CHRIS25 - 2-6-2012 at 13:06
I was looking for ways to control HCL content within copper chloride solution and discovered someone saying the following: That 453.6 grams of copper
needs 3.78 litres of 8m HCL to dissolve (forget the H2O2), the litres were converted from the US gallon not the UK gallon.
So therefore 1 gram of copper needs 12.45mils of 12m HCL to dissolve.
Now I have no idea how or where to check this sort of thing up. Needless to say I understand that solubility figures on Wikpedia and the like refer
only to water. So does anyone know if this can be verified by some database or has anone any knowledge please.
I say forget about the equation and need for oxygen via air pump because that is a variable where no regulation is required. (I am not using H2O2
anymore).
Thankyou
nezza - 3-6-2012 at 03:33
Copper is (usually) Divalent so a simple equation would be :-
Ignoring the requirement for an oxidiser.
Cu + 2H+ > Cu2+ + H2
So 1 gFormula weight of Copper needs 2 gFormula weight of acid.
The Mwt of Copper is 63.5
A 1M solution of a salt is 1 gFormula weight dissolved in 1 Litre.
63.5g Cu would need 2 Litres of 1M HCl or 1/8 (250ml of 8M HCl)
453.6g Cu would therefore need 453.6/63.5 x 250 ml of 8M HCl
This is 1.78l, not 3.78l.
CHRIS25 - 3-6-2012 at 04:59
Hi Nezza, this is where I am going wrong when it comes to calculating acids into mils. Take the original equation:
Cu+2HCL = CuCl2 +2H2o Ignore the second half of the equation,
Stoichemetry is : 63.5g + 72g (2x36g) Density of HCL is: 1.18g/cm3.
Therefore since one measures in Mils this needs to be converted to: 72/1.18 = 61mils which is ridiculous obviously
However I know that this must be wrong because: learning from your example I worked out that 63 gram copper needs 166mils of 12m solution to
dissolve. Obviously there is a conflict here so I assume that one can never ever take the molecular weight of a solution and convert to mils which is
a pain since I am fed up with reading solution requirements in grams which is very very irritating.
In the case of HCL I can only assume that since 12m would have about 64% water in it I would have to add the molecular weight of water to the 2HCL?
PS I am slow with maths, don't have the right brain waves for seeing things easily, so I am appreciating your explanations - thankyou.
[Edited on 3-6-2012 by CHRIS25]
unionised - 3-6-2012 at 07:16
If the density of the HCl solution is 1.18 then it's not 8M.
1 g of copper is about 1/63.5 moles
that takes 2/63.5 moles of HCl ( plus an oxidiser).
0.0314 moles
Commercial conc HCl has a density of about 1.18 g/ml and is about 12M
so it contains 1 mole in 1/12 litres (i.e. 83 ml).
1g of copper needs the Cl from 2.6 ml of HCl to make CuCl2
However the question is a bit academic if you plan to blow air through it, since some of the HCl will be lost by evaporation.
[Edited on 3-6-12 by unionised]
CHRIS25 - 3-6-2012 at 13:53
===1 g of copper is about 1/63.5 moles
that takes 2/63.5 moles of HCl ( plus an oxidiser).
0.0314 moles
Commercial conc HCl has a density of about 1.18 g/ml and is about 12M
so it contains 1 mole in 1/12 litres (i.e. 83 ml).
1g of copper needs the Cl from 2.6 ml of HCl to make CuCl2====
Well I arrived at exactly the same answer as you when I said above 63 grams copper needs 166mil (166/63 = 2.63). However I followed your maths 10
times and still could not work out your reasoning on the last sentence "so it contains 1 mole in 1/12 litres (i.e. 83 ml).
1g of copper needs the Cl from 2.6 ml of HCl to make CuCl2" I do not see how you are working out things here? I get the answer but not the method that
you are mentally using, Sorry. And why do you say" "...that takes 2/63.5 moles of HCl...."? Why are you dividing 2 by 63.5, you mention 0.0314...but
where is this figure used in your maths?
the point is that following Nezza's method I understood it. i was trying to ascertain why there would be a molecular weight given to a solution
since it plays no part in all this maths. I understand that you add up the chlorine and hydrogen weights due to the H and CL, but I was simply asking
that this is irrelevant when it comes to solutions? Hence my original mistakes of taking the HCL and using the weight of 36 in stoichemetry - which
now I see to be wrong.
Your last comment I agree with, "However the question is a bit academic if you plan to blow air through it, since some of the HCl will be lost by
evaporation." But on the other hand this is relevant because I need guidelines.
[Edited on 3-6-2012 by CHRIS25]