octanol? 

• Since 99.99% must be removed the decimal fraction equivalent of this is  0.9999.  This leaves 0.0001 in the aqueous phase.  Since we have equal volumes then Vr is 1.00.

• We are able to find from the Interactive Analysis Web site that K for Aspirin is 35.5.  We plug these values into the q equation and the power is the unknown.

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Solution

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What if our compound can dissociate or participate in some other equilibrium?

• A compound such as aspirin is a carboxylic acid. We can represent this as HA.

• Do we expect the ion A- to be very soluble in the organic phase???

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Dissociation

• So if we have dissociation then less will go into the organic phase.

• Kp is the ratio of concentration of aspirin (in the un-dissociated form) in each phase. This ratio will always be the same.

• How do we account for the ion formation?

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Distribution Coefficient

• Where C is the formal

concentration of the species.

• Ca = [HA] + [A-]

• Dc will vary with conditions

• For this compound what is that condition?

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Dc

• Since the ion is not very soluble in the organic phase then we may assume that the dissociation will not happen in that phase.

• This gives us the expression to the right.

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Acid Equilibria

• What is the equilibrium?• Ka

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With a little algebra

• So if you know Kd and Ka

then you can determine Dc

as a function of H+ (pH)

• However if [H+] is much larger than Ka then Dc will equal Kd. If the [H+] is close in value to Ka then D will be related to the pH

• Plotting this we get.

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So What, Why is this useful.• Well we can now move a solute (analyte) from one phase to another. 

This can be very useful when extracting a compound that has significant chemical differences from other compounds in solution.  As a matter of fact this has been used as an interview question for prospective co-ops when I worked in industry.

• The question would go like this.  You have carried out a series of reactions and it is now time to work up the product which currently sits in an organic solution (methylene chloride).  Your expected product is a primary amine.  Which of the following solutions would you extract this methylene chloride solution with to isolate your amine.  

Your choices are: A)   Toluene. B)   0.1 N NaOH (aq) C)    0.1 N HCl (aq) D)    I never wanted to work here anyhow.  

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Separation

• So far we can tell how one compound moves from one phase to another. What if we are try to separate two compounds, A and B

• Well we might just suspect that if we find a solvent system that has different values of Dc

for each compound we could end up with most of one compound in one phase and the other compound in the opposite phase. It is not that simple.

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Example

• System I • Da = 32 Db = 0.032 (A ratio of 1000)

• Vr= 1

• Let's recall our equations • q (fraction in aqueous)  =  1 /  (DVr + 1)

• p (fraction in organic)       =   DVr / (DVr + 1)

• Vr (volume ratio)            =   Vo / Va

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Case I

• pa  =  32*1 / (32*1 + 1)  =  0.97

• pb  =  0.032*1/ (0.032*1 + 1)  =  0.03

• If we assume that we have equal moles of A and B to start then what is the purity of A in the Organic Phase?

• Purity =  moles A /  (moles A + moles B) • Purity =  0.97 / (0.97 + 0.03)  =  0.97 or 97 %

 

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Case II

• Da = 1000 Db = 1 VR = 1 (Ratio is still 1000)

• pa  =  1000*1 / (1000*1 + 1*1)  =  1000/1001  =  0.999  

• Aha! we got more a into the organic, as we would expect with a higher D value.

• Now • pb =  1*1 / ( 1*1 +1) =  1/2  = 0.5

• oh-oh • What do we get for purity of compound a now? • purity =  0.999 / (0.999 + 0.50)  =  0.666

• Yuck!

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How can we get around this issue?

• Once we have selected the solvent and pH,  then there is little that we can do to change D.     What else do we have in our control?????

• Let's look • p  =  DVr / (DVr + 1)

• Not much here except Vr  and in fact that is the key to this

problem.  Is there an optimum Vr value for the values of D that

we have?  Yes! • Our equation for this is      V r(opt)  =  (Da*Db)

-0.5

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Revisit the two cases

• So let us look at our two cases and see which will give us the optimum values.

• Case I • Da  =  32   and Db  =  0.032

  • V r(opt) =  (32 * 0.032)-0.5   =   ( 1 )-0.5  = 1

• So we were already at the optimum.

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Case II Revisited

• Case II • Da = 1000 and Db = 1

• Vr (opt)  =  (1000*1)-0.5  = 1000-0.5  =  0.032

• Which mean that when we do our extraction we will extract _______ mL of organic for each _______ mL of aqueous.

 

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Purity for Case II

• What is our purity for this system? • pa  =  1000*0.032 / (1000*0.032 + 1)  =  32/33  =  0.97

• and • pb  =  1*0.032 / (1*0.032 + 1)  =  0.032/1.032 = 0.03 

• Purity of a then is 0.97/ (0.97 + 0.03)

• Which will give us the 97% purity we had for Case I with with the Vr of

1.  

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Can we improve this purity?• If we were to extract again then we would just

remove the same proportions. We would get more compound extracted but it would be the same purity.

• What if we were to take the organic phase and extract it with fresh aqueous phase. We know that one of the two compounds will end up mostly in that aqueous phase so we should enhance the purity of the other compound in the organic phase.

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Back Extraction

• Called that since you are extracting back into the original phase.

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Back Extraction Case I Example

• Let's look at the numbers. • Da = 32

Db = 0.032 Vr = 1

• pa = 0.97 pb = 0.03

•qa = 0.03

qb = 0.97

• Let’s prepare a table. 

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Initial conditionsprior to starting back extraction

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Before Shaking Amount A Amount B

Organic Phase 0.97 0.03

Fresh Aqueous Phase 0 0

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Now we extract – shake shake shake

• How much goes to the Aqueous phase • q        which is 0.03 for A and 0.97 for B

• How much goes to the Organic phase • p        which is 0.97 for A and 0.03 for B

     

After Shaking Amount A Amount B

Organic Phase (0.97)(0.97) (0.03)(0.03)

Aqueous Phase (0.97)(0.03) (0.03)(0.97)

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Now what is the purity for A in the organic phase???  

• Purity = Amount A / (Amount A + Amount B)  = 

• 0.97*0.97 / (0.97*0.97 + 0.03*0.03) =

• 0.94/(0.94 + 0.0009) = 99.9%

• What is the yield of A (fraction of the total amount that we started with)

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Let’s do it again – Can we improve purity even more?

After second Back Extraction Amount A Amount B

Organic Phase 0.94*0.97 0.0009*0.03

Aqueous Phase 0.94*0.03 0.0009*0.97

 

Purity A   =  0.913 / (0.913 + 0.000027) =  99.997%

But our yield has dropped to 91.3%,    there is a price to pay for the added purity.

 

 

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Can We Expand This?Why Would We Want to?

• Such multiple extraction systems have been developed. • Still a viable option for preparative work.• For separations it has been replaced by HPLC

• Called Craig Counter Current Extraction.• Special glassware is used.

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Craig CCE

• Equal amounts of organic (red) and aqueous (blue) solvents with the analyte(s) are added to the A arm of the tube via port O. Fresh Aqueous Solvent is added to each of the tubes down the apparatus.

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Craig CCE

• Rock the system back and forth and to establish equilibrium.

• Allow the system to stand for the layers to separate.

• Rotate the apparatus counter clockwise about 90o to 100o.

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Craig CCE

• Rotate Back to Horizontal

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Starting Conditions

        Tube# 0 1 2 3 4

Organic Phase

0        

Aqueous Phase

1 0 0 0 0

After One Equilibrium

         Tube# 0 1 2 3 4

Organic Phase p        

Aqueous Phase q 0 0 0 0

Transfer Step 1

        Tube# 0 1 2 3 4

Organic Phase 0 p      

Aqueous Phase q 0 0 0 0

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Now here is what is in each tube/phase after equilibrium is reached.

         Tube# 0 1 2 3 4

Organic Phase

pq pp      

Aqueous Phase

qq qp 0 0 0

Now we do Transfer 2

 

         Tube# 0 1 2 3 4

Organic Phase 0 pq pp    

Aqueous Phase q2 pq 0 0 0

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Now here is what we have in each tube after the next equilibrium.

The total in each tube times either p or q as appropriate.

         Tube# 0 1 2 3 4

Organic Phase

pq2 p*2pq p3    

Aqueous Phase

q3 q*2pq qp2 0 0

We transfer again.

 

Transfer Step 3

 

         Tube# 0 1 2 3 4

Organic Phase 0 pq2 2p2q p3  

Aqueous Phase q3 2pq2 p2q 0 0

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Shake Again  Equilibrium 4

 

 

        Tube# 0 1 2 3 4

Organic Phase

pq3 p*3pq2 p*3p2q p4  

Aqueous Phase

q4 q*3pq2 q*3p2q q*p3 0

Transfer 4

 

 

        Tube# 0 1 2 3 4

Organic Phase 0  pq3 3p2q2 3p3q p4

Aqueous Phase q4 q*3pq2 3p2q2 p3q 0

 

See a trend????

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Craig CCE

• How about a binomial expansion? • (q  +  p)n  =  1

• Powers of the two terms in each tube will add up to n• Coefficients will be found from Pascal Triangle

1 1     1

1     2     1 1     3     3     1

1     4     6     4     1 1     5     10     10     5     1

1     6     15     20     15     6     1 1     7     21     35     35     21     7     1

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Craig CCE

• Or the formula

• Fr,n  =  n!/((n-r)!r!) pr q(n-r)

• n is the number of transfer and r is the tube number. You start counting at zero!

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Craig CCE

• Let's look at and example for a four tube system.• Da   = 3 p =  0.75 q = 0.25 • Db  = 0.333 p = 0.25 q = 0.75

  • What would be the purity and yield of Compound A if collected from the last in our above example.

  • Amount of A               p4   or  0.754   =   0.3164

Amount of B               p4   or  0.254   =   0.0039 • Purity of A         0.3164 / (0.3164 + 0.0039)  =  0.9878    or 98.78%

Yield of A          We collect a fraction of 0.3164  or 31.64%

• Horrible Yield!

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Craig CCE

• What if we collect the last two tubes?? • Amount of A      p4   and  4p3q  or  0.754 +  4*(0.75)3(0.25)  =   0.3164 + 0.4219 = 0.7383

Amount of B      p4   and  4p3q   or  0.254 +  4*(0.25)3(0.75)  =   0.0039 + 0.0469 = 0.0508 • Purity of A    (0.3164 + 0.4219) / (0.3164 + 0.4219 + 0.0039 + 0.0469 )  =  0.9356 or 93.56%

• Yield of A              We collect a fraction of 0.3164  +  0.4219  =  0.7383 or 73.83 % • Purity still ok and yield is much better.

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Craig system n= 200 transfers.  Da of 2.0 and Db of 4.0 pa of 0.666 pb of 0.800.

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Final Formulas(1)

• rmax =   np   = nDVr/(DVr +1)

• To find the separation between two peaks we would use. • Drmax = (rmax)a - (rmax)b  =  n(pa-pb)

  • The Gaussian distribution approximation for our binomial expansion would be (when n>24) • Fr,n =  (2p)-0.5(npq)-0.5 exp-[((np-r)^2)/2npq]

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Final Formulas(2)

• The width of the distribution through the system would be: • w = 4s = 4(npq)0.5 • Resolution would be • R = Drmax/w = Drmax/4 s• or • R = nDp/(4(npq)0.5)  =  n0.5 Dp / 4(pq)0.5