Zircon (ZrSiO4) is a mineral found abundantly in placer deposits located in the central coast of Vie — Inorganic Chemistry — Solid State Chemistry Question
Radiochemistry
Zircon (ZrSiO4) is a mineral found abundantly in placer deposits located in the central coast of Vietnam. Besides being widely utilized in the ceramic industry, zircon is also used as a raw material for the manufacture of zircaloy which is used to build fuel rods that hold the uranium dioxide (UO2) fuel pellets in nuclear reactors. Zircon ore contains a trace amount of uranium, and it is not a viable source of uranium in practice. However, zircon crystals make a perfect storage medium to avoid the loss of uranium and lead (Pb) isotopes because of its stable crystal structure. This allows developing uranium-lead dating method.
There are three naturally occurring decay series:
- The thorium series begins with 232Th and ends up with 208Pb.
- The uranium series (also referred to as the uranium-radium series) is headed by 238U. The half-life (t1/2) of 238U is 4.47 ⋅ 109 years.
- The actinium series is headed by 235U with the half-life of 7.038 ⋅ 108 years.
Four stable isotopes of Pb exist in nature: 204Pb, 206Pb, 207Pb, and 208Pb. The natural abundance of each isotope is shown in the following table.
An analysis of a zircon mineral sample gives the following mass ratios of U and Pb isotopes:
m(238U) : m(235U) : m(206Pb) : m(204Pb) = 99.275 : 0.721 : 14.30 : 0.277
Indicate the stable isotope of Pb which is not involved in the above decay series.
Model Answer
204Pb 206Pb 207Pb 208Pb
x
Determine the mass ratio of 238U to 235U when the zircon mineral was first formed. Assume that the mineral already contained natural Pb right at the onset of its formation.
Model Answer
Assume that the mineral initially contained n1,0 moles of 238U, n2,0 moles of 206Pb, and n3 moles of 204Pb; and at present, it contains n1 moles of 238U, n2 moles of 206Pb, and n3 moles of 204Pb (this isotope is not generated by the decay of 238U and 235U). The age of the zircon mineral is usually very large, and we can consider that the century equilibrium for the decay process has been reached (i.e. loss of 1 mole of 238U will lead to formation of 1 mole of 206Pb). By conservation of mass, we have the following equation:
n1 + n2 = n1,0 + n2,0 (1)
Dividing (1) by n3:
n1/n3 + n2/n3 = n1,0/n3 + n2,0/n3
→ n2/n3 = n1,0/n3 – n1/n3 + n2,0/n3 (2)
In addition, we have n1,0 = n1e λt, where λ is the decay constant of 238U, and t is the age of the mineral,
→ n2 / n3 = n1e λt / n3 – n1 / n3 + n2,0 / n3 = (n1 / n3)( eλt – 1) + n2,0 / n3 (3)
→ (n2 / n3) – (n2,0 / n3) = (n1 / n3)(eλt - 1)
→ eλt = 1 + [(n2/n3 - n2,0/n3) / (n1/n3)]
→ t = (1/λ) ln(1 + [(n2/n3 - n2,0/n3) / (n1/n3)]) (4)
According to the data given:
n2/n3 = (14.30 / 206) / (0.277 / 204) = 51.12; n2,0/n3 = (24.10 / 206) / (1.4 / 204) = 17.05
n1/n3 = (99.275 / 238) / (0.277 / 204) = 307.19
t = (4.47 × 10^9 / 0.693) ln (1 + (51.12 – 17.05) / 307.19) = 6.78 × 10^8 years
m0( 235U) = 0.721 × e^(0.693 × 6.78 × 10^8 / 7.038 × 10^8) = 1.406 g
m0( 238U) = 99.275 × e^(0.693 × 6.78 × 10^8 / 4.47 × 10^9) = 110.28 g
m0( 235U)/ m0( 238U) = 1.406 / 110.28 = 0.0127
Production of uranium from low-grade will encounter many difficulties, notably large concentration of impurities and low concentrations of uranium in leach solutions. Various technological advances have been applied to overcome the aforementioned problems; these include fractional precipitation, liquid-liquid extraction, or ion exchange methods.
In an experiment to extract uranium from sample of low uranium content using diluted H2SO4, in the preliminary treated leach solutions, the concentration of uranyl sulfate (UO2SO4) is 0.01 mol dm-3 and the concentration of iron(III) sulfate (Fe2(SO4)3) goes up to 0.05 mol dm-3. The separation of uranium from iron and other impurities can be carried out by the fractional precipitation method.
Calculate the pH necessary to precipitate 99% of Fe3+ without losing uranium ions. Assume that the adsorption of uranium onto Fe(OH)3 is negligible. Under the experimental conditions, the solubility product values for UO2(OH)2 and Fe(OH)3 are 1.0 ⋅ 10–22 and 3.8 ⋅ 10–38, respectively.
Model Answer
After 99% of Fe3+ precipitated, the concentration of the remaining Fe3+ in the solution is:
[Fe3+] = 2 × 0.05 ⋅ 10−2 = 1 ⋅ 10−3 mol dm-3
The concentration of hydroxide ions necessary to maintain a Fe3+ concentration of 1 ⋅10−3 mol dm-3 in the solution is:
[OH−] = (Ks(Fe(OH)3) / [Fe3+])^(1/3) = (3.8 ⋅ 10−38 / 10−3)^(1/3) = (38)^(1/3) ⋅ 10−12 mol dm-3
Thus, the pH value of the solution can be calculated as follows:
pH = – log{10-14 / [(38)^(1/3) ⋅ 10−12]} = 2 + (1/3) log 38 = 2.53
At this pH, the reaction quotient of the dissociation of UO2(OH)2 in 0.01 mol dm-3 of solution is:
[UO2 2+][OH−]2 = 0.01 × [(38)^(1/3) ⋅ 10−12]2 = 1.13 ⋅ 10−25 < 10−22
Since the ionic product is much smaller than the solubility product of UO2(OH)2, we can conclude that uranium cannot precipitate under these conditions.
One of the proper methods to obtain a rich uranium solution is the liquid-liquid extraction with the organic phase containing the extracted agent of tributylphosphate (TBP) diluted in kerosene. When extracting uranium in the form of uranyl nitrate (UO2(NO3)2) under appropriate conditions, the relationship between the concentrations of uranium in water and organic phases is given by:
Distribution coefficient: D = corg. / caq. = 10
where: corg and caq are the equilibrium concentrations (mol dm-3) of UO2(NO3)2 in organic and aqueous phases, respectively.
Calculate the mole percentage (in comparison with the initial concentration) of UO2(NO3)2 remaining in the aqueous phase after extracting 1.0 dm3 of the solution (with an initial concentration of 0.01 mol dm-3) with 500 cm3 of organic solvent.
Model Answer
Volume ratio of the two phases: Vaq / Vorg = 1000 : 500 = 2
Let x represent the equilibrium concentration of UO2(NO3)2 in the aqueous phase.
Let co represent the initial concentration of UO2(NO3)2 in the organic phase.
The equilibrium concentration of UO2(NO3)2 in the organic phase is calculated as follows:
corg = (Vaq / Vorg) ( c0 – x)
D = corg / caq = (2(c0 - x)) / x = 10
x : co = 1 : 6 = 16.67 %.
Propose a scheme to extract 96 % of UO2(NO3)2 from 1.0 dm3 of the aqueous phase into 500 cm3 of the organic phase. Assume that the distribution coefficient remains constant throughout the extraction process (D = 10).
Model Answer
500 cm3 of organic solvent may be divided into n equal portions for extraction.
Volume ratio of the two phases: Vaq / Vorg = 1000 : (500/n) = 2 n
- After the first extraction:
D = corg / caq = (2n(co - x1)) / x1 = 10
→ x1 = 2n / (D + 2n) * co
- For the second extraction, the initial concentration of the aqueous phase is x1, while the equilibrium concentration is x2. Using equation (8), we replace x2 with x1, and x1 with Co to obtain the following expression:
x2 = 2n / (D + 2n) * x1 = (2n / (D + 2n))^2 * c0
- After n extractions, the concentration of UO2(NO3)2 remaining in the aqueous phase is:
xn = (2n / (D + 2n))^n * c0
% UO2(NO3)2 remaining in the aqueous phase after n extractions is:
(xn / c0) * 100% = (2n / (D + 2n))^n * 100%
n = 1: 16.67%
n = 2: 8.16%
n = 3: 5.27%
n = 4: 3.9%
n = 5: 3.1%
n = 6: 2.63%
n = 5 → (xn / c0) * 100% = (10 / 20)^5 * 100% < 4%
Thus, the optimal approach is to divide 500 cm3 of solvent into 5 portions and extract 5 times.
Other schemes are acceptable, if all calculations and justifications are reasonable.