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Ramium
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Thanks!
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Ramium
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Quote: Originally posted by blogfast25  |
If you use fairly dilute NH3 and make sure not to exceed the NH3 stoichiometry of:
Cu2+ + 2 NH3 + 2 H2O === > Cu(OH)2(s) + 2 NH4+
... then only cupric hydroxide will form.
But with strong ammonia in excess then:
Cu(OH)2(s) + 4 NH3 === > [Cu(NH3)4]<sup>2+</sup>(aq) + 2 OH-
... will occur.
So keeping the molar ratio NH3/Cu2+ below 2 should result in Cu(OH)2 only. |
Thank you for this explanation. I now need to know how many grams of CuSO4 and how many ml of ammonia (which is 18gm/litre) i should use to make a
decent amount of Cu(OH)2 (s). I would appreciate any help.
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blogfast25
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For every 250 g of CuSO4.5H2O you need 70 g of 'ammonium hydroxide'.
Work it out from there and I will check your calculations.
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Ramium
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| Quote: Originally posted by blogfast25 | For every 250 g of CuSO4.5H2O you need 70 g of 'ammonium hydroxide'.
Work it out from there and I will check your calculations. |
So if I had 250 gm of the CuSO4.5H2O
Would i weigh out 70gm of ' ammonium hydroxide' ?
Or if it says on bottle that it has 18g/l would that be 70g divided by 18 = 3.88 litres?
If it was 3.88 litres i would want to scale down those quantities
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CHRIS25
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I will give you an example and maybe you could work out your own calculation? A standard bottle of Ammonia gas in water (some call it ammonium
hydroxide but won't get into that), is 28%. You appear to have only 1.8% (18g/L). So when I do this to convert grams of gas per litre to mL liquid
this is my calculation:
I need say 60 g of NH3 in my reaction.
I look at a chart that tells me Ammoniia solution has a Molecular weight of 35 g/mole (I am rounding everything off for convenience). This chart also
tells me that this is 28% and has a specific gravity of 0.88 (leaving aside temperature fluctuations as this is not so important for most of us). So
I need to do some maths:
60 g / 0.88 = 68 g Then 100% / 28% = 4
4 x 68 g = 243
This 243 is the number of mL that will contain around 60 g of ammonia gas.
[Edited on 28-1-2015 by CHRIS25]
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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blogfast25
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| Quote: Originally posted by Ramium |
So if I had 250 gm of the CuSO4.5H2O
Would i weigh out 70gm of ' ammonium hydroxide' ?
Or if it says on bottle that it has 18g/l would that be 70g divided by 18 = 3.88 litres?
If it was 3.88 litres i would want to scale down those quantities |
Yes, 3.88 L for 250 g of CuSO4.5H2O.
Bear also in mind that 250 g of CuSO4.5H20 would have to be dissolved in about 1 L of water.
So scale down: 25 g of CuSO4.5H20 in 100 mL of water. Slowly and with constant intense stirring add 388 mL of your ammonia solution.
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blogfast25
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Nope. An ammonia solution doesn't really have a molecular weight ('molar mass' or MM is a better term) because it's a solution comprised of 2
components: ammonia (NH3) and water (H2O). The MM of ammonia is 18 g/mol.
35 g/mol is the MM of the hypothetical 'ammonium hydroxide', NH<sub>4</sub>OH.
[Edited on 28-1-2015 by blogfast25]
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CHRIS25
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| Quote: Originally posted by blogfast25 |
Nope. An ammonia solution doesn't really have a molecular weight ('molar mass' or MM is a better term) because it's a solution comprised of 2
components: ammonia (NH3) and water (H2O). The MM of ammonia is 18 g/mol.
35 g/mol is the MM of the hypothetical 'ammonium hydroxide', NH<sub>4</sub>OH.
[Edited on 28-1-2015 by blogfast25] |
I know that ammonia gas dissolved in water is not ammonium hydroxide, but I never ever took the weight of 17g/mol in any of my reactions because I
was not obviously using the gas per se, using the solution I always used the 35g/mol. So are you saying that we should never use this latter figure?
[Edited on 28-1-2015 by CHRIS25]
[Edited on 28-1-2015 by CHRIS25]
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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blogfast25
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No, not really but you are at risk of confusing things.
Take the precipitation reaction we're talking about here. It can be written in two equivalent ways:
Cu2+ + 2 NH3 + 2 H2O === > Cu(OH)2(s) + 2 NH4<sup>+</sup>
or:
Cu2+ + 2 NH4OH === > Cu(OH)2(s) + 2 NH4<sup>+</sup>
In the first case you would need 2 x 18 g of NH<sub>3</sub> per mol of CuSO4, in the second case 2 x 35 g of
NH<sub>4</sub>OH per mol of CuSO4. Needless to say, these amounts are perfectly equivalent to each other.
If your ammonia solution, as is the general case, is expressed in % NH<sub>3</sub> then it makes more sense to do the
calculation in g NH<sub>3</sub> solution needed to conduct your specific reaction.
[Edited on 28-1-2015 by blogfast25]
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Ramium
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| Quote: Originally posted by blogfast25 | | Quote: Originally posted by Ramium |
So if I had 250 gm of the CuSO4.5H2O
Would i weigh out 70gm of ' ammonium hydroxide' ?
Or if it says on bottle that it has 18g/l would that be 70g divided by 18 = 3.88 litres?
If it was 3.88 litres i would want to scale down those quantities |
Yes, 3.88 L for 250 g of CuSO4.5H2O.
Bear also in mind that 250 g of CuSO4.5H20 would have to be dissolved in about 1 L of water.
So scale down: 25 g of CuSO4.5H20 in 100 mL of water. Slowly and with constant intense stirring add 388 mL of your ammonia solution.
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Thank you for that. I tried the experiment and ended up with a dark blue solution and a small amount of blue precipitate. I'm guess the precipitate is
copper hydroxide and the solution is a copper complex. Is that so?
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blogfast25
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| Quote: Originally posted by Ramium |
Thank you for that. I tried the experiment and ended up with a dark blue solution and a small amount of blue precipitate. I'm guess the precipitate is
copper hydroxide and the solution is a copper complex. Is that so? |
This indicates that you probably added too much ammonia solution. The dark blue is indeed the copper tetrammine - [Cu(NH3)4]<sup>2+</sup>
- complex.
The precipitate is Cu(OH)2.
The colour of the solution is deceptive though because the colour of the complex is very, very intense and so low concentrations of it can cause
significant blueing of a solution.
Try again, reducing the amount of ammonia used by 20 to 40 %.
[Edited on 29-1-2015 by blogfast25]
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gdflp
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Yes, the blue precipitate is copper hydroxide. It sounds like you may have used too much ammonia, copper hydroxide is soluble in ammonia as it forms
the water soluble complex [CuNH34H2O2](OH)2, also known as Schweizer's Reagent. If
you decant off the deep blue solution, and add a dilute acid to it slowly until you see a precipitate forming, you might be able to recover some more
copper hydroxide.
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blogfast25
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[CuNH34H2O2](OH)2, also known as:
[Cu(NH<sub>3</sub> <sub>4</sub>(H<sub>2</sub>O)<sub>2</sub>](OH)<sub>2</sub>
But adding HCl will not work here.
By adding HCl to [Cu(NH<sub>3</sub><sub>4</sub>(H<sub>2</sub>O)<sub>2</sub>](OH)<sub>2</sub> the strongest base, the 2
OH<sup>-</sup>, would be neutralised first to [Cu(NH<sub>3</sub><sub>4</sub>(H<sub>2</sub>O)<sub>2</sub>]Cl<sub>2</sub>.
Adding further HCl will convert that to CuCl2 plus NH4Cl.
[Edited on 29-1-2015 by blogfast25]
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Ramium
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I tried again with less ammonia and got a much higher yield
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blogfast25
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Well, there you go then!
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