Question:

Except moles of univalent strong base is only identical to moles of acid for mono-acids.  Since most of the acids in wine are di-acids, it makes some sense (not a huge amount, granted, but some), to express the unit in terms of an agreed upon di-acid.

I think it would be best to express it in terms of moles of available protons, rather than mass of some acid which may or may not be the major species in your wine. It even requires less calculations. The Mad Alchemist http://members.xoom.com/madalch

Response:

Well, if the OH- is any indication of H+, then it is a sensible arraignment. Except moles of univalent strong base is only identical to moles of acid for mono-acids.  Since most of the acids in wine are di-acids, it makes some sense (not a huge amount, granted, but some), to express the unit in terms of an agreed upon di-acid. I think it would be best to express it in terms of moles of

available protons, rather than mass of some acid which may or may not be the major

species in your wine. It – Hide quoted text — Show quoted text – even requires less calculations. The Mad Alchemist http://members.xoom.com/madalch

Response:

I hope, you know, you make perfect sense. You don’t want to express the acidity in terms of moles of NaOH or mass of NaOH because all you’re interested in is the OH- part of the molecule. You could use any salt of OH- that you wanted. Mass of NaOH would be just as bad as describing acidity as

mass of a particular acid.  Moles of NaOH, however, woudl be the most sensible way to measure things, since it would be identical to the moles of acid (and it

doesn’t matter if it’s moles of KOH, NaOH, LiOH, CsOH, or tetramethylammonium

hydroxide- a mole of a – Hide quoted text — Show quoted text – univalent strong base is a mole of a univalent strong base, although I wouldn’t titrate anything with methyllithium). However, winemakers don’t seem to like using units that only chemists understand.  Regrettable. The Mad Alchemist http://members.xoom.com/madalch Except moles of univalent strong base is only identical to moles of acid for mono-acids.  Since most of the acids in wine are di-acids, it makes some sense (not a huge amount, granted, but some), to express the unit in terms of an agreed upon di-acid.  Since tartaric is the most prevalent acid in wine, it doesn’t seem entirely unreasonable to use this as the standard. Sulfurous acid as a standard, on the other hand, I totally don’t understand. Dave

Dave Breeden

Response:

Ok, maybe you’re the guy to talk to. What about the phenolphthalein? I have powdered variety. How much of the phenolphthalein should I add? – Hide quoted text — Show quoted text – So I get the idea of titration back to neutrality with NaOH to determine the level of available acid in the must, but what is the unit definition and what is the standard procedure if you intend to use a pH meter and NaOH standard? According to Jeff Cox’s book, you take 15ml juice and add 0.1N NaOH until pH 7.0 is reached. The number of ml added NaOH for 15ml juice is divided by 2 to give parts per thousand which is multiplied by 100 to express in % or grams/100ml but what is this really? Wouldn’t it make more sense to exprese the mass of NaOH or moles of NaOH required to achieve neutrality?  Where is the unit of mass coming from in gm/100ml, or for that matter where is 100ml coming from. Anybody got any insights? Jeff Cox may be a good writer and a great winemaker but he is clearly no chemist (yes… I am, but I’m still too stupid to figure this out… damn state schools) Hi, You don’t want to express the acidity in terms of moles of NaOH or mass of NaOH because all you’re interested in is the OH- part of the molecule. You could use any salt of OH- that you wanted. Anyway, the basic idea is that you keep track of how much NaOH (how many moles) you’ve added.  For fun, let’s say 30 mL.  If you’ve added 30 mL of 0.1 M NaOH, that’s 0.003 moles NaOH (30 mL/1000mL times 0.1). If Jeff had actually done it right, by having you neutralize to pH 8.2 (the phenolphthalein endpoint, at which point ALL the acid is neutralized), you would have neutralized all the acid at this point.  Let’s pretend you had.  Since the North American unit for acidity in wine is g tartaric acid per 100 mL, and tartaric acid is a di-acid (two acid groups per molecule), that means you’ve neutralized 0.003/2, or 0.0015 moles of tartatic acid.  If you’ve got 0.0015 moles tartaric acid, and the molecular weight of tartaric acid is 150 g/mole, then that represents 0.225 g of acid in your 15 mL of must (or wine, etc.).  That means in 100 mL of wine you’d have 0.225 times 100/15 = 1.5 g/100 mL. Needless to say, this wine is **WAY** too acid. But that’s the idea, anyway.  The reason why you can divide by 2 has to do with the amount of wine you’ve added, the concentration of NaOH, and the molelcular weight of tartaric acid. Hope this helps. Dave

Dave Breeden

Response:

Ok, maybe you’re the guy to talk to. What about the phenolphthalein? I have powdered variety. How much of the phenolphthalein should I add?

You’ll need to make up a solution, as it wouldn’t be practical to add a weighable amount of phenolphthalein (you use way too little to be able to weigh it). I’ve used 1% phenolphthalein in ethanol solutions.  Aldrich (a standard chemical company) sells 0.5% in 50% EtOH/H20.  For the first you’d just use 1 g in 100 mL of ethanol, for the second use 0.5 g in 50 mL ethanol, then add 50 mL water. When I use 1% solutions, I normally add 3 drops or so. Hope this helps. Dave P.S.  Gram-wise, 3 drops of 1% phenolphthalein would be 0.0015 g!

Response:

In defense of Jeff Cox, it way my false assumption that phenophthalein goes through a color change at pH 7. Mr Cox does write that this indicator should be used but he does not specify what the pH is at the end-point. Speaking in terms of chemistry, by definition, all acids should be neutralized at neutral pH shouldn’t they? Maybe the titratable acidity is actually measuring total buffering capacity between the must pH and 8.2. The transition between the two pHs is a reflection of how much deprotonation is required to reach the target pH, but the total measurement is not really titratable acidity in the conventional sense if the target is 8.2, is it?

Response:

In defense of Jeff Cox, it way my false assumption that phenophthalein goes through a color change at pH 7. Mr Cox does write that this indicator should be used but he does not specify what the pH is at the end-point. Speaking in terms of chemistry, by definition, all acids should be neutralized at neutral pH shouldn’t they? Maybe the titratable acidity is actually measuring total buffering capacity between the must pH and 8.2. The transition between the two pHs is a reflection of how much deprotonation is required to reach the target pH, but the total measurement is not really titratable acidity in the conventional sense if the target is 8.2, is it?

I haven’t actually done the calculations (and don’t actually have any clue how to do them, or even if it can be done, given all the stuff floating around in wine), but my distinct impression is that at neutral pH, there will still be a non-negligible amount of undissociated organic acids, since the acids in question are so weak.  In order for the calculation in my previous response to be correct (in particular the calculations that follow from tartaric being di-acidic), you have to be sure both protons of tartaric are titrated. In order to really figure it out, you’d have to know what kind of metal ions were in solution, how much of them there were in solution, and how to adjust your use of Henderson-Hasselbach accordingly. If Lum is out there, he might know for sure. Dave P.S.  The phenolphthalein endpoint is variously given as pH 8.2 to 8.4, for what that’s worth.

Response:

You don’t want to express the acidity in terms of moles of NaOH or mass of NaOH because all you’re interested in is the OH- part of the molecule. You could use any salt of OH- that you wanted.

Mass of NaOH would be just as bad as describing acidity as mass of a particular acid.  Moles of NaOH, however, woudl be the most sensible way to measure things, since it would be identical to the moles of acid (and it doesn’t matter if it’s moles of KOH, NaOH, LiOH, CsOH, or tetramethylammonium hydroxide- a mole of a univalent strong base is a mole of a univalent strong base, although I wouldn’t titrate anything with methyllithium). However, winemakers don’t seem to like using units that only chemists understand.  Regrettable. The Mad Alchemist http://members.xoom.com/madalch

Response:

You don’t want to express the acidity in terms of moles of NaOH or mass of NaOH because all you’re interested in is the OH- part of the molecule. You could use any salt of OH- that you wanted. Mass of NaOH would be just as bad as describing acidity as mass of a particular acid.  Moles of NaOH, however, woudl be the most sensible way to measure things, since it would be identical to the moles of acid (and it doesn’t matter if it’s moles of KOH, NaOH, LiOH, CsOH, or tetramethylammonium hydroxide- a mole of a univalent strong base is a mole of a univalent strong base, although I wouldn’t titrate anything with methyllithium). However, winemakers don’t seem to like using units that only chemists understand.  Regrettable. The Mad Alchemist http://members.xoom.com/madalch

Except moles of univalent strong base is only identical to moles of acid for mono-acids.  Since most of the acids in wine are di-acids, it makes some sense (not a huge amount, granted, but some), to express the unit in terms of an agreed upon di-acid.  Since tartaric is the most prevalent acid in wine, it doesn’t seem entirely unreasonable to use this as the standard. Sulfurous acid as a standard, on the other hand, I totally don’t understand. Dave

Response:

P.S.  You really do want to titrate to pH 8.2, and NOT pH 7.0.  Not all of the acids (in must and wine, all weak organic acids) will be neutralized at pH 7.0, so you’ll get an artificially low reading.  I’m not sure why he says to do it this way. Maybe he says do it to pH 7 because that is the way so many of the european countries do it. In fact the Official Methods of Analysis of the European Union still say titrate to pH 7. It is a struggle trying to persuade them this is incorrect. Little Podge

Hi Podge, Thanks!  I had no idea. I would think a quick explanation of dissociation constants and the Henderson-Hasellbach would have been all that was required.  :-) Dave

Response:

P.S.  You really do want to titrate to pH 8.2, and NOT pH 7.0.  Not all of the acids (in must and wine, all weak organic acids) will be neutralized at pH 7.0, so you’ll get an artificially low reading.  I’m not sure why he says to do it this way.

Maybe he says do it to pH 7 because that is the way so many of the european countries do it. In fact the Official Methods of Analysis of the European Union still say titrate to pH 7. It is a struggle trying to persuade them this is incorrect. Little Podge

Response:

<snip Hi,

<snip Anyway, the basic idea is that you keep track of how much NaOH (how many moles) you’ve added.  For fun, let’s say 30 mL.  If you’ve added 30 mL of 0.1 M NaOH, that’s 0.003 moles NaOH (30 mL/1000mL times 0.1).  If Jeff had actually done it right, by having you neutralize to pH 8.2 (the phenolphthalein endpoint, at which point ALL the acid is neutralized), you would have neutralized all the acid at this point.  Let’s pretend you had.  Since the North American unit for acidity in wine is g tartaric acid per 100 mL, and tartaric acid is a di-acid (two acid groups per molecule), that means you’ve neutralized 0.003/2, or 0.0015 moles of tartatic acid.  If you’ve got 0.0015 moles tartaric acid, and the molecular weight of tartaric acid is 150 g/mole, then that represents 0.225 g of acid in your 15 mL of must (or wine, etc.).  That means in 100 mL of wine you’d have 0.225 times 100/15 = 1.5 g/100 mL. Dave

P.S.  You really do want to titrate to pH 8.2, and NOT pH 7.0.  Not all of the acids (in must and wine, all weak organic acids) will be neutralized at pH 7.0, so you’ll get an artificially low reading.  I’m not sure why he says to do it this way. And you’ll find you spend a LOT less time and waste a lot less NaOH if you titrate 5.0 mL of must/wine, rather than 15.  There’s no good reason I can think of to titrate all that extra must/wine.  I normally titrate 5 mL of wine in 100 mL water, to give me enough volume for my pH electrode and stirbar and burette tip.  Diluting the wine doesn’t affect the amount of acid, of course.

Response:

So I get the idea of titration back to neutrality with NaOH to determine the level of available acid in the must, but what is the unit definition and what is the standard procedure if you intend to use a pH meter and NaOH standard? According to Jeff Cox’s book, you take 15ml juice and add 0.1N NaOH until pH 7.0 is reached. The number of ml added NaOH for 15ml juice is divided by 2 to give parts per thousand which is multiplied by 100 to express in % or grams/100ml but what is this really? Wouldn’t it make more sense to exprese the mass of NaOH or moles of NaOH required to achieve neutrality?  Where is the unit of mass coming from in gm/100ml, or for that matter where is 100ml coming from. Anybody got any insights? Jeff Cox may be a good writer and a great winemaker but he is clearly no chemist (yes… I am, but I’m still too stupid to figure this out… damn state schools)

Hi, You don’t want to express the acidity in terms of moles of NaOH or mass of NaOH because all you’re interested in is the OH- part of the molecule. You could use any salt of OH- that you wanted. Anyway, the basic idea is that you keep track of how much NaOH (how many moles) you’ve added.  For fun, let’s say 30 mL.  If you’ve added 30 mL of 0.1 M NaOH, that’s 0.003 moles NaOH (30 mL/1000mL times 0.1).  If Jeff had actually done it right, by having you neutralize to pH 8.2 (the phenolphthalein endpoint, at which point ALL the acid is neutralized), you would have neutralized all the acid at this point.  Let’s pretend you had.  Since the North American unit for acidity in wine is g tartaric acid per 100 mL, and tartaric acid is a di-acid (two acid groups per molecule), that means you’ve neutralized 0.003/2, or 0.0015 moles of tartatic acid.  If you’ve got 0.0015 moles tartaric acid, and the molecular weight of tartaric acid is 150 g/mole, then that represents 0.225 g of acid in your 15 mL of must (or wine, etc.).  That means in 100 mL of wine you’d have 0.225 times 100/15 = 1.5 g/100 mL. Needless to say, this wine is **WAY** too acid. But that’s the idea, anyway.  The reason why you can divide by 2 has to do with the amount of wine you’ve added, the concentration of NaOH, and the molelcular weight of tartaric acid. Hope this helps. Dave

Response:

So I get the idea of titration back to neutrality with NaOH to determine the level of available acid in the must, but what is the unit definition and what is the standard procedure if you intend to use a pH meter and NaOH standard? According to Jeff Cox’s book, you take 15ml juice and add 0.1N NaOH until pH 7.0 is reached. The number of ml added NaOH for 15ml juice is divided by 2 to give parts per thousand which is multiplied by 100 to express in % or grams/100ml but what is this really? Wouldn’t it make more sense to exprese the mass of NaOH or moles of NaOH required to achieve neutrality?  Where is the unit of mass coming from in gm/100ml, or for that matter where is 100ml coming from. Anybody got any insights? Jeff Cox may be a good writer and a great winemaker but he is clearly no chemist (yes… I am, but I’m still too stupid to figure this out… damn state schools)

Response: