blogfast25
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Recrystallisation calculation…
I’ve been having some trouble with getting erroneous results with various hot-to-cold recrystallisations, in particular with the amounts of
recrystallites obtained.
So I decided to look into how to calculate the amounts of crystals obtained for a generic hydrate, KaAn.nH<sub>2</sub>O.
Definitions:
H: amount of hydrate crystals formed after recrystallisation (expressed as gram of KaAn.nH<sub>2</sub>O).
A: amount of salt (KaAn) dissolved in supernatant water (expressed as gram of KaAn, NOT as KaAn.nH<sub>2</sub>O).
W: amount of supernatant water, in gram.
p: water content of KaAn.nH<sub>2</sub>O, expressed as a percentage (%).
s: solubility of KaAn in water at final temperature, expressed as gram KaAn per 100 g of water.
Calculation:
To focus attention, I’ll take the example of 100 gram of hydrate to which w gram of pure water is added. The mixture is then taken to a temperature
where all the hydrate dissolves, then cooled to the temperature where solubility of KaAn is s (gram KaAn per 100 ml of W).
At that temperature, solid hydrate (recrystallised) H, dissolved salt A and liquid water W coexist. To find the respective amounts three independent
equations are needed.
Firstly, the solubility relationship dictates A = (s/100) x W ….. (eq. 1)
Secondly, the mass balance: we started off with 100 g of hydrate and w gram of water and ended up with H gram of hydrate, A gram of dissolved salt and
W gram of liquid water. So mass conservation dictates:
H + A + W = 100 + w ….. (eq. 2)
Thirdly, water conservation. We started off with w gram of water, plus 100 x (p/100) of water contained in the 100 gram of hydrate and ended up W
gram of water and (p/100) x H in the recrystallised hydrate. So:
(p/100) . H + W = p + w ….. (eq. 3)
Plugging eq. 1 into eq. 2 gives:
H + [(s/100) + 1] . W = 100 + w ….. (eq.4)
Eq. 3 and eq. 4 then constitute a couple of simultaneous equations in the same variables H and W and on solving one gets:
W = f . w
With f = [100 . (100 - p)] / [10,000 - p. (100 + s)]
H = 100 - {[(100 + s) / 100] . f - 1}. w
A = (s/100) x W
Examples:
For alum (KAl(SO<sub>4</sub> <sub>2</sub>.12H<sub>2</sub>O) the water content p = 45.6 % and the solubility of KAl(SO<sub>4</sub><sub>2</sub> at 20 C is 12.0 g per 100 g of water.
Calculate f to be 1.11 and thus H = 100 - 0.243 . w
Example 1: w = 10 gram, then:
W = 11.1 g; H = 97.57 g; A = 1.33 g. Sum = 110 g
Example 2: w = 31.8 g, then:
W = 35.3 g; H = 92.27 g; A = 4.23 g. Sum = 131.8 g
Note how W > w because some of the hydrate remains dissolved. For alum, adding about 30 % of water to the crude to be recrystallised seems about
right: only 8 % is ‘lost’ to the supernatant liquid.
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m1tanker78
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Is p defined as a constant or is it defined as the % of water before or after crystallization? In other words, does p = the % of water in the compound
to be dissolved or is it the % of water in the recrystallized compound.
Also, please define f.
Tank
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blogfast25
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Tank:
Here it is assumed (reasonably) that the compound (hydrate) one starts off from is also the same one ends up with after recrystallisation. So e.g.
alum dodecahydrate to alum dodecahydrate or sodium carbonate decahydrate to sodium carbonate decahydrate or oxalic acid dihydrate to oxalic acid
dihydrate. This is the case for most recrystallisations. p is then the water content (in %) of the hydrate. It is indeed a constant: a material
constant like molecular weight, specific to that specific material.
f is simply an intermediate factor that makes it easier to write the relationships between W and w and between H and w. It’s an algebraical device,
without special meaning. One could also say that f is the factor by which the initially added amount of water (w) has to be multiplied to obtain the
amount of liquid water (W) obtained after recrystallisation took place (hence W = f . w).
[Edited on 17-6-2011 by blogfast25]
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m1tanker78
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It's funny that you mention Na2CO3. I began trying to plug values into the equations and quickly realized how complex the relationships are between
concentration, temperature, and 'p'. When [Na2CO3]'s parameters are graphed, it resembles a damned Picasso!
That aside, it makes sense to treat p as a constant if you're beginning and ending at the same temperature. I imagine it gets trickier if you weigh
out your salt at, say, 37*C (the average temperature in my garage, ughh), dissolve/heat, then take it down to 20*C to recrystallize. I suppose it
depends on the individual salt as well. IIRC, Na2CO3.7H2O is favorable at ~37*C but has a very narrow temp window within a narrow water content.
One other thing. This may well be a dumb question. What do you mean by "f . w" or "0.235 . w" I'm not familiar with the dot notation you're using.
Tank
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blogfast25
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Yes, it relies on the hydrate you start from being the same as the one you end up with. For most intents and purposes practical, that is the case.
Solubility curves become complicated when there are multiple hydrates, each existing in their own temperature range, as is the case with Na2CO3 (also
Na2SO4). But if you start from the soda decahydrate (the stable Na2CO3 form at RT), the calculation is entirely valid. Starting from other hydrates
(or from anhydrous) a relatively simple correction needs to be made to the formulas.
f . w = f x w = fw: mean f times w; all are modern acceptable algebraic notations of the same thing.
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