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blogfast25
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A batch of contaminated La chloride hydrate, crystallised:
Crystallising this chloride hydrate is not easy, due to its high solubility. Boiling in the solution eventually results a syrupy liquid that may or
may not solidify on cooling, depending on how much water you evaporated off. Yesterday one such syrupy liquid was poured into a silicone baking mould
and I watched it slowly crystallise:
You should be able to see the areas from which crystal growth emanated.
pH selective precipitation of Pr?
Due to the lanthanide contraction, higher REEs have slightly smaller atomic and ionic radii than the lower ones. For La<sup>3+</sup> the
ionic radius is 103 pm, for Pr<sup>3+</sup> it’s 99 pm (both Wiki).
As smaller ions (all other things being equal) emit a stronger central electrical field, the deprotonation reaction:
Ln[(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup> + OH<sup>-</sup> === >
Ln[(H<sub>2</sub>O)<sub>5</sub>(OH)]<sup>2+</sup> + H<sub>2</sub>O
… as well the two subsequent deprotonation steps, can be expected to take place at slightly lower pH for the higher REEs. In plain English, the
higher lanthanide hexaqua cations are slightly more acidic.
I prepared a 75 ml solution that was about 0.5 M in La (contaminated) and 3 M in NH<sub>4</sub>Cl, the latter with the aim of buffering
the solution somewhat. 1:1 NH3 33%:water was then added from a burette while magnetic stirring and measuring the pH with a calibrated pH meter. Setup:
Starting pH was 2.3 but after just adding 1 ml of NH3 solution it had already gone up to 7.5 and the first permanent precipitate had formed. A few
more drops of NH3 took the pH to 7.6 and I stopped adding NH3.
The first precipitate (call it A) was then Buchnered off and the ‘titration’ continued on the filtrate. At about 4 ml NH3 added the pH was 8.5 and
no more precipitate appeared to form. This precipitate (call it B) was also Buchnered off.
Precipitate A was the dissolved in a minimum quantity of strong HCl, a few drops only. I expected to see green there but that didn’t happen [edit:
the solutions of A and B turned out indistinguishable]. Had A been enriched in Pr with respect to B, a much stronger green would have been observed.
Precipitate B was dissolved somewhat more strong HCl and dissolved to a familiar very light greenish solution.
Conclusion: no separation, however partial, appeared to have been achieved.
[Edited on 26-10-2014 by blogfast25]
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Oscilllator
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Why do you think that no separation has been achieved? It seems to me that getting two different coloured solutions means that at least some
separation occured
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blogfast25
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Quote: Originally posted by Oscilllator  | | Why do you think that no separation has been achieved? It seems to me that getting two different coloured solutions means that at least some
separation occured |
Because the solution prepared from precipitate B contained all the La AND also Pr (green) and the solution from A was the same colour. Had the
solution from B been colourless and the solution from A green (Pr(III)) then separation had occurred. Alas, no.
I may not have made that abundantly clear in the post which has been lightly edited to that effect.
[Edited on 26-10-2014 by blogfast25]
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