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Magpie
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Quote: Originally posted by Magpie  |
Dilute the acid first, very carefully, by adding the acid slowly to water, with stirring. Use an ice bath if necessary.
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You could also dissolve the NaOH first, adding a bit at a time, in water with a lot of stirring. Use an ice-bath to keep the container cool. Then
add the two components together, slowly, with cooling. By using an ice bath and going slowly you can keep the temperature under control. Use a
thermometer as a stirring rod to regulate your addition rate. Use evaporation and patience to dry your NaHSO4 to a white powder - probably a hydrate.
Wear good gloves (at least mid-arm length) and eye protection (goggles as a minimum).
[Edited on 8-4-2012 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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CHRIS25
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Thanks Peach, doesn't diluting the acid change the g/mole calculations? Logic suggests not, after all there is still the same amount of acid in the
reaction, but just to make sure.....? I might try the Sodium bicarbonate first to be honest, but caustic soda here works out 5 times cheaper than the
bicarbonate.
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peach
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You can dilute them down fine yep. So long as you have the same amount of concentrated sulphuric and the same amount of NaOH there - as you say. That
will be easier in terms of the heat and making sure they fully react, because (with enough water) it'll always be a solution. Whereas, doing it with
concentrated reagents, the product (bisulphate monohydrate) is a solid at room temperature.
The only draw back (if you want the bisulphate on it's own as a solid) is having to remove all the water afterwards. The bisulphate will dissolve in
it, so you need to warm it up and put a fan over it to dry it down to the monohydrate solid.
I would suggest you skip all that. You're going to make a pickling bath with it, which I would assume is made by dissolving the bisulphate in water.
Rather than bothering to get the bisulphate as a solid, just keep it in solution. Provided you know how much is in there (by measuring in the
sulphuric and NaOH), it's still the same thing.
Say you want to make 200g of bisulphite:
That much bisulphate dissolves in about 400ml of water at 0C to make a saturated solution (can't dissolve any more).
It doesn't matter whether you add the water to the sulphuric, to the NaOH or both. Personally, I would add 200ml to the NaOH (to turn it into a
solution and dilute it) and 200ml to the acid (to dilute that down).
When you add them together, you will end up with 200g of bisulphate dissolved in 400ml of water.
Do you have a recipe or guide for what is in the pickling baths besides bisulphate and how much of each to add to how much water? It may be that you
can add considerably more to the sulphuric and naoh so, instead of making solid bisulphate only to then redissolve it, you set the bath up, fill it
with water, add some sulphuric and then the NaOH. That'd be much safer and easier.
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weiming1998
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No, the concentration of the acid doesn't change the g/mol calculation. The g/mol is determined by the combination of the atomic weights of the atoms
that makes up the molecule. Dilution and concentration doesn't change the atomic weight of the atoms or the amount of atoms that makes up the molecule
in most cases, and thus has no effect on the g/mol calculations.
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CHRIS25
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Peach amd Welming, you are both being very helpful, this part of chemistry goes a long way now to cementing all those random pieces of information
that I have been learning but now it is all coming together nicely with clarity.
I will be keeping it in a solution so no need for the solid, and that bit about adding more water certainly makes common sense. I do not have a
recipe, simply learned from various sources that pickling solutions are essentially and mainly sodium bisulpate. I have been using Acetic acid and
salt warmed up for my de-oxidising, but this is not ideal for silver. Also Not ideal for continuous repetitive de-oxidising on copper - hence the
need for the more proper stuff so to speak.
I have to also make Ferric nitrate. (Etching silver and putting dark patinas in copper grooves) I have the formula - 2Fe+8Hno3 = 2Fe(no3)3 + 2No +
4H2o
So my two questions here are: 1. the 2 and the 8 mean that I add two times and 8 times the g/mole amount respectively?
and 2. I have some old rusty horseshoes from about 60 years old. If I clean off the rust thoroughly can I use this Iron (obviously cut it to size and
weight)? Any better solution would be welcome but this seems pretty straight forward at the moment--Mmm.
Kind regards, chris
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CHRIS25
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2Fe + 8Hno3 = 2Fe(no3)3 + 2 No + 4H2o
2x56 + 8x63 = 2x242 (or 404) + 2x30 + 4x18
That's 616 = 616
make 100 grm ferric nitrate:
484/100 = 0.21
0.21x112 (Fe) and 0.21x504 (Nitric acid) is 23.52 and 106 respectively.
So 24 grams of Iron and 106 grams of Nitric acid, outside and bobs your uncle, you have ferric nitrate.
I think now I have it? Yes.
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weiming1998
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Yes, that is correct, with some calculations being 1 or 2 off, but that doesn't really matter in this setting. So now you have learnt it.
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CHRIS25
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Thanks Welming for your help, and Peach also, also magpie.
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AndersHoveland
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I suppose you could just bubble SO2 and air into sodium carbonate solution.
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CHRIS25
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too awkward for my setup, plus I would get out of breath blowinh air down a long rubber tube!. Plus when it rains here 360 days of the year and a
blue sky can turn cloudy before you've had time to finish breakfast, i have to limit the need for outdoor work.
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CrEaTiVePyroScience
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Well I don't know if you are making this to explore the science or if you are making this because you need it for another application (for chemistry).
But anyhow if you are needing it for another application or to make somethingelse like sulfuric acid, it is availble at hardware stores as a pool
pH-lower. Check your local store!
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