# Wine-making Experiment Suggestions and Help

1. Jun 29, 2014

### FredericChopin

I am currently in Year 12 (high school) and have been doing a wine-making experiment for chemistry. This has involved crushing grapes, taking their juices (removing the skins), pouring them into flasks and leaving them to ferment with yeast.

The aim of our experiment was to see how the amount of tartaric acid added to the juices (before fermentation) affected the amount ethanol produced at the end of fermentation.

In our experiment, we had six flasks each containing the juices; one control where no tartaric acid was added, and then in the other five flasks, tartaric acid added to them in increments of 5 ml.

Our hypothesis was that because yeast has acid tolerance levels, there were be an optimum amount of acid to add to the juices which would produce the most amount of ethanol; too much acid, however, would result in less ethanol, as it is killing the yeast. We also had a null hypothesis, where we said that adding no tartaric acid to the juices would result in the most ethanol produced, meaning that the tartaric acid that was found naturally in the grapes was sufficient. The teacher made it a condition, however, the juice had to contain tartaric acid.

Our teacher said that throughout his years of teaching, the results of almost all Year 12 wine-making experiments have contradicted their team's hypothesis (probably due to experimental error), and so he said if that was the case, we were allowed to tweak our results. The results, therefore, are not that important.

There are two problems I am facing:

1. I have searched on the internet for resources that discuss and explain the effect of adding tartaric acid to grape juices before fermentation, yeast acid tolerance levels, and the amount of ethanol produced in fermentation, but I haven't found anything. Do you know of any resource that discusses and explains the chemistry behind something similar to our experiment?

2. To get a good grade, we need to include something in our assignment that we haven't been taught in class (something relevant to chemistry). This means any first-year or second-year university-level chemistry. Do you have any suggestions?

Thank you very much.

2. Jun 30, 2014

### Staff: Mentor

Discuss calculation of pH in the solution of a multiprotic weak acid with relatively similar Ka values (tartaric being a great example for that).

3. Jun 30, 2014

### FredericChopin

Thank you for the suggestion, but we have done calculating the pH of a polyprotic acids.

I'm sorry, I should have given you more background information.

We have done the following in class:

* Chemical equilibrium and Le Chatelier's Principle
* Acids (strong and weak) and bases (only strong)
* pH calculations of acids (strong and weak) and bases (only strong)
* Indicators (nothing technical, just knowing when to use which indicators and how they work)
* Buffer solutions (how they work and how to calculate their pH)
* Titrations (strong acid vs. strong base, weak acid vs. strong base, how to calculate concentrations, mid-points and equivalence points)

In our experiment, we titrated the wine (tartaric acid) against sodium hydroxide, titrated the wine for sulphur dioxide, used a pH probe, refractometer to find the potential alcohol, and ebulliometer to find the concentration of ethanol.

I'm sorry to ask, but do you have any other suggestions (something really challenging? )

4. Jul 1, 2014

### Staff: Mentor

I don't see of a weak, multiprotic acid on your list, but I will take your word for it.

Still, even if you did, you did some simplified cases. Many of the simplifications fail for the tartaric acid. I doubt you did a full, complete attack that yields a 4th degree polynomial.

5. Jul 1, 2014

### FredericChopin

Ah... I see what you mean.

Thank you for the suggestion.

6. Jul 4, 2014

### FredericChopin

Hello again,

I tried obtaining a 4th degree polynomial but I'm a little stuck. Here's what I've done:

Tartaric acid is a diprotic weak acid. If "T" represents the tartarate ion, then tartaric acid undergoes the following dissociations:

H2T <=> HT- + H+

, and then:

HT- <=> T2- + H+

Since it dissociates twice, there would be two Ka equations:

Ka1 = [H+]*[HT-]/[H2T]

, and:

Ka2 = [H+]*[T2-]/[HT-]

For the first equilibrium, before dissociation, there would be some concentration of H2T and a concentration of 0 H+ and HT-. At equilibrium, x amount of H2T would have been used to produce x amount of H+ and HT-. So the Ka equation for the first dissociation would be:

Ka1 = x*x/([H2T] - x)

For the second equilibrium, before dissociation, there would be x concentration of HT-, as well as x concentration of H+ from the first dissociation. At equilibrium, y amount of HT- would have been used to produce y amount of H+ and T2-. It can't be an x amount again, because that would imply that the same amount of H+ and T2- is produced, but that's only valid if the Ka values for the two dissociations are the same. So the Ka equation for the second dissociation would be:

Ka2 = (x + y)*y/(x - y)

Hm... I'm not really sure where I'm going.

7. Jul 5, 2014

8. Jul 5, 2014

### FredericChopin

Ah, ok.

You don't necessarily have to put it into polynomial form though, do you? You can just solve using a system of equations, right?

9. Jul 6, 2014

### Staff: Mentor

No, you don't have to.

Yes. But if you try to solve it by substitution you will get to the polynomial. This is described on other pages of the same site I linked to earlier.

10. Jul 7, 2014

### FredericChopin

I see.

This type of calculation is called systematic chemical equilibrium, right?

Can it take into account the effect of buffers?

11. Jul 8, 2014

### Staff: Mentor

No idea if it has any specific name - but yes, it is definitely a systematic approach.

It automatically takes into account everything. Finding pH of buffer, or finding buffering capacity of the solution is nothing else but solving simplified (incomplete subset) system of equations describing the solution. If you take all equations into account, solutions to all these simplified cases are automatically covered.

12. Jul 8, 2014

### FredericChopin

Oohh! Cool!

I'll try it and see how it goes.

Thank you.

13. Jul 10, 2014

### FredericChopin

I did it, and I think it's working!

Have a look! They're in order of page number.

Ok, now my next task is to include the buffer!

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14. Jul 11, 2014

### FredericChopin

Ok, I've attempted the buffer part, but I don't know how to do the mass balance equation for potassium hydrogen tartrate (KHT), which is the buffer naturally found in grapes.

What should I do?

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15. Jul 11, 2014

### Staff: Mentor

Looks OK to me, but I admit I have just skimmed.

One thing that caught my attention: just because you have 5 equations and 5 unknowns doesn't guarantee the system will have a solution. It is a necessary, but not a sufficient condition.

Edit: you posted the buffer part in the meantime, so what I wrote here is about an earlier post.

16. Jul 11, 2014

### FredericChopin

I see. I think it's working then.

So for the buffer, [KHT] = [H+] + [HT-] is valid as a mass balance equation? (The post after the calculation without the buffer).

EDIT: Oops. I should have been patient.

17. Jul 11, 2014

### Staff: Mentor

1. There is no such thing as Ka for KHT dissociation. 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.

2. For being very precise, you should also include equilibrium for KOH dissociation. pKb=0.5 (see http://www.chembuddy.com/?left=FAQ).

18. Jul 11, 2014

### FredericChopin

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? Also, that would leave 6 unknowns and 5 equations, unless...

... I should include the equilibrium for KOH dissociation?

19. Jul 11, 2014

### FredericChopin

EDIT:

What if I don't have the concentration of K+? Am I doomed?

20. Jul 11, 2014

### Staff: Mentor

It dissociates completely.

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.

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.

21. Jul 11, 2014

### FredericChopin

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.

22. Jul 11, 2014

### FredericChopin

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/Chem422A/Organic%20acids%20in%20wine.pdf [Broken]). 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/chem1110-20/044 Unknown Acid Ka MM.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?

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23. Jul 12, 2014

### Staff: Mentor

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).

24. Jul 12, 2014

### FredericChopin

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.

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25. Jul 12, 2014

### Staff: Mentor

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

$$[K^+][HT^-] \le K_{sp}$$