• A process developed to treat minerals containing 4000 ppm of tetravalent uranium (UO
2
) is based on the following conditions
• Consumption of H
2
SO
4
: 25 kg/t
• Consumption of NaClO
3
= 2 kg/t
• Volume of attacking solution/ mineral mass
= 1 m 3 /t
• This treatment is used to solubilize 96% of uranium content.

• You are asked to
• Write the oxidation equation of U(IV) by
ClO
3
ions
• Determine the efficiency of the oxidizing treatment
Number .
of .
necessary .
ClO
3

.
moles
E

Number .
of .
added .
moles
• Determine the composition in uranium and
H
2
SO
4
, of the solutions after attack (by making the following hypothesis that the variation of [H + ] is negligible during the reaction period)

• The purification of uranium is performed by chromatography on anion exchange resins columns ( with an exchange capacity for one monovalent ion =
3 eq/kg)
• Determine, by using Figure 1, the distribution coefficient of U(VI) between the attacking solution and the resin ( consider that an excess of resin is used per unitary volume of solution)
• Determine the volume of uraniferous solution that you can transfer on a column composed of 100 kg of resins where we want to saturate to 70% maximum the exchange sites, knowing that the extraction reaction is
4 R
4
NHSO
4

UO
2
SO
3
4

( R
4
N )
4
UO
2
( SO
4
)
3

4 HSO
4

• The extraction of 1 U(VI) is saturating 4 sites of the resin

Concentration (NH
4
)
2
SO
4 or H
2
SO
4 in Mol/L

• Write the oxidation equation of U(IV) by
ClO
3
ions
UO
2
6 H

UO
2
2


ClO

3


6 e

2 e

3 UO
2

6 H
 
ClO

3
OXIDATION
Cl
 
3 H
2
O
3 UO
2
2
 
Cl

REDUCTION

3 H
2
O

• Number of moles of U solubilized
4000 ppm of U is equal to 4000 g/t of U
Knowing that the molar mass M of U is 238 g / mole, 4000g of U represents 4000/238 =
16.8067 moles/t. Since 96% of U is solubilized, we will have 0.96*16.8067 = 16.13 moles/t of U solubilized.
• Number of moles of NaClO
• M of NaClO3 is 106.5 g/mole
3 used
• Consumption is 2 kg/t, = 2000g/t = 2000/106.5 =
18.78
moles/t

• Number of moles of NaClO
3 to the process necessary
After Eq 1, 1 mole of NaClO
3 moles of UO
2 reacts with 3
If we have 16.13 moles of U, we need
16.13/3 = 5.37
moles of NaClO
3
The efficiency will be 5.37
/ 18.78 = 0.286

• Uranium
Mass of U solubilized
4000 g * 96% = 3840g/t or 3.840g/L, knowing that the volume is 1 m 3 = 1000L.
[U] is 3.840/238 = 0.016M
/L
• H2SO4
25kg/t = 25000g/t or 25 g/L, knowing that the volume is 1 m 3 = 1000L.
M = 98g/mole
[H2SO4] = 25/98= 0.255moles/L

H2SO4 = 0.25M
Kd U(VI) = 200

4 R
4
NHSO
4

UO
2
SO
3
4

( R
4
N )
4
UO
2
( SO
4
)
3

4 HSO
4

• Exchange capacity of the resin
• 3 eq/kg * 100 kg of resin = 300 eq/kg in monovalent ion
• In uranium we will have 300/4 = 75 moles of
U/column
• The saturation is 70% so 75*70% = 52.5
moles of U
The total volume of the uraniferous solution is
52.5
moles/ 0.016
= 3254.8 liters