Question

one stage of nuclear reprocessing involves dissolving the spent fuel rods in nitric acid to produce uranyl nitrate. An organophosphorus liquid, tributyl phosphate, is then used to extract the uranyl nitrate from the water. The two liquids are highly immiscible (1 mL tributyl phosphate/ 165 mL water). We would like to remove 90% of the uranyl nitrate from 2 kg of a 20 wt% aqueous solution. How much tributyl phosphate is required for thie process if we use 3 stages in a cross-current congifuaration? How much is needed if the three stages are operated in a counter-current configuration? why is there a difference in the amount of solvent needed for the two configurations? You can assume the distribution coeffiecent on a solute-free basis is 5.5

Answer 1

Mass of aq solution=2kg

Mass of uranyl nitrate=0.20*2kg =0.4 kg (20% by wt)

Mass of water=2.0-0.4=1.6 kg

Density of water=1g/ml

So volume of water=V=mass/density=1.6 *1000 g/1g/ml=1600 ml=1.6 L

Let the amount of tributyl phosphate used be=S

For multistage cross-current extraction,

E=[V/DS+V]^n    where, E =fraction of solute remaining in the solvent after n extractions

For 90% extraction , E=100-90=10% or 0.10

D= distribution coeffiecent on a solute-free basis is 5.5

0.90=[1.6 L/5.5*S+1.6L]^3

0.96=[1.6 L/5.5*S+1.6L]

0.96(5.5*S+1.6L)=1.6L

5.28*S+1.54L=1.6L

S=0.0113L(amount of tributyl phosphate used)

For counter-current,

E=[V/DS+V] (for every stage of extraction)

E%= *(^n-1)/^(n+1)-1      where =extraction factor=D *(V/S)

=D *(V/S)=5.5 *(1.6L/S)=8.8/S

0.90=8.8/S* [(8.8/S)^3 -1]/[((8.8/S)^4 -1]

Solving this,

S^3-0.1S^4=81.77

S=0.0048L

Volume used is more in cross current as the feed containing the solute is contacted with fresh solvent for each extraction.The extracts may be collected separately or combined later.

In counter current ,the feed in each stage is contacted with solvent from the preceeding stage. It results in gradual enrichment of the solute in the solvent phase throughout the process of extraction