Calculation

1) Using the volume of NaOH to reach the end point study full calculations for the titratable acerbity of the untreated drink. Include full calculations for the centre of CaCO3 to be added to reach a titratable acidulousness of 5g/L. Ans. A) good sulkiness by Titration with standardised NaOH pH of fuddle at 30sec = 3.45 pH of deionised water = 9.07 Final burette translation = 8.8 ml renewal into litres = 8.8/ gigabyte = 0.0088 litres Moles of NaOH added = 0.008x0.1= 8.8 x 10-4 moles Moles of tartaric acid = Moles of NaOH 2 Moles of tartaric acid 8.8 x 10-4 /2 = 4.4 x 10-4 moles T.A = Moles of tartaric acid x molecular weight of tartaric acid batch of wine summate Acidity = (4.4 x 10-4 x 150.09) / (0.01) nitty-gritty Acidity = 6.6 g/L Change in titratable acidity ? TA (g/l ) = 6.6 5 = 1.6 g/L Mass of CaCO3 (g) to be added to 100ml of wine = 1.6 x 0.667 x 0.1 = 0.107 g/L instalment of wine to be treated (F) = ?TA TAi-2 F = 1.6/ 6.6-2 = 0.348 g/L Volume of wine to be treated is F x 100ml = 34.8ml B) Deacidification using the Double Salt Method pH of wine + CaCO3 = 5.14 pH of deionized water = 9.

3 Total acidity of wine treated with CaCO3 titration with standardized NaOH Final burette reading = 7.1 Titrated nourish = 7.1/1000 = 0.0071 Moles of tartaric acid = Moles of NaOH ! 2 Moles of tartaric acid 0.0071/2 = 0.00355 moles T.A = Moles of tartaric acid x molecular weight of tartaric acid Volume of wine Total Acidity = (0.00355 x 150.09) / (0.01) Total Acidity = 5.3 g/L Total acidity of wine by double table salt method using CaCO3 = 5.3g/L C) Deacidification of wine using straight addition of CaCO3 and KHCO3 Deacidification with KHCO3...If you want to get a full essay, identify it on our website: OrderEssay.net

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