Bis(2-ethylhexyl) hydrogen phosphate (di-(2-ethylhexyl)phosphoric acid, DEHPA) is used in the extrac — Organic Chemistry Chemistry Question
Uranyl extraction
Bis(2-ethylhexyl) hydrogen phosphate (di-(2-ethylhexyl)phosphoric acid, DEHPA) is used in the extraction of uranyl ions from an aqueous solution to an organic solvent. This water-to-kerosene extraction is known as the “Dapex process”.
DEHPA (HA):
i) Is a weak acid that is partially dissociated in water, with a dissociation constant
HA A− + H+
ii) It can be extracted to karosene with a distribution constant:
iii) It forms a hydrogen-bonded dimer in non-polar organic solvents, with a dimerization constant
HA (HA)2
iv) When dissociated in an aqueous solution, it forms a neutral compound with the uranyl ion in a ratio of 2:1 (Note: In real systems, the structure of the neutral compound can vary).
2 A– + UO2 2+ UO2A2
This neutral compound can be extracted to kerosene with a distribution constant
Assume that:
• The concentration of DEHPA before the extraction: cHA,org,0 = 0.500 mol dm-3 and cHA,aq,0 = 0.000 mol dm-3.
• cUO2+ ≪ cHA, therefore it is possible to omit the concentration of UO2A2 in the mass balance of HA in both the aqueous and organic phase.
• The volume ratio is Vorg/Vaq = 1.00.
Uranyl ions also form hydroxo complexes
UO2 2+ + i OH– [UO2(OH)i] 2– i where i = 1 - 4
(Note: For clarity square brackets as a symbol for a complex were omitted in the numerator.)
Decimal logarithms of the overall complexation constants are as follows:
log β1 = 10.5, log β2 = 21.2, log β3 = 28.1, log β4 = 31.5.
Considering that the pH of the aqueous phase after reaching equilibrium equals to 1.7, calculate the yield of uranyl ions extraction.
Hint: First, calculate the concentration of DEHPA in the organic phase after reaching the equilibrium with the aqueous phase, i.e. calculate [HA]org. To do this, use the mass balance of HA. Consider the different forms of DEHPA in both the organic and aqueous solution.
Model Answer
First, [HA]org is calculated:
cHA,org,0 = 2 [(HA)2]org + [HA]org + [HA]aq + [A−]aq
The concentration of UO2A2 is omitted as recommended in the introductory text.
From the definition of Kp,HA, KD,HA and Ka,HA, [HA]org can be obtained by solving the quadratic equation
i.e.
Considering that the proton concentration corresponds to the analytic concentration of HNO3, [H+]aq = 10−pH = 2.00×10−2 mol dm−3, we get [HA]org = 3.41×10−3 mol dm−3.
Next, the uranyl ion distribution ratio, Dc,UO2+ is expressed as:
Using β2,UO2A2, KD,UO2A2 and βi for [UO2(OH)i]2-i complexes, Dc,UO2+ can be expressed as
The concentration of hydroxyl ions is obtained from the concentration of protons,
For [H+]aq = 2.00×10-2 we get [OH-]aq = 5×10-13
Casting this value, [HA]org = 3.41×10−3 and all the necessary constants into the expression for the distribution ratio, we obtain
Dc,UO2+ = 5.61
Then the yield R defined as
can be calculated, providing the final result of
Considering that the pH of the aqueous phase after reaching equilibrium equals to 10.3, calculate the yield of uranyl ions extraction.
Hint: Use the same procedure as in task 12.1.
In both cases, consider only the equilibria that have been mentioned so far.
Model Answer
For the conditions of [H+] = 10−pH = 5.01×10-11 and using the same calculation procedure, we get
[HA]org = 1.50×10−5
Dc,UO2+ = 1.22×10−4
and the yield R = 0.0122 %.