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THE RADIOCHEMISTRY OF PLUTONIUM - Sciencemadness.org

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a<br />

ml<br />

0<br />

-1<br />

5<br />

0<br />

4<br />

3<br />

2<br />

-1<br />

-2<br />

-3<br />

~- x<br />

I I<br />

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I<br />

-1 0 1,<br />

Log HN03 CONCENTRATION w)<br />

Fig. 12. Extraction of various actinides into 0.5<br />

—<br />

M HDEHP (isooctane diluent) from<br />

HN03 solutions.238<br />

curve at high acid concentrations is due to disproportionation of Np(V ) into Np(lY ) and<br />

Np(VI), both of which are more extractable than Np(V). The minimum in the Am(lII)<br />

curve at high acid concentrations is probably due to nitrate completing of the extracted<br />

Am species.<br />

Horner and Coleman 187 get a different result for the extraction of Pu(lV ) by<br />

HDEHP. As shown in Fig. 13, the extraction curve is concave downward with increasing<br />

HN03 concentration using 0.01 ~ HDEHP in Amsco 125-82. The magnitude of distribution<br />

coefficients are very much larger than those of Kosyakov ~.,<br />

238 ~ the second<br />

power dependence on the extractant concentration is taken into account. Another dif-<br />

ference is the decreased distribution coefficient at 0.1 ~ acid. Homer and Coleman<br />

ascribe this decrease to hydrolysis of the Pu(IV). Horner and Coleman prepared Pu(IY)<br />

by reduction to Pu(IIf) with hydroxylamine nitrate, and reoxidation and stabilization<br />

238<br />

with sodium nitrite, while Kosyakov et al. dld not state their method of preparation of<br />

Pu(lv).<br />

Dreze and Duyckaerts<br />

\<br />

121 investigated the extraction of Pu(IY) by di - n-butyl<br />

phosphoric acid (HDBP) from nitrate solutions as a function of nitric acid concentration,<br />

nitrate concentration, HDBP concentration, and ionic strength. Representative results<br />

are shown in Figs. 14 and 15. These experimenters were able to invoke the known<br />

stability constants of Pu(IV ) nitrate complexes to fit the various fwctional dependencies.<br />

They consider the extraction reaction to be<br />

PU+4 + 2(N03 )- + 2(HDBP)2(0) - PU(N03)2 (H(DBP)2)2(0) + 2H+ (2)<br />

for which they calculate the equilibrium constant to be (1.7 + 0.3) X 109 (m/1 )2.<br />

43<br />

I

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