Calculation

1) Using the volume of NaOH to reach the end point say full calculations for the titratable acidity of the untreated drink. Include full calculations for the total of CaCO3 to be added to reach a titratable acidity of 5g/L.
Ans. A) good tartness by Titration with standardised NaOH
pH of fuddle at 30sec = 3.45
pH of deionised water = 9.07
Final burette reading = 8.8 ml
Conversion 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
Volume of wine
Total Acidity = (4.4 x 10-4 x 150.09) / (0.01)
Total 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
Fraction 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 salt method using CaCO3 = 5.3g/L

C) Deacidification of wine using direct addition of CaCO3 and KHCO3

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