solid effort
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50%w/w NaOH solution - what should it look like?
Hi everyone, hoping you can help me with this very basic question.
I am trying to make a 50% w/w solution of NaOH. I am using AR grade reagent. I am aware that it is near the saturation limit, so may be the reason
for what I am seeing.
From what I have read, this solution should be milky white. However, I am getting a solution with lots of flaky, precipitate - it's definitely not
all completely dissolving.
I have run this on the magnetic stirrer for hours. I've added heat (after all the NaOH has almost dissolved and cooled off) and stirred. I've tried
to add water to the solute, and solute to the water, still getting the same result.
Is there something I am doing wrong, or is this just what it looks like? I haven't made a solution this concentrated before so not sure what to
expect. The NaOH I am using is probably 10yrs old, but I doubt that would be the reason.
Any comments or suggestions would be greatly appreciated.
S.E
[Edited on 28-5-2013 by solid effort]
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DraconicAcid
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A solution of NaOH should look liek water. If it's milky or white, it's not a solution, but a suspension. If there's flaky precipitate that just
isn't dissolving, that's probably sodium carbonate. NaOH absorbs CO2 from the air to form sodium carbonate and water- if it's ten years old, it's
probably absorbed a fair amount.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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solid effort
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Thanks for the reply DA. I had thought that the solution should be clear, but after having this trouble looked around online for a description and
found an MSDS for 50% solution that stated it was milky white.
Sodium carbonate sounds like a plausible explanation - thanks for the tip. Luckily NaOH is cheap so buying in some fresh stuff is no issue.
always learning 
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DerAlte
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Have a look at http://www.solvaychemicals.com/Chemicals%20Literature%20Docu...
A 50% concentration precipitates nearly all the carbonate - this is one way of purifying it of carbonate...
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solid effort
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So assuming all the NaOH goes into solution, i could make up a 50%w/w solution, decant off all the liquid and weigh the precipitate. Knowing how much
originally went in, and the mass of precipitate i could then calculate the actual concentration of the solution?
Sounds easy enough, or am I missing something?
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elementcollector1
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That should be it - mind your stoichiometry!
Elements Collected:52/87
Latest Acquired: Cl
Next in Line: Nd
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solid effort
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If I am basing it purely on a w/w % how will stoichiometry be an issue?
If I had 50g starting material ('contaminated' NaOH) + 50g water in emulsion/solution, then collected 10g of solid precipitate(for example), then
wouldn't it simply be 40/90*100 = ~44% NaOH (there will obviously be some losses of water and NaOH in the removed precipitate)?
I'll probably end up just buying some fresh stuff, but am trying to gain a better understanding of what is going on here.
Thanks for the help everyone
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