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Rochelles salt

CHRIS25 - 18-6-2014 at 04:41

All I am wanting to do is to understand how this reaction works:
I have made 3 batches and crystals, (now am proceeding to make the biggest crystal I can)
Short results: Sodium Carbonate + Potassium Hydrogen tartrate:
Consistent results reveal that you need a 2:1 (Mole) ratio Carbonate to Bitartrate ratio to achieve full dissolution of tartrate.
No matter how hard I try to balance this equation I can not. the closest I get is this: All balanced apart from carbon.
4xK
4xNa
18xC (20 on the right side)
20xH
30xO
Looking out the atomic structures on wolfram alpha I see that 4 Oxygens are lost and 2 hydrogens, This can be translated into the following equation:
4KC₄H₅O₆+4Na₂CO₃=4NaKC₄H₅O₆+2CO₂+H₂

However if I follow the ratios above this is what I get:
KC₄H₅O₆+2Na₂CO₃=2?NaKC₄H₅O₆+2CO₂+H₂
Maybe somebody could just direct me at this point to understand how this reaction proceeds since it appears that ions are "disappearing". But particularly how the hydroxide ion is torn away from the bitartrate molecule to make it dissolve, still on wobbly feet when it comes to dissecting the atom level of reactions.

[Edited on 18-6-2014 by CHRIS25]

[Edited on 18-6-2014 by CHRIS25]

[Edited on 18-6-2014 by CHRIS25]

blogfast25 - 18-6-2014 at 05:11

Try and write it like this:

KHTr + Na2CO3

...where Tr is the tartrate ion. It then becomes a simple acid/base reaction:

2 KHTr + Na2CO3 === > 2 NaKTr + CO2 + H2O

Tr = C4H4O6<sup>2-</sup>, with one Na+ and one K+ that becomes neutral, as it should be.

[Edited on 18-6-2014 by blogfast25]

hyfalcon - 18-6-2014 at 05:29

Quote: Originally posted by blogfast25

Try and write it like this:

KHTr + Na2CO3

...where Tr is the tartrate ion. It then becomes a simple acid/base reaction:

2 KHTr + Na2CO3 === > 2 NaKTr + CO2

Isn't there a water in there somewhere to balance things?

2 KHTr + Na2CO3 === > 2 NaKTr + CO2 + H2O

aga - 18-6-2014 at 05:32

Quote: Originally posted by CHRIS25

KC₄H₅O₆+2Na₂CO₃=2?NaKC₄H₅O₆+2CO₂+H₂

I think there's one too many hydrogens in the salt formula.

KNaC₄H₄O₆

source : http://en.wikipedia.org/wiki/Potassium_sodium_tartrate

Edit: I dunno why, but evolving two gasses seemed wrong.

2 KC₄H₅O₆ + Na₂CO₃ = 2 KNaC₄H₄O₆ + CO₂ + H₂O

[Edited on 18-6-2014 by aga]

hyfalcon - 18-6-2014 at 05:51

This would be done in aqueous conditions so the additional H2O would only add to the solvent here.

aga - 18-6-2014 at 05:54

The h20 makes the equation balance if the salt formula nis H4 not H5, and if it's an acid-base reaction, water must be a product no ?

CHRIS25 - 18-6-2014 at 06:29

@Aga, embarrassed, you spotted well.Ugh mm!
@Blogfast but the reality is we need always 2 moles carbonate to 1 mole bitartrate. I see that this makes sense: 2 KHTr + Na2CO3 === > 2 NaKTr + CO2 + H2O and is balanced. But that leaves the fact that we have a terrible inconsistency between the theoretically balanced formula and the actual practical reality.

This is the practical, but not balanced: and it simply can not be balanced unless we re-write chemical notation with a 'Tr', I simply can not see that this is correct.
Sodium ions are disappearing, if you write a 2 on the right side, then you have less potassium ions, and so this cycle is endless and there seems no resolution. Someone has the answer....
KC₄H₅O₆ + 2Na₂CO₃ = KNaC₄H₄O₆ + H₂O + CO₂

Nicodem - 18-6-2014 at 07:25

Quote: Originally posted by CHRIS25

@Blogfast but the reality is we need always 2 moles carbonate to 1 mole bitartrate. I see that this makes sense: 2 KHTr + Na2CO3 === > 2 NaKTr + CO2 + H2O and is balanced. But that leaves the fact that we have a terrible inconsistency between the theoretically balanced formula and the actual practical reality.

I don't know what you talk about here. There is no inconsistancy whatsoever. There cannot be any when you are just doing stoichiometry by balancing equations. It is a mathematical method so the theory equals the experimental (stoichiometry is one of the few laws of chemistry). Even if you write an equation based on the rational formulas, you still get the same result:

2HOOC-CH(OH)CH(OH)-COOK(aq) + Na₂CO₃(aq) ---> 2NaOOC-CH(OH)CH(OH)-COOK(s) + H₂O + CO₂

The way you write the equation is completely irrelevant for the final outcome of the stoichiometry. The only thing that matters is finding out which bond form and which break (or as in this case also which ions are incorporated in the crystal lattice). You could just as well write the equation as:
2AX + B₂CD -> 2AB + CX₂ + D ... and obtain the same stoichiometry.

aga - 18-6-2014 at 07:33

If the forumlae are correct, you cannot rewrite the tartrate as Tr as there is loss of 1 hydrogen atom somewhere in the reaction isn't there ?

That would then be Tr -> Tr - H at some stage.

I gotta say that the idea is appealing, as it's hard to keep track of the the H C and O.

[Edited on 18-6-2014 by aga]

aga - 18-6-2014 at 07:43

Quote: Originally posted by CHRIS25

we need always 2 moles carbonate to 1 mole bitartrate.
KC₄H₅O₆ + 2Na₂CO₃ = KNaC₄H₄O₆ + H₂O + CO₂

Most likely that the concentration of one of your reactants is out by a factor of two, or maybe four.

The balanced equation says it's 2 mol tartrate to 1 mol carbonate.
Maybe the pH affects the reaction, needing a huge excess of carbonate ?

I knew the Al Sulph horror would come in handy one day !

[Edited on 18-6-2014 by aga]

CHRIS25 - 18-6-2014 at 07:57

Quote: Originally posted by Nicodem

Quote: Originally posted by CHRIS25

I don't know what you talk about here. There is no inconsistancy whatsoever. There cannot be any when you are just doing stoichiometry by balancing equations.

I detect your irritation at my ignorance, with a simple lab and kitchen stool and no formal education I do my best. Let me explain please, and maybe you could address my ignorance:
In Your perfectly reasonable formula and the one that I also expected to see; there is an inconsistency that I do not understand undoubtedly due to lack of experience education knowledge, simple for you but bloody difficulty for me to see: In the formula You have 2 K ions and 2 Na ions producing the same in the reaction. In the actual process of making the crystals you have to use 2 moles of Sodium carbonate, this is 4 Na ions to 2 K ions (from the Tartaric acid salt) in order to achieve dissolution and make rochelles salt. This is what I am seeing, so what do I not understand in order to resolve this please.

[Edited on 18-6-2014 by CHRIS25]

blogfast25 - 18-6-2014 at 08:30

Ooopsie... I did forget a H2O there. My bad.

Chris:

Assuming you're making Rochelle's salt by mixing sodium carbonate with potassium bitartrate you need to mix 2 mol of potassium bitartrate with 1 mol of sodium carbonate (or different quantities in that ratio), both as solutions. On evaporation you then obtain the solid salt (a double salt of K2Tr and Na2Tr, I believe).

Using different molar ratios will not work to produce Rochelle's salt, unless it is poorly soluble at low temperature, for instance.

What method precisely are you using?

[Edited on 18-6-2014 by blogfast25]

Nicodem - 18-6-2014 at 09:19

Quote: Originally posted by CHRIS25

In the actual process of making the crystals you have to use 2 moles of Sodium carbonate, this is 4 Na ions to 2 K ions (from the Tartaric acid salt) in order to achieve dissolution and make rochelles salt.

Let me explain it in simple words: No, you do not need 2 equivalents of sodium carbonate in order to prepare potassium sodium tartrate from potassium bitartrate.

Since you produced no reference up to now, it is obvious that you made a faulty assumption based on who knows what. In general, when you are having troubles understanding a problem, always check all your assumptions first. Oftentimes it is a simple matter of checking the literature for even those assumptions that you firmly trust.

I'm sorry, if I appeared irritated and if this appearance irritated you. It wasn't my intention.

CHRIS25 - 18-6-2014 at 09:26

Oh? All batches are made the same way. The last batch went as follows: Measured out 0.1 mol Potassium Bitartrate and put into 100 mLs water and heated to 35 c. Insoluble naturally, then added sodium carbonate until fizzing stopped and solution goes clear, at this point you know that all the bitartrate has now dissolved and is in solution as the tartrate ion. 28.2g bitartrate took 21.86g of Na Carbonate to dissolve. All batches eventually crystallize, the crystals conform in appearance to images I know from chemistry sites. Was that simple.
(the previous solution took 78.7g carbonate to dissolve 94g bitartrate)

[Edited on 18-6-2014 by CHRIS25]

Etaoin Shrdlu - 18-6-2014 at 09:59

Chris, are you sure you don't have either your reagents or your molar calculations flipped?

blogfast25 - 18-6-2014 at 10:37

Quote: Originally posted by CHRIS25

28.2g bitartrate took 21.86g of Na Carbonate to dissolve. [Edited on 18-6-2014 by CHRIS25]

Which type of sodium carbonate did you use: anhydrous (zero water)? Decahydrate (commercial 'washing soda' is mostly Na2CO3.10H2O)? Did you add it as a solution or as a solid?

28.2 g potassium bitartrate is 28.2/188.177 = 0.15 mol and would require 0.075 mol of sodium carbonate to neutralise it.

Adding carbonate 'until the fizzing stops' is highly subjective...

[Edited on 18-6-2014 by blogfast25]

CHRIS25 - 18-6-2014 at 10:40

Quote: Originally posted by Etaoin Shrdlu

Chris, are you sure you don't have either your reagents or your molar calculations flipped?

No. My first reaction was 5.3g 0.05 mol carbonate + 28.2 g 0.15 mol bitartrate and so should it be mol ratio wise. However as stated previously I always almost have to add double the moles carbonate to dissolve the bitartrate, and no it's not sodium bicarbonate, it is washing soda crystals.................................damn it......................contains sodium carbonate decahydrate more than 30%.....suddenly, just as I was writing I decided to check the Irish packet. Assumption assumption...........

@blogfast Decahydrate (commercial 'washing soda' is mostly Na2CO3.10H2O)? Did you add it as a solution or as a solid? as a solid, and yes, the heat ....it was 0.15, typo, no excuse.

[Edited on 18-6-2014 by CHRIS25]

blogfast25 - 18-6-2014 at 11:49

You've probably made Rochelle's salt and it crystallised out, leaving behind (in solution) any excess sodium carbonate that was inadvertently added.

You can obtain anhydrous Na2CO3 by heating sodium bicarbonate (baking soda) for about an hour at 220 C. Also, washing soda will dehydrate in those conditions to anhydrous sodium carbonate.

2 NaHCO3 === > Na2CO3 + CO2 + H2O

[Edited on 18-6-2014 by blogfast25]

CHRIS25 - 18-6-2014 at 13:24

I let the whole solution evaporate in the hot sun. There was no wet solid or solution left, just hard crystals. So wherever there was extra sodium carbonate it must have vanished just like that iron

I admit that the washing soda is the problem, being not completely sodium carbonate, which was my fault, maybe I should not have taken so many peoples' references to washing soda, they obviously do not know this themselves. Minimum 30% means 70 percent of what else?

@Nicodem, you are right, I made an assumption. That washing soda was sodium carbonate.

[Edited on 18-6-2014 by CHRIS25]

aga - 18-6-2014 at 13:43

"Assumption is the mother of all fuckups" - some Steve Seagull film.

@blogfast25: does your oven ever smell of roasting onions, garlic, etc ?
Dehydrating stuff in there is ok once in a while i suppose ...

[Edited on 18-6-2014 by aga]

blogfast25 - 19-6-2014 at 04:51

Quote: Originally posted by CHRIS25

Minimum 30% means 70 percent of what else?

Mostly other, lower hydrates because Na2CO3.10H2O isn’t stable in air and loses water (‘effloresces’, as the term goes) over time. In addition, commercial washing soda often contains small amounts of an anti-caking agent, like MgO, which can only be removed by dissolving the soda and filtering off the MgO.

From crude washing soda quite pure anh. Na2CO3 can be obtained with simple purification methods.

Quote: Originally posted by aga

@blogfast25: does your oven ever smell of roasting onions, garlic, etc ?

Dehydrating stuff in there is ok once in a while i suppose ...

I use an old cooking gas oven (converted to propane) in my lab, so I don’t have that problem. But drying non-toxic stuff in a kitchen oven is not a problem…

[Edited on 19-6-2014 by blogfast25]

pesco - 29-3-2016 at 10:51

Quote: Originally posted by aga

If the forumlae are correct, you cannot rewrite the tartrate as Tr as there is loss of 1 hydrogen atom somewhere in the reaction isn't there ?

That would then be Tr -> Tr - H at some stage.

I gotta say that the idea is appealing, as it's hard to keep track of the the H C and O.

[Edited on 18-6-2014 by aga]

In some above posts potassium bitartrate was quoted as KC₄H₅O₆. Better version would be KHC₄H₄O₆.

Having that in mind the below is perfectly sound :

Na₂CO₃ + 2 KHTr + 7 H₂O = 2 KNaTr*4H₂O + CO₂

I have a question.
Is NaHCO₃ not reacting like Na₂CO₃ ?
If it does react then step with drying NaHCO3 wont be needed.
Have anyone tried ?

On paper both work, but paper is not a beaker ...

pesco - 30-3-2016 at 10:11

OK, tested it last night.

NaHCO₃ works just as well as Na₂CO₃, so cooking bicarbonate is not necessary.
Save time and money/energy.

Warming up the mixture is also not necessary, but it speeds up reaction a lot. I mean A LOT.

I've grown whole bunch of tiny (at first) crystals by saturating solution NaKTr while still warm and then cooling and testing piezo effect of grown crystals.

Below one more important note. More of past experience, common sense and a hunch combined. No scientific test carried out to verify its validity. Comments very much welcome.

To avoid any (bi)carbonate to remain in solution after reaction I add a bit of an excess of KHC₄H₄O₆. Its low solubility + common ion effect should result in (nearly) all of it being removed during filtration after reaction ended.

Regards

[Edited on 30-3-2016 by pesco]

chemrox - 30-3-2016 at 13:10

Solutions of Rochelle salt are sometimes used for quenching reactions. I don't know why. I think I read about in context of C = C (alkene) reduction. Is this at all familiar to anyone? Glyoxylamide reduction?

Dwarvensilver - 22-10-2016 at 14:21

Greets all,

I synthesized Rochelle salt using Sodium carbonate and Potassium Bitartrate in a boiling water bath.
I know you can purchase more "Cream of tartar" but I have access to a lot of sodium L-tartrate, Can this be used with say Potassium carbonate or other potassium salt to make Rochelle salt?
Cheers,

Dwarven
Dang I wish I remember more about my basic chemistry! ah well off to the books.
ok Apparently it will work with K2CO3 if I start with Monosodium Tartrate "2 C4H9NaO8 + K2CO3 + 3 H2O = 2 KNaC4H4O6*4H2O + CO2" now how to remove an Na from sodium tartrate to get
" Monosodium tartrate
Maybe this is not a great idea, looks like I purchase some Cream of tartar

[Edited on 2016-10-23 by Dwarvensilver]

[Edited on 2016-10-23 by Dwarvensilver]

DJF90 - 23-10-2016 at 09:45

Quote: Originally posted by chemrox

Solutions of Rochelle salt are sometimes used for quenching reactions. I don't know why. I think I read about in context of C = C (alkene) reduction. Is this at all familiar to anyone? Glyoxylamide reduction?

Occasionally used to work up reactions involving Al or Ti species, as it forms soluble complexes. LiAlH4 reduction is probably the most common application as a quench reagent, but also useful for others.