Library / Solution chemistry, equilibria and kinetics / SCH-35
Solution chemistry, equilibria and kinetics
Recovery after repeated solvent extractions
Percentage of a solute taken into the organic phase after N extractions, each with a fresh portion of volume Vorg, from a fixed aqueous volume and a constant distribution ratio D.
- Ref
- SCH-35
- Rev
- v1
- Field
- Solution chemistry, equilibria and kinetics
- Result
- recovery [%]
After one extraction the fraction left in the aqueous phase is Vaq/(Vaq + D Vorg). With N fresh portions the fractions multiply, so the fraction remaining is [Vaq/(Vaq + D Vorg)]ᴺ and the recovery is 100 × (1 − that). The calculation uses a log form so that it stays accurate for small fractions. Several small portions beat one large portion of the same total volume.
Percentage of a solute taken into the organic phase after N extractions, each with a fresh portion of volume Vorg, from a fixed aqueous volume and a constant distribution ratio D.
Calculator
- Fraction left in the water0.125
- Recovery from one portion50 %
- Recovery from one extraction with all the solvent at once75 %
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Three 25 mL portions recover 87.5 % of the solute, leaving 12.5 % in the water.
- The same 75 mL used in one go would recover only 75 %, so splitting the solvent pays off.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (%) | Local step | Change in result |
|---|---|---|---|---|---|---|
| D Distribution ratio (organic/aqueous) | 0.01 to 1,000 | 0.7463 | 100 | 99.25 | +10% | +1.95 % |
| Vorg Volume of each organic portion | 1 to 1,000 mL | 11.1 | 100 | 88.9 | +10% | +1.95 % |
| Vaq Volume of the aqueous phase | 1 to 1,000 mL | 24.87 | 100 | 75.13 | +10% | −2.14 % |
| N Number of extractions | 1 to 10 | 50 | 99.9 | 49.9 | +1 | +7.14 % |
Sweep of Number of extractions
| Number of extractions | Solute recovered (%) |
|---|---|
| 1 | 50 |
| 2.125 | 77.07 |
| 3.25 | 89.49 |
| 4.375 | 95.18 |
| 5.5 | 97.79 |
| 6.625 | 98.99 |
| 7.75 | 99.54 |
| 8.875 | 99.79 |
| 10 | 99.9 |
Sweep one variable
Variables and how each one contributes
Solute recovered
What the equation gives you. Everything else feeds it.
Distribution ratio (organic/aqueous)
The distribution ratio, organic over aqueous. A larger D takes more solute into each portion, so the recovery rises. Raising it by 10 % (to 4.4) moves the result by +1.95 %, from 87.5 to 89.2 %.
Volume of each organic portion
The volume of each organic portion. A larger portion takes more solute each time, so the recovery rises. Raising it by 10 % (to 27.5) moves the result by +1.95 %, from 87.5 to 89.2 %.
Volume of the aqueous phase
The volume of the aqueous phase, in the same unit. A larger aqueous volume leaves more solute behind, so the recovery falls. Raising it by 10 % (to 110) moves the result by −2.14 %, from 87.5 to 85.63 %.
Number of extractions
The number of extractions with fresh solvent. More extractions raise the recovery, with falling returns. Raising it by 10 % (to 3.3) moves the result by +2.68 %, from 87.5 to 89.85 %.
Worked example with the values above
- Fraction left after one extraction: Vaq/(Vaq + D Vorg) = 100 / (100 + 100) = 0.5.
- After N = 3: (0.5)^3 = 0.125.
- Recovery = 100 × (1 − 0.125) = 87.5 %.
Assumptions and limits
- A 1:1 complex, or a single neutral species partitioning, with constants in concentration units.
- Equilibrium reached in each step, and no side reactions unless the conditional constant includes them.
Common mistakes
- Using the total ligand concentration for the free ligand when binding depletes it.
- Using the absolute EDTA constant at a pH where much of the EDTA is protonated.
- Mixing the units of the two phase volumes.
Statistics and uncertainty
Formation constants depend on the medium and the temperature, and a quoted constant for a different ionic strength can be off by a factor of two or more. Extraction results are sensitive to pH near the pKa.
Where it is used
- Designing complexometric titrations and metal-binding experiments.
- Planning solvent extractions and pH-controlled separations.
Maths you need
- Formation constant: Kf = [ML]/([M][L])
- Partition coefficient: the ratio of a species in two phases
- Mass balance: total equals free plus bound