I am looking at an older text from the forum library "Fundamental Processess of Dye Chemistry" which is chock full of wonderful preparative procedures
richer in details than almost anything in Vogel.
However in the one I am interested in at the moment, the procedure calls for preparing a hypochlorite solution from 40-degree Baume NaOH solution in
water.
Now, I know that the Baume scales (light and heavy, the latter in this case) are hygrometer scales. There is a formula for converting Be (heavy) to
density at 60 F (yes, F!) which is 145/(145-deg Be)
and in this case that comes out to 1.381 g/ml
So I googled for a chart of NaOH density vs concentration at specific temperatures, and Dow conveniently has one for %w/w at 20 C among other
temperatures, and that is close enough to 60 F for my purposes. The chart gives densities for 34 and 36% NaOH at 20 C that bracket 1.38 very nicely so
it would appear that 40 Be (heavy) NaOH in water is very close to 35% w/w.
Furthermore, there were no entrues for lower temperatures above 34% and my preliminary interpretation is that 35% at 20 C is about saturated. I now go
to look at Merck and Ullmann's (which are on another computer, downstairs) to see if this can be confirmed. Is 40 Be NaOH a saturated soln at 60 F? If
so it would have saved me some legwork if the author had just said so!
Well, Merck confirmed that 35% appears to be d 1.38, they gave 1.33 for 30% and 1.43 for 40%. The Dow chart was more precise. It said 1.37 for 34% and
1.39 for 36%.
I was wrong about saturation, though; Merck gives physical data for aqueous NaOH solutions up to 50%.
So until better data turns up I will use 35% w/w as equivalent to 40 Be NaOH. This will allow me to rewrite the anthranilic acid prep from this
excellent old book into something useful without having to purchase a set of Baume hygrometers.
A liter of 35% w/w NaOH weighs 1381 g so 35% of that is NaOH content, right? 483.4 g/L or 48.34 g per 100 ml. Or have I buggered this calculation?
Arithmetic after midnight is not my strong suit.
[Edited on 30-3-2008 by Sauron]Nicodem - 29-3-2008 at 09:22
Additionally, the MSDS for NaOH lists the solubility in water, at 20°C: 109 g / 100 ml (which is about 52%).
As for the Baume "units", well, yes today they look stupid but I guess there was a reason for using that scale in past. Unfortunately I do not have my
lab manuals at home so I can't check the conversion tables to verify if what you got is the same value.Sauron - 29-3-2008 at 09:36
Yes, thanks. I certainly did not mean to ridicule the Baume scales, they still pop up from time to time, although in general they simply seem a little
quaint.
As you can see, my visit to Merck CD clarified the saturation issue. But thanks anyway.
and I double checked the formula for Be to density at 60 F which is indeed as stated in my post, as well as the result for 40 Be. 1.381 g/ml. The
difference between 60 F and 68 F (20 C) I think I can ignore for my purposes.
[Edited on 30-3-2008 by Sauron]Rosco Bodine - 29-3-2008 at 10:11
Merck gives solubility for NaOH
as 109.2 grams at 20C in 100 grams H2O .
at 20C density 1.55 concentration 52.2%Sauron - 29-3-2008 at 10:19
Well, that seems to be wrong. Because a liter at d = 1.55 is 1550 g and 52.2% of 1550 g is certainly not 1090 g is it?
The density data is correct.
The solubility data, however, appears dubious.
The Dow chart is from "International Critical Tables" Volume III, first ed., p.79
Yet another chart also from rhodium gives the density of a 47% soln NaOH as 1.5, not very precise, and molarity as 17.6. The MW being 40, it is very
clear that the mass of NaOH is <800 g and that if in fact 1092 g would dissolve in a liter of water, the molarity would be >27 wouldn't it?
(Not allowing for correction for volume after dissolution.)
I'm not trying to be a smart ass, I am just questioning Merck's number on this. It does not check out.
Most likely the explanation is this:
The density figures are correct as those from 4-5 different sources are mutually supportive.
When 1092 g NaOH dissolves in 1 liter of water, the resulting solution weighs 2092 g, correct?
The weight of a liter of that solution is 1550 g, so therefore, the VOLUME of the solution resulting from the dissolution of 1092 g NaOH in 1 L of
water is more like 1350 ml. at 20 C.
So Merck is correct, it merely required a little thought to correlate the information.
The NaOH content of a liter of such saturated solution is 52.2% of 1550 g = 808 g and molarity is therefore 20.2.
As for my original problem, I just need to weigh out 483.4 g NaOH and dissolve it in sufficient water to make up 1 L after cooling down to 20 C and
this will have d = 1.381 and be approx 40 Be and have a molarity of 12.09 or so.
Furthermore the amount of water required to make up a L after dissolution of 483.4 g ought to be 1381 g - 483.4 g = 867.6 g water (everything at 20
C).
I really shouldn't do arithmetic at 2 am.
[Edited on 30-3-2008 by Sauron]
Merck Lab Tools attached
Rosco Bodine - 29-3-2008 at 12:15
At below 60C the NaOH forms a monohydrate
so this is likely to skew any density and volume compression figures above 60C , there's no exact formation temperature given so say "sixtyish" on
that one
for that 1.5 d Merck shows 47.33% and 17.75 molarity
It is faster to do Baume-related calculations by looking at a table made from the ICT, and a place to do this is old CRC's. All questions seem to have
been answered, but since the thread is here, this page is from 1939, when the list price was $4.
[hint]The entire ICT as released by Knovel since 2003 is available BTW.[/hint]
[Edited on 30-3-2008 by S.C. Wack]
Attachment: naoh.pdf (24kB) This file has been downloaded 1045 times
Sauron - 29-3-2008 at 20:20
Thanks. Our confidence in reference tables has been reasserted.
I am going to rewrite the anthranilic acid prep from "Fundamental Processes of Dye Chemistry" and replace the 40 Be NaOH with the solution calculated
above, from 483.4 g NaOH/L Also, replace chlorine with bromine (as used in Vogel's version of this procedure) so that sodium hypobromite is the
oxidant. Then I will try this prep with commercial phthalimide and report results.bio2 - 30-3-2008 at 17:52
The solution calculator in this free mole/solution calculator
gives the density to 4 decimal places for common acids and bases over the full range of concentrations as well as a other common chemicals.
Click on the little beaker on the calculator to open the solution calculator. It's handier than using cumbersome tabular data.
It gives 1.3799 as the density of 35% NaOH. I assume this
is at standard conditions.
If anyone can figure out how to add to the chemical library
that would be helpful. An example is given but I have never been able to figure it out.Sauron - 30-3-2008 at 19:01
1.3799, 1.381 close enough for gubbiment work, as we used to say.
It's just for a prep, not for quantitative analysis.
Thanks anywayNicodem - 30-3-2008 at 22:55
Quote:
Originally posted by bio2
If anyone can figure out how to add to the chemical library
that would be helpful. An example is given but I have never been able to figure it out.
I also always use that little calculator for any chemical calculations. The nicest thing in it is that one can use formulas in brackets instead of the
numerical value of the molecular weight ( for example "(CH3COOH)" instead of "60", etc.). Anyway, I never added anything in the chemical properties
base, but it looks simple to add basic chemical data. You just open the "Chemicals.txt" file (in the program folder) and add a new entry as
instructed. However, I don't think you can also add the empirical density equation. The default entries do not have it saved in that same file.bio2 - 8-4-2008 at 18:32
.....You just open the "Chemicals.txt" file (in the program folder) and add a new entry as instructed..........
Yes, it is a very handy program.
I've tried as you said and that's what I couldn't figure out as it will never accept my entry.
Guess it's time to try again maybe the instructions will make sense this time, lol.MagicJigPipe - 8-4-2008 at 18:58
I would like to comment that, in my experiences, at least 95g of NaOH (ACS Reagent) will dissolve in 100g/mL of dH2O at about 19C. It could possibly
dissolve more but it takes A LONG TIME. That's just a little less than 50%. I think the Merck data is correct.
Just to make sure I am going to try to dissolve 10g of NaOH in 10mL of dH2O tonight (that's 50%). Then I will add NaOH slowly until no more will
dissolve.
That should settle it, don't you think?
EDIT
Sorry, I didn't realize until now that the saturation issue has already been settled.
