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I'm doing an investigation into the total acidity of wine, but I’m confused because I've got the amounts of NaOH needed to titrate the acid (I’m using tartaric acid to represent all the acids in wine) and from this I can calculate the number of moles of acid used. However is the number the titratable acidity or are there further calculations needed?
Some sites say the TA is proportional to the cc used e.g. 1 cc = 0.1% TA, is this correct?
Also I am planning on doing a steam distillation to find the volatile acids. However I cannot find anything that tells me the end point for this experiment?
After I have titrated the distillate against NaOH I am unsure on which calculations to do. Do I find the concentration of acids again, but surely everything that was distilled off is acidic?
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Igloo writes ...
I presume that the first paragraph of your question refers to titrations carried out with original wine samples (perhaps decolourised first). If so, all the acids present, some volatile and the remainder non volatile, will be neutralised. Your titration reading then gives a measure of all the acids present. You quite rightly mention tartaric acid as being a suitable one to choose to represent all the acids present, so I suggest that you carry out your calculations on this basis. Your titration reading plus a knowledge of the concentration of the NaOH will enable you to calculate the number of moles of NaOH used. This value, divided by two, will give you the number of moles of tartaric acid present. Knowing the volume of wine used for each titration you can easily calculate the concentration of acids in mol dm-3 in terms of tartaric acid.
Your steam distillation will enable you to distil over all the volatile acids present in a known volume of the wine, leaving the non volatile ones in the distillation flask. Steam distillation needs to be carried out until drops of distillate no longer turn pH paper red. After making up the distillate to a known volume in a volumetric flask, portions can once again be titrated with the NaOH solution, and calculations can then be carried out to estimate the number of moles of volatile acid present in 1 dm3 of wine. In this case the representative volatile acid normally chosen is ethanoic acid.
In both cases it’s best to leave concentrations in terms of mol dm-3. Converting to g or mg per dm3 is pretty meaningless since you’re dealing with a complex mixture in each case.
I hope that my answer covers all that you need to know. If not, have a look at the wine tutorial on the React website to get further information.
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updated: 18 March 2009
