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Author: Subject: Copper carbonate - controlling the colour
blogfast25
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[*] posted on 24-12-2014 at 09:48


Chris:

What you should do is run one of your tests in such a way that you can measure what I wrote above:


Quote: Originally posted by blogfast25  


Quote-

Malachite:

2 Cu<sup>2+</sup>(aq) + 2 CO<sub>3</sub><sup>2-</sup>(aq) + H<sub>2</sub>O(l) === > Cu<sub>2</sub>CO<sub>3</sub>(OH)<sub>2</sub>(s) + CO<sub>2</sub>(g)

So for every 2 mol of carbonate 1 mol is shed.

Azurite:

3 Cu<sup>2+</sup>(aq) + 3 CO<sub>3</sub><sup>2-</sup>(aq) + 2 H<sub>2</sub>O(l) === > Cu<sub>3</sub>(CO<sub>3</sub>;)<sub>2</sub>(OH)<sub>2</sub>(s) + CO<sub>2</sub>(g)

So for every 3 mol of carbonate 1 mol of CO2 is shed. (Of course Azurite cannot be prepared that way).

Unquote-

By weighing everything carefully and measuring the weight loss after reaction, the molar ratio of carbonate used/CO2 lost can be established.

Mass balances don't lie: if the molar ratio was 2 (or close), what you precipitated was Malachite, it simply cannot be otherwise stoichiometrically speaking.

In any case, if that paper on M/A stability shows us one thing it's that in ordinary precipitation conditions pH will ALWAYS be (way) too high for Azurite to form.

Reading that paper reinforces my belief that Debray had the right idea and that it's me who is doing it wrong.

[Edited on 24-12-2014 by blogfast25]




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[*] posted on 26-12-2014 at 09:19


Quote: Originally posted by kmno4  

Reactions between CuSO4 and Na2CO3 were investigated and tested long time ago by several authors, for example:



Of course. We're not pretending doing anything new here.

The emphasis of this thread has shifted to the rather elusive second copper basic carbonate, Azurite.




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[*] posted on 31-12-2014 at 08:48


I'll keep this short. After 8 titrations over two weeks and two today I can safely say this: Corn starch is a complete waste of time when titrating for Cu2+ ions. Both the copper chloride and todays copper nitrate do exactly the same thing. Dirty brown/ (sand colour depending on dilution levels of analyte) turns to dirty white, then milky white on addition of thiosulphate. No matter how much corn starch (1g in 50 mL boiled water, filtered and cooled) you add, it Never changes the endpoint colour and believe me you can add one drop or half a gallon. So having said that, I titrated and considered the endpoint simply to be the nice clean milk white point. And the short of it here is that I have nothing once again, except 2 wrong readings: Second one I considered to be more accurate...
0.1M solution Copper nitrate. (0.01 mol in 100 mL water)
0.02 KI added dry
0.1 M Thiosulphate (0.01 mol in 100 mL water)

10 mL was then extracted from this solution and titrated.
11.5 mL thiosulphate to suspected endpoint.

Maths:
0.0115 L x 0.1 M = 0.00115
0.00115 / 0.01 L (what was extracted) = 0.115 mol/L

If this 10 mL contained 0.115 mol/L then obviously since it was taken from a 100 mL 0.1 M solution the whole thing is way off.




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[*] posted on 31-12-2014 at 09:14


Quote: Originally posted by CHRIS25  

0.1M solution Copper nitrate. (0.01 mol in 100 mL water)
0.02 KI added dry
0.1 M Thiosulphate (0.01 mol in 100 mL water)

10 mL was then extracted from this solution and titrated.
11.5 mL thiosulphate to suspected endpoint.

Maths:
0.0115 L x 0.1 M = 0.00115
0.00115 / 0.01 L (what was extracted) = 0.115 mol/L

If this 10 mL contained 0.115 mol/L then obviously since it was taken from a 100 mL 0.1 M solution the whole thing is way off.


Let me be clear: you weighed of 0.01 mol of what we assume to be Cu(NO3)2.6H2O, that is 2.955 g and dissolved them to a volume of 100 ml? Did you use a beaker or a volumetric flask (VF)?

You then titrated 10 ml of this analyte solution after adding the KI. You needed 11.5 mL of 0.1 M Na2S2O3 titrant solution.

That contains 0.00115 mol thiosulphate and means the 10 mL of analyte solution also contained 0.00115 mol Cu2+. In 10 mL volume that is 0.115 M Cu2+.

That's not 'way off', it's 11.5 % off our assumption. Inaccuracy in weighing the copper nitrate, use of a beaker instead of a VF, inaccuracy in preparing the titrant solution and poor end-point detection (w/o starch) can EASILY explain the discrepancy. Accurate titrations are NOT easy to do.

Re. your starch woes, I'm totally at a loss. I never have these problems.

[Edited on 31-12-2014 by blogfast25]




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[*] posted on 31-12-2014 at 09:33


Yes absolutely - what you said is what I measured exactly, once again it is my Maths to be blame, and frustratingly so. And yes, I used a beaker and despite having a weigh scales to 2 decimal points there is always discrepancies in small amounts. And on top of that I can not guarantee my concentrations and such like to be absolutely accurate for titrating, (but very good for normal chemistry though). My first titration came out at 0.113 M, and presumed this to be innacurate as well due to a bad mood at corn starch. But the maths lets me down. Anyway, at least I am assured that I have the 6 H2O and not the deca or trihydrate.

I have dissolved the Limestone and converted to calcium carbonate. It is drying at the moment and will then be able to determine the percentage of calcium carbonate in this Limestone, the one I did a few days ago (test tube size) was 45% but that sounds too wrong, so I repeated this one with a larger amount and will post when ready.

[Edited on 31-12-2014 by CHRIS25]
Ah Problem....the dilution factor was 9 not 10 since I added 90 mL water to the 10 mL sample from the mother solution. therefore this would be 0.00115 / 0.009 = 0.127

[Edited on 31-12-2014 by CHRIS25]




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[*] posted on 31-12-2014 at 09:55


Quote: Originally posted by CHRIS25  
Anyway, at least I am assured that I have the 6 H2O and not the deca or trihydrate.

[Edited on 31-12-2014 by CHRIS25]


Unfortunately, due to this measuring error, you can not make this statement with any degree of certainty.

But I believe it is the hexahydrate too and fairly small error on that should not affect the result of your experiment too much.




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[*] posted on 1-1-2015 at 07:44


Quote: Originally posted by CHRIS25  

[Edited on 31-12-2014 by CHRIS25]
Ah Problem....the dilution factor was 9 not 10 since I added 90 mL water to the 10 mL sample from the mother solution. therefore this would be 0.00115 / 0.009 = 0.127

[Edited on 31-12-2014 by CHRIS25]


No, no, no. In this case there was no dilution factor. The copper you titrated, i.e. 0.00115 mol, was still contained in 10 mL (0.01 L), so the molarity of that sample was still 0.115 M. That you added some water prior to the titration of the 10 mL is irrelevant here.

And just to give you an idea of just how inaccurate the use of beakers is, in this instance.

My 250 mL beakers have an internal diameter of 6.7 cm. That means that the surface area of the meniscus is Pi x 3.35<sup>2</sup> = 35.2 cm<sup>2</sup> (Pi is 3.14).

If I misread the level of liquid in the beaker by no more than 0.5 mm (0.05 cm) then I'm also misreading the volume by 35.2 x 0.05 = 1.8 cm<sup>3</sup> = 1.8 mL. That's a relative measuring error of 1.8 / 100 x 100 % = 1.8 %. By contrast a Class A VF would give about 0.02 / 100 x 100 % = 0.02 %.

And you're making this same error twice: once for the analyte solution, once for the titrant solution.


[Edited on 1-1-2015 by blogfast25]




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[*] posted on 1-1-2015 at 11:35


CHRIS25, what is the source of the Starch that you're using as indicator ?



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[*] posted on 1-1-2015 at 11:47


@Blogfast - See it, got it. Why on earth I did that ?
@Aga - Corn starch. It worked for titrating Ferric chloride - I had no problem achieving clarity for the endpoint. Anyway, due to you know what I am confident of no more 'unnecessarily silly' mistakes in 2015.




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[*] posted on 2-1-2015 at 05:05


Chris, have a look at this.

Scroll down to about 3/4. Is that brown CuI/I<sub>2</sub> slurry a bit like what you call a 'mess'?

http://www.csudh.edu/oliver/demos/hh-cubr/hh-cubr.htm

[Edited on 2-1-2015 by blogfast25]




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[*] posted on 2-1-2015 at 08:51


Quote: Originally posted by blogfast25  
Chris, have a look at this.

Scroll down to about 3/4. Is that brown CuI/I<sub>2</sub> slurry a bit like what you call a 'mess'?

http://www.csudh.edu/oliver/demos/hh-cubr/hh-cubr.htm

[Edited on 2-1-2015 by blogfast25]

The image that is 5 photos up from the bottom of the web page on the left side both on the left and right side - a dirty brown - exactly what I get, which I believe is normal. Changes eventually to milk white bit at this point it becomes subjective as to when the white first starts to appear over shooting is easy at tis point. My only gripe is that the corn starch does nothing, no matter how much or little you use, and believe me, I tested it. Anyway, I am sure this will be re-solved.

I am ready finally to start the two pressure bottles, I tested another batch of Limestone and after drying now have the final composition:
14.57 g Limestone dissolved in HCl. Then converted to Calcium carbonate, washed several times and dried in an oven. Yield is 11.51 g. This means that this Limestone is composed of 78.9 % carbonate. Now with knowing both the stoichiometric values of the limestone and the nitrate I can proceed less blindly.




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[*] posted on 2-1-2015 at 09:34


Your point about the limestone containing 78.9 % calcium carbonate is slightly contentious because during conversion you probably lost small amounts of calcium carbonate, for instance stuck on the filter.

But 80 % sounds in the right ballpark, definitely. Limestone's composition probably also varies a bit by provenance.

For starch I used to use 'ironing starch', that worked well. Now I use a reagent grade which is marketed as water soluble. That's what I'll send you.

[Edited on 2-1-2015 by blogfast25]




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[*] posted on 2-1-2015 at 11:03


Quote: Originally posted by blogfast25  
Your point about the limestone containing 78.9 % calcium carbonate is slightly contentious because during conversion you probably lost small amounts of calcium carbonate, for instance stuck on the filter.

But 80 % sounds in the right ballpark, definitely. Limestone's composition probably also varies a bit by provenance.
[Edited on 2-1-2015 by blogfast25]

Yes I agree, and probably lost something. However I did apply myself on this one to strict washing with a syringe to get every last grain down the sides of the filter paper and then dried the compound inside the filter paper for 48 hours (by the open fire:P) and then scraped every last bit off the dried filter paper into the oven, I am confident that any discrepancy will be Ireland's geological fault.

UPDATE
Like a twit I forgot that I should have prepared enough nitrate for a 2 litre bottle. So it is all in a 1 litre bottle now and not the quantities I wanted but it should be enough anyway for a preliminary test:
102 g Copper nitrate hexahydrate
166 mL carbonated water
100 g Limestone pieces between 1/3 and one full inch size.
Now just wait till next christmas.

[Edited on 2-1-2015 by CHRIS25]




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[*] posted on 6-1-2015 at 09:50


Update

Posting an image of the reaction, this was started on 1 january, and by day 2 the malachite was beginning to form. The hope here is that by keeping the CO2 under pressure this will maintain a slightly acidic environment and thereby maintaining enough H+ ions through dissociation of the HCO3 to create the carbonates.

The goal of course is to form the azurite.
PS, Can someone tell me please how to type reversible arrows to indicate a reaction is reversible. I can not work it out for this forum. And neither the superscript button nor the HTML for the superscript work anymore. Anyone else this problem?


IMG_1827.jpg - 247kB




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[*] posted on 2-2-2015 at 17:43


A while back I found this very interesting article on the determination of the Solubility Product of Malachite, by JF Scaife, first search result (*.pdf) of this Google search:

https://www.google.co.uk/?gws_rd=ssl#q=solubility+product+of...

It outlines the methodology as well as the result: pK<sub>s</sub> = 31.9, which indicates extremely low solubility (much, much lower for instance than CaCO<sub>3</sub>;).

Also a method for the preparation of highly pure Cu<sub>2</sub>CO<sub>3</sub>(OH)<sub>2</sub> is presented and used.

The article further confirms that by simple precipitation methods, Malachite is the only 'copper carbonate' that can reasonably be expected to form.

[Edited on 3-2-2015 by blogfast25]




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[*] posted on 3-2-2015 at 05:09


I saw this too, weeks ago, but it was such a hard read that I left it, besides, I am still none the wiser on what it is supposed to be saying, even reading through it again. (ex trucker, not ex Phd). Yes also, I saw not one reference to Azurite except in the introduction?



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[*] posted on 3-2-2015 at 08:11


The pK<sub>s</sub> is the negative logarithm of the K<sub>s</sub>, the so called solubility product. Here K<sub>s</sub> = 1.26 x 10<sup>-32</sup>, an EXTREMELY small number.

It helps explain why Cu2CO3(OH)2 precipitates and not CuCO3: the solubility limit of Cu2CO3(OH)2 is reached very easily.




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[*] posted on 3-2-2015 at 10:44


Quote:
It outlines the methodology as well as the result: pKs = 31.9, which indicates extremely low solubility (much, much lower for instance than CaCO3).

Rubbish.
Even my grandma knew the difference between solubility and product solubility and why product solubilities cannot be compared if different kinds of salts are taken into accout. Different means ex. 2-2 like PbSO4 and 1-2 like PbCl2... etc .
In another words: very large pS (pK if you like) value does not maen very small solubility (like in case of your stupid Cu2CO3(OH)2).




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[*] posted on 3-2-2015 at 11:17


Quote: Originally posted by kmno4  
Rubbish.
Even my grandma knew the difference between solubility and product solubility and why product solubilities cannot be compared if different kinds of salts are taken into accout [sic]. Different means ex. 2-2 like PbSO4 and 1-2 like PbCl2... etc .
In another words: very large pS (pK if you like) value does not maen [sic] very small solubility (like in case of your stupid Cu2CO3(OH)2).


Listen you dipshit, I'm starting to resent your tone and your presumptuousness here.

I KNOW very well what is solubility and what is a solubility product.

Calculate, as I have, the solubility S (mol/l) of CaCO3 and Cu2CO3(OH)2 in comparable conditions of pH and you'll see that Malachite is indeed much less soluble than CaCO3. Not only that, it is quite relevant re. the method of Azurite preparation we're considering here.

Want me to show you? Ask politely, otherwise go f*ck your grandma.

And what's 'stupid' about Malachite??

[Edited on 3-2-2015 by blogfast25]




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[*] posted on 3-2-2015 at 11:36


Quote: Originally posted by blogfast25  

Listen you dipshit,... blabla.....

In general, I do not discuss with idiots and their invectives. Post your calculations there, in "comparable conditions of pH", whatever it means.




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[*] posted on 3-2-2015 at 12:27


Quote: Originally posted by kmno4  
In general, I do not discuss with idiots and their invectives. Post your calculations there, in "comparable conditions of pH", whatever it means.


You can't discuss with an 'invective', imbecile.

A few weeks ago I performed the calculation of the solubility of CaCO3 and Malachite in function of CO2 pressure.

That involves setting up a neutrality balance, the relevant equibria for CO2/H2CO3/HCO3-/CO3(2-), the solubility equilibria, the auto dissociation of water equation and relevant mass balances.

From this set of equations the concentrations of all ionic species, including (Ca/Cu)<sup>2+</sup> in function of CO2 pressure can be calculated. I did not use activity coefficients in this case (but that could be easily rectified).

This calculation isn't particularly easy or difficult, more 'finnicky' because quite long algebraical expressions result from all that substituting to reduce the set to one single expression.

One could make things easier by means of certain assumptions but I chose to do it ab initio.

The excel spread sheet I used is on a computer that is currently down (damned malware!). It should be back up in a few days and I will then post it here. If you don't believe me on that point then that says more about your presumptuousness than my understanding of solubility and solubility products.

The bottom line, off the top of my head, is that Malachite's solubility is much smaller than CaCO3's, across the entire range of pH.


[Edited on 3-2-2015 by blogfast25]




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[*] posted on 3-2-2015 at 14:19


Quote: Originally posted by blogfast25  


It helps explain why Cu2CO3(OH)2 precipitates and not CuCO3: the solubility limit of Cu2CO3(OH)2 is reached very easily.

Is there any chance you could put this explanation into plain English and leave out the maths and notations please?




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[*] posted on 3-2-2015 at 16:00


Let's another example: calcium hydroxide (Ca(OH)2). This is a 'sparingly soluble' compound with a listed solubility limit at 20 C of 0.173 g/100 g of water (Wiki solubility table).

Literally this value means that at 20 C you cannot dissolve more than 0.173 g of Ca(OH)2 for every 100 g of water: any excess would simply sink to the bottom as a wet solid.

Now take 1 L of a solution that is 1 M in CaCl2. To it we add twice the volume of a sodium hydroxide solution of 1 M.

This would imply that 1 mol of Ca(OH)2 would be in solution, that is about 40 + 2 x 17 = 74 g. The total volume by then is 3 L or about 3000 g.

Divide 74 g by 3000 g and multiply by 100 g = 2.47 g Ca(OH)2/100 g water, which is way above the solubility limit of Ca(OH)2, so by definition solid Ca(OH)2 must leave the solution as precipitate (as indeed it does).

Similarly Cu2CO3(OH)2 has a very low solubility limit (but much, much lower than Ca(OH)2), so when we mix a solution of CuCl2 (for instance) with a solution than contains both OH<sup>-</sup> and CO<sub>3</sub><sup>2-</sup> ions, the solubility limit is exceeded very quickly and most (basically 100 %) of the Cu2CO3(OH)2 precipitates out.

Had the solubility limit of CuCO3 been even lower it would have been reached first and CuCO3 would precipitate out.


[Edited on 4-2-2015 by blogfast25]




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[*] posted on 4-2-2015 at 03:12


Ok, so I need to know what chemistry principle is at work (maybe a new topic to learn?) that allows calcium carbonate to be precipitated without OH ions and yet copper carbonate is stuck with those OH ions? Drawing with my limited understanding I can not see what prevents the OH ion doing the same thing with calcium carbonate. The carbonate molecule is trigonal planar, the Cu and Ca atoms both have 2+ charge, the 2 single oxygen bonds coming out of the carbon atom are happy to have some attraction to the the copper and calcium atoms though this is not ionic and obviously not covalent, it just seems to be an attraction since the copper has a slight positive and the oxygen has a slight negative and the oxygen has no need of any electrons since its shell is complete with the carbon (any special term for this)?. They both also have negligible solubilities.

Why (according to principles I have been learning) does copper even have a 2+ charge possible since the 3d block is full and the 4s orbital has only one electron and is a higher energy level; this question only arose because I was wondering what was so different about copper that it should have the wish to attract the OH ion.

I know I know, I can learn and learn read and read study study, but if I don't ask these questions I will never fully understand what I have just learned:)




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[*] posted on 4-2-2015 at 06:23


Compare calcium and magnesium in that respect: Mg does form a basic carbonate, i.e. Mg2CO3(OH)2, Ca does not.

Now if you look at the solubility limit of Mg(OH)2: it is much lower than that of Ca(OH)2.

So we have a bit of a pattern here: Mg and Cu both have very insoluble hydroxides and form basic carbonates, Ca(OH)2 is 'sparingly soluble' and Ca does not form a basic carbonate.

As regards the existence of Cu(II), despite that it requires making the 4d orbital only part filled, I have no off-the-cuff explanation.


[Edited on 4-2-2015 by blogfast25]




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