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Wine-making Experiment Suggestions and Help

by FredericChopin
Tags: experiment, suggestions, winemaking
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FredericChopin
#19
Jul11-14, 02:21 AM
P: 57
EDIT:

Quote Quote by Borek View Post
The only thing you have to add to your system is the concentration of K+ - once you plug it into the charge balance, you will have a new system of equations. Just solve.
What if I don't have the concentration of K+? Am I doomed?
Borek
#20
Jul11-14, 03:07 AM
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Quote Quote by FredericChopin View Post
Is that because; 1, KHT is a salt and not an acid, and 2, KHT dissociates completely (because it's a salt) so there isn't a dissociation equation?
It dissociates completely.

Also, that would leave 6 unknowns and 5 equations, unless...
No. Amount of K+ (or amount of KOH) has to be given. Just like you have Ca, you need Cb. Note that technically introducing KOH you introduce two species: K+ and KOH.

Quote Quote by FredericChopin View Post
What if I don't have the concentration of K+? Am I doomed?
Yes.

Note, that the system is a mixture of tartaric acid and KOH. Its final pH depends on the composition - so you need both concentrations of acid and base to be able to calculate pH.

But I think showing how the system works and how it can be used to calculate pH should be enough - even if you don't have enough data.

Alternatively, starting with pH you can write system of equations and solve it for Cb.
FredericChopin
#21
Jul11-14, 03:42 AM
P: 57
Ok then. Too bad, but I guess it's not a disaster because, as you said, I can show the concentration of the base (I have the pH data, but needed to verify it mathematically).

Thank you for helping me.
FredericChopin
#22
Jul11-14, 10:42 PM
P: 57
Quote Quote by Borek View Post
It dissociates completely.
Actually, there is one thing I need to address. It is a quote from our textbook "Chemistry In Use: Book 2" (Deb Smith, Sue Monteath, Mark Gould, Roland Smith):

"A consequence of adding tartaric acid may be a reaction between potassium ions (K+) present in the must and hydrogen tartrate ions (HC4H4O6-). The reaction produces the acid salt, potassium hydrogen tartrate (KHC4H4O6 - cream of tartar), which precipitates as white crystals in the must:

K+ (aq) + HC4H4O6- (aq) <=> KHC4H4O6 (s)

This reaction may occur regardless of whether extra tartaric acid is added.
"

This quote suggests that the precipitation or dissociation of potassium hydrogen tartrate is an equilibrium reaction. Thinking about it, it does kind of make sense since this reaction is grape's natural buffer (see the bottom of Page 3: https://people.ok.ubc.ca/neggers/Che...0in%20wine.pdf). If there is too much hydrogen ion concentration, then the equilibrium will reduce it by reacting the hydrogen ions with the potassium ions to form potassium hydrogen tartrate. If there is a deficiency in hydrogen ion concentration, then the potassium hydrogen tartrate can dissociate to produce more hydrogen ions and potassium ions. What I'm saying is that while it is not Ka, could there just an equilibrium constant "K"? Which may have been the 21,739.1 I used in my calculation? (I researched the value. It said the pKa of potassium hydrogen tartrate, so I just converted it into Ka and then took its reciprocal to reverse the reaction. See the table on Page 3: http://apbrwww5.apsu.edu/robertsonr/...%20Ka%20MM.pdf)

In our experiment, we did add tartaric acid to the must, and we observed the formation of crystals in the wine, so that must have been what was happening.

If there exists this equilibrium reaction, then I would have one more equation but one more variable to solve for, which then I might come up with another equation to solve? Or maybe not. I'm not sure... What do you recommend I do?
Borek
#23
Jul12-14, 01:57 AM
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Do you know what solubility product is?

You will always have enough equations to describe the system (otherwise the system won't be able to find the equilibrium, you can think about it as if the system was a "solver" for the given system of equations).
FredericChopin
#24
Jul12-14, 03:45 AM
P: 57
Quote Quote by Borek View Post
Do you know what solubility product is?
After reading your post, I looked up what solubility product is, and so now, I only have a superficial understanding of it.

But it was enough for me to manage to find the Ksp value for potassium hydrogen tartrate and find up with an equation.

I did a system of equations, but something went wrong... (again )

The details in are the attachments.
Attached Thumbnails
!!!!!.jpg   !!!.jpg   !!.jpg  
Borek
#25
Jul12-14, 04:40 AM
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This is tricky. Ksp is conditional. Solid appears once the reaction quotient for the dissolution reaction gets higher than Ksp. So the correct way of writing it is in this case

[tex][K^+][HT^-] \le K_{sp}[/tex]
FredericChopin
#26
Jul12-14, 06:27 PM
P: 57
Hm... I tried putting that into two different system of equations solvers but they couldn't calculate the answer.

It must be because of the inequality sign. In which case, how would you solve it by hand?

Also, why doesn't [KHT] doesn't appear in the set of equations? Is there a mass balance equation missing?
Attached Thumbnails
Does Not Compute.jpg  
Borek
#27
Jul13-14, 02:05 AM
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KHT should be present in two mass balances - one for potassium, one for tartaric acid.

No idea what solver you use (and can't say I have a lot experience with them), but can't you add a condition that all variables are greater than 0 and lower than what mass balances define?
FredericChopin
#28
Jul18-14, 09:02 PM
P: 57
Hello again,

I submitted my assignment yesterday, but I managed to do the systematic equilibria calculations including the buffer.

Even though I've already handed it in, can you check if it is correct?

Thank you.
Attached Files
File Type: pdf Done 2.pdf (141.8 KB, 2 views)
Borek
#29
Jul28-14, 04:43 AM
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Sorry for not getting to you earlier, I was away.

First thing I don't like is [KHT] - there is no such thing as KHT concentration, as it is a solid. That makes mass balances more difficult, as you need to take into account the initial volume and number of moles of solid present.

If you are still interested, I can look into details later.


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