Sciencemadness Discussion Board

why is my sodium chloride precipitate blue?

CHRIS25 - 9-6-2012 at 06:50

Trying desperately to follow woelen's explanations (which I do appreciate) about colour changes in NaOH solution thread I find I still can't understand what is happening, there is a lot of theory which I find difficult to relate to in the actual practical processes at this moment. For the moment I can only understand CuCl1 and CuCl2 which everybody seems to refer to them as. It's like trying to discover how to type by learning where the history of the alphabet began. I need to know what is going on and ONLY then can I dig deeper, it does not work the other way around for me, so I hope someone can understand this please.

I have 50mil of water, I add 1mil of pale green copper chloride solution (mixed with small but unknown amount of Hydrochloric acid)it turns very very pale blue, When I add a 1mil amount of NaOH I get a milky blue/white cloudy precipitate this I presume is the copper (1) hydroxide; I then add an Excess amount of 1mole solution of Sodium Hydroxide, namely 5 mils. The solution turns bright beautiful sky blue, I allow to settle and out precipitates a very fine blue powdery precipitate (settling at the bottom) leaving the rest of the solution completely clear.

Do I have copper hydroxide and sodium chloride (which can be the only result I presume) but the sodium chloride must must be impregnated with copper? Namely like so: Cu(H2O)6]2+, which means that this precipitate is not pure NaCl? But what complicates my ability to work it out is the presence of HCl in the solution. So the whole thing is confusing right now.



[Edited on 9-6-2012 by CHRIS25]

99chemicals - 9-6-2012 at 07:25

You are precipitating copper hydroxide not sodium chloride. NaCl is to souluble to precipitate that easily.

This is what you are doing:

HCl+NaOH->NaCl +H2O Since the HCl is gone the copper ions go blue because the soultion is not acidic any more.
Watch this video https://www.youtube.com/watch?v=pJSQq494oV4&feature=rela...

Then: CuCl2+NaOH->Cu(OH)2+2NaCl

The soulution you have will be sodium chloride with the copper hydroxide on the bottom.
Similar to this video https://www.youtube.com/watch?v=-7ecs0Mavzc
But he added the Copper soulution into the NaOH. That way you get a compelex of copper hydroxide in water which is dark blue. If you added the NaOH to the copper solution you get copper hydroxide which is pale blue.

Hope that helps

CHRIS25 - 9-6-2012 at 08:45

Right, way off then. Thankyou very much for that. rest assured I will watch these videos.

When you said that since the HCl is gone, the copper ions go blue, logic dictates (my logic is usually wrong but here goes) logic tells me that when I add that first 1 mil of NaOH to the 1 mil copper chloride in water and the water stays clear or very very pale blue after stirring, then the HCl is still present in solution, preventing copper hydroxide from precipitating?

99chemicals - 11-6-2012 at 15:39

Yes I would thinks so.

woelen - 11-6-2012 at 23:29

Quote: Originally posted by CHRIS25  
Trying desperately to follow woelen's explanations (which I do appreciate) about colour changes in NaOH solution thread I find I still can't understand what is happening, there is a lot of theory which I find difficult to relate to in the actual practical processes at this moment. For the moment I can only understand CuCl1 and CuCl2 which everybody seems to refer to them as. It's like trying to discover how to type by learning where the history of the alphabet began. I need to know what is going on and ONLY then can I dig deeper, it does not work the other way around for me, so I hope someone can understand this please.

I have 50mil of water, I add 1mil of pale green copper chloride solution (mixed with small but unknown amount of Hydrochloric acid)it turns very very pale blue, When I add a 1mil amount of NaOH I get a milky blue/white cloudy precipitate this I presume is the copper (1) hydroxide; I then add an Excess amount of 1mole solution of Sodium Hydroxide, namely 5 mils. The solution turns bright beautiful sky blue, I allow to settle and out precipitates a very fine blue powdery precipitate (settling at the bottom) leaving the rest of the solution completely clear.

Do I have copper hydroxide and sodium chloride (which can be the only result I presume) but the sodium chloride must must be impregnated with copper? Namely like so: Cu(H2O)6]2+, which means that this precipitate is not pure NaCl? But what complicates my ability to work it out is the presence of HCl in the solution. So the whole thing is confusing right now.

The only thing you get in this experiment is more or less pure copper(II) hydroxide. You do not get a precipitate of NaCl! Copper(II)hydroxide has a beautiful blue color, if it is very finely divided is is pale blue. You apparently have such a low concentration of chloride ions that hardly any chloride co-precipitates with the hydroxide. Otherwise your precipitate would be more greenish.

Copper(I) plays no role in this experiment, not at all. It is all copper(II) chemistry.

CHRIS25 - 12-6-2012 at 00:54

Quote Woelen:=====You apparently have such a low concentration of chloride ions that hardly any chloride co-precipitates with the hydroxide.======

So adding copper oxychloride to this solution would be the answer - I will try that. But I want to focus on that word "Co-Precipitates" The green would be the Cl(2)? that would be the co-precipitant?



[Edited on 12-6-2012 by CHRIS25]

[Edited on 12-6-2012 by CHRIS25]

Could you help me ? Both The Initial and final etchant molarity is determined by my calculation: Amount of mils of NaOH added to the 30mils of water and 1mil of CuCl in solution until copper hydroxide will not re-dissolve. Is this then the molarity of acid in solution?

volume of concentrated hydrochloric acid (liters) = VT × (CF - CI) / ( C (acid)
- CF )
where
Vt = initial etchant volume (liters)
Ci = the initial etchant acid molarity
Cf = the final etchant acid molarity
C (acid)= the molarity of concentrated hydrochloric acid being added.

I seem to have cross-posted here on this and I apologise, I did not intentionally want two posts ending up at the same point - which is what seems to have happened.

[Edited on 12-6-2012 by CHRIS25]

woelen - 12-6-2012 at 02:20

The symbol Cl(2) has no meaning. There is Cl2, which is a pale green very toxic gas. There also are Cl(-) ions, which are colorless ions in solution. These Cl(-) ions are the things you have in solution.

Your pale blue solution contains the following ions:
[Cu(OH2)6](2+) from the CuCl2
Cl(-) from the CuCl2 and the HCl
H3O(+) from the acid HCl

When you add sodium hydroxide to your solution you first have reaction between the hydroxide ions and the hydronium ions:

H3O(+) + OH(-) ---> 2H2O

If all acid is used up, then you get a precipitate of copper(II) hydroxide.

A simplified equation for this is: Cu(2+) + 2OH(-) --> Cu(OH)2
In reality the situation is somewhat more complicated, because you also still have the water molecules attached to the copper(II) ion and in reality the precipitate does not only consist of Cu(2+) and OH(-) ions, but also contains water molecules.

If the concentration of chloride ions is large enough, then part of the OH(-) ions is replaced by Cl(-) ions and you get a precipitate, which consists of Cu(2+) ions, OH(-) ions and Cl(-) ions. The more Cl(-) ions in this precipitate the more greenish the color of the precipitate.

CHRIS25 - 12-6-2012 at 03:32

Woelen, I can grapple with what you are saying, just, I have to read some more on these electrons and ions though because the theory is interesting but I don't see ions and electrons I see mixtures colours and reactions, I am keen to understand - don't mistake me, but at the moment it's a bit like learning to use a camera by first reading about LAB and RGB colour spaces and optical parameters within aperature fields - when actually how to take a photo and what you need to do comes first - then the underbody work comes second. I do appreciate your elaborations otherwise I could be days even weeks stumbling onto that page in a book that throws light.

Regarding the equation above though, I followed this and did my experiments. The Acid molarity of my CuCl solution was 2.0 upon beginning, and then 1.5 on finishing. My Acid is 11.8 g/moles and the volume of solution was 1 litre exactly.

Therefore: 1 x (1.5 - 2.0) / 11.8 - 1.5 (According to the equation above, these are my measurements)

Therefore: - 0.5 divided by 10.3 = - 0.0048. As you can see, how the h**** can I have a minus in the answer when LOGIC says that the acid has been used?

This is the problem with reading stuff on your own, I miss the demonstrations and explanations, that's why I am so busy trying to figure out what could be learned in 5 simple minutes in a classroom. Thanks though for correcting me as no doubt I need.