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Author: Subject: Prep. of dilute sodium isopropoxide from NaOH
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[*] posted on 24-2-2012 at 16:21
Prep. of dilute sodium isopropoxide from NaOH


An experiment I'd like to perform calls for a very dilute solution (0.07M) of sodium isopropoxide in iPrOH. Ideally I'd use plain ol' sodium hydroxide in absolute isopropanol which translates to slightly less than 0.28g NaOH per 100ml iPrOH.

iPrOH + NaOH <-----> [iPrO-] + [Na+] + H2O(g)

The water produced is distilled off with the isopropanol, driving the reaction to the right.

I've managed to get this "working" as evidenced by the fact that the fairly small amount of NaOH does indeed dissolve in boiling isopropanol (and doesn't precipitate out when the solution cools.) Any thoughts on how I can confirm this is what's happening? My tools are all "kitchen scale," though I do have a set of 19/22 glass with only a few parts broken.

Thanks in advance.
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[*] posted on 26-2-2012 at 03:11


Quote: Originally posted by scientician  
Any thoughts on how I can confirm this is what's happening? My tools are all "kitchen scale," though I do have a set of 19/22 glass with only a few parts broken.

It is not really possible to measure the amount of H2O in such a solution using chemical methods (you are dealing with a dynamic equilibrium here, so for every molecule of H2O consumed, another one forms by the reaction of hydroxide with isopropanol). Spectroscopic and perhaps some voltametric methods can give you the answer to such a difficult question, but these are not "kitchen scale" tools. Have you checked the literature if the pKa values of isopropanol and H2O in isopropanol are already known? It would be stupid to reinvent the wheel, just because you don't like to search the literature.

What can be measured by the chemical methods is the combined amounts of H2O and hydroxide. This value is in any case much more useful for some applications of such solutions where the amount of H2O is actually irrelevant, but the combined hydroxide and water is important. For example, a chemical mean would be to react your solution with a known amount of isopropyl benzoate and measure the ratio of sodium benzoate vs. isopropyl benzoate. From this, the amount of hydroxide and H2O can be calculated (provided that the isopropyl benzoate was dry). To measure the extent of hydrolysis you would need to develop some analytical method which would either require an acidic quench and HPLC analysis or some other more or less reliable method. Again, not really kitchen friendly.

The only kitchen friendly method that I can think of, is a conductometric measurement. This does not require any particularly expensive equipment (nowadays, fairly reliable multimeters can be bought for a few dozens euros). I'm not sure if it would give reliable results (this depends on the measurement error and the influence of the solution composition on the conductance), but another major problem is that you would need a standard solution of sodium isopropoxide in isopropanol to create a calibration curve. This standard solution could be made by careful distillation of the water-isopropanol azeotrope from a solution of NaOH in isopropanol, but you would need some special equipment like a very efficient distillation column and a precise thermometer to find out when there is no more water in the distillate (constant T). You would also need to determine the exact volume of the remaining isopropoxide solution to get a reliable concentration. The calibration curve is then done by measuring the conductance in dependence to the amount of added water. In any case, you would need an analytical balance, precise volumetry, a reliable multimeter and a fixed conductometric cell (at least this can be easily self-made). The most expensive "non-kitchen" equipment is therefore the analytical balance (leaving the exact composition of the starting NaOH as the major source of error would, however, make the precision of the balance pretty obsolete).
(can you see why physical chemistry is so unpopular?)
Quote:
I've managed to get this "working" as evidenced by the fact that the fairly small amount of NaOH does indeed dissolve in boiling isopropanol (and doesn't precipitate out when the solution cools.)

I don't understand how you come to the conclusion that is evidence for whatever. NaOH is relatively soluble in isopropanol.
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[*] posted on 26-2-2012 at 15:31


Vogel confirms this actually works. :D
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