21-Mat-A3 Structure and Characterization of Materials · December 2014
Question 5 of 7: Hydrometallurgy / Solvent Extraction (20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2014 — 10-Met-A3, Metal Extraction Processes. Three hours, closed book, one approved calculator (Casio or Sharp). Seven problems of 20 marks each; the rubric asks for any five, and only the first five in the answer book are marked. All seven are solved here, because this set is a study resource rather than an exam script.
Note on the exam title. The printed exam header reads 10-Met-A3, Metal Extraction Processes. The content is extractive metallurgy — mineral processing, roasting, refining, hydrometallurgy and aluminum production — and is answered as such.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
F. Habashi, Textbook of Pyrometallurgy — roasting, smelting, mass and heat balances.
F. Habashi, Textbook of Hydrometallurgy, 2nd ed. — solvent extraction, distribution ratios, phase-ratio effects.
T. Rosenqvist, Principles of Extractive Metallurgy, 2nd ed. — refining, molten-salt electrolysis, vacuum metallurgy.
B. A. Wills and J. Finch, Wills' Mineral Processing Technology, 8th ed. — comminution, classification, flotation, thickening.
D. R. Gaskell, Introduction to the Thermodynamics of Materials — heat capacities, transformation and reaction enthalpies.
W. G. Pfann, Zone Melting, 2nd ed. — zone refining theory and the normal-freezing distribution equation.
ASM Handbook, Vol. 2 (Nonferrous Alloys) and the CIM / Canadian Institute of Mining practice literature for Canadian smelter and refinery practice.
(i) Extraction. Extraction is the transfer of a solute (typically a metal ion, via a complex it forms with a selective reagent) from an aqueous phase into an immiscible organic phase brought into contact with it, driven toward the phase in which the complex is more stable. At equilibrium the split is described by the distribution ratio $D=[\text{solute}]_{org}/[\text{solute}]_{aq}$.
(ii) Stripping. Stripping is the reverse operation: the loaded organic phase is contacted with a fresh aqueous phase whose chemistry (usually a strong acid, or a shift in pH or complexing-ion concentration) is chosen to favour the aqueous side, transferring the metal back out of the organic and regenerating the organic for reuse while producing a concentrated strip liquor ready for electrowinning or precipitation.
(iii) Phase ratio. The phase ratio is the ratio of the volumes (or volumetric flow rates) of the two contacted phases in a mixing stage or circuit, commonly written aqueous:organic (A:O). Together with the distribution ratio $D$, it fixes the fraction of solute extracted (or stripped) per stage, and it is the operating variable a plant adjusts to control extraction efficiency without changing chemistry.
(iv) Extractant. The extractant is the active organic reagent, dissolved in a diluent, that selectively complexes or chelates the target metal ion at the aqueous–organic interface — for example, an aldoxime/ketoxime blend (e.g. LIX-type reagents) for copper, an organophosphorus acid (D2EHPA) for zinc/cobalt separation, or tributyl phosphate (TBP) for uranium.
(v) Diluent. The diluent is the bulk, chemically inert organic liquid — typically an aliphatic or mildly aromatic kerosene-type hydrocarbon — that dissolves the extractant and carries it, controlling the organic phase's viscosity, density and phase-disengagement (settling) behaviour without itself participating in the extraction chemistry.
Part (b) — enrichment factor.
Given. Equal initial amounts of ions A and B are present in the aqueous feed. Their distribution ratios are $D_A=18$ and $D_B=3$. The aqueous-to-organic phase ratio $R=V_{aq}/V_{org}$ is 2 in (i) and 12 in (ii).
Given data
Quantity
Symbol
Value
Distribution ratio, ion A
DA
18
Distribution ratio, ion B
DB
3
Aqueous:organic ratio, case (i)
R
2
Aqueous:organic ratio, case (ii)
R
12
Find. The enrichment factor — the A:B ratio in the organic extract relative to the 1:1 ratio in the feed — at each phase ratio.
Approach. Write the fraction of each ion extracted into the organic phase in one equilibrium contact as a function of its own distribution ratio and the phase ratio, then take the ratio of the two extracted fractions, since the feed starts 1:1.
Write the single-stage extraction fraction. For an ion with distribution ratio $D_i$ contacted at aqueous:organic ratio $R=V_{aq}/V_{org}$, a mass balance on the solute (starting amount $n_0$, all in the aqueous phase) gives the fraction extracted into the organic phase
$$E_i=\frac{D_i}{D_i+R}$$
Because $D_A$ and $D_B$ differ, $E_A$ and $E_B$ differ at the same $R$, and that difference is what produces enrichment.
Define the enrichment factor from equal starting amounts. With $n_{A,0}=n_{B,0}=n_0$, the amounts reporting to the organic phase are $n_{A,org}=n_0E_A$ and $n_{B,org}=n_0E_B$. The A:B ratio in the extract relative to the feed's 1:1 ratio is
$$\mathrm{EF}=\frac{n_{A,org}/n_{B,org}}{n_{A,0}/n_{B,0}}=\frac{E_A}{E_B}=\frac{D_A\,(D_B+R)}{D_B\,(D_A+R)}$$
Evaluate part (i), $R=2$.
$$\mathrm{EF}=\frac{18\,(3+2)}{3\,(18+2)}=\frac{18\times5}{3\times20}=\frac{90}{60}$$
$$\boxed{\mathrm{EF}(R=2)=1.50}$$
Evaluate part (ii), $R=12$.
$$\mathrm{EF}=\frac{18\,(3+12)}{3\,(18+12)}=\frac{18\times15}{3\times30}=\frac{270}{90}$$
$$\boxed{\mathrm{EF}(R=12)=3.00}$$
Check the limiting behaviour. As $R\to0$ (mostly organic phase), $E_A,E_B\to1$ and $\mathrm{EF}\to1$ — everything is extracted regardless of $D$, so there is no enrichment. As $R\to\infty$, $E_i\to D_i/R$ and $\mathrm{EF}\to D_A/D_B=6$, the maximum attainable in a single equilibrium stage. Both computed values, 1.50 and 3.00, sit between these bounds and rise with $R$ as expected — a larger aqueous:organic ratio makes the single organic contact more selective, at the cost of extracting less of A overall.
