Question

The acceptable range of titratable acidity is 0.7-0.9% grams tartaric acid per 100 mL. After titrating 5 mL of wine sample with 0.1N NaOH. You determine that 3 mL of NaOH are needed to fully neutralize the acids present in wine.

What is your recommendation for the quality of the product?

Answer 1

normality of NaOH = 0.1 N

volume of NaOH = 3 mL = 0.003 L

equivalents of NaOH = (normality of NaOH) * (volume of NaOH in liters)

equivalents of NaOH = (0.1 N) * (0.003 L)

equivalents of NaOH = 0.0003 eq

equivalents of tartaric acid = equivalents of NaOH

equivalents of tartaric acid = 0.0003 eq

mass of tararic acid = (equivalents of tartaric acid) * (equivalent mass of tartaric acid)

mass of tararic acid = (0.0003 eq) * (75.0435 g / eq)

mass of tararic acid = 0.0225 g

% tartaric acid = (mass of tararic acid / volume of wine) * 100

% tartaric acid = (0.0225 g / 5 mL) * 100

% tartaric acid = 0.45 %

The given wine product is less acidic than acceptable range of titratable acidity