Wine Analysis

Derivation of Titratable Acidity Calculation Equation - Titratable Acidity

Titratable acidity (g/L as tartaric acid) = 0.75 x Titre ml (of 0.10M NaOH)


Calculating the reaction ratio between NaOH and tartaric acid (H₂Ta)

It requires two (2) molecules of NaOH to react with one (1) molecule of H₂Ta (tartaric acid).



H2Ta(aq)

< = >

2H+(aq)

+

HTa-(aq)

(see full equations)




H₂Ta + 2NaOH < == > 2Na⁺ + Ta⁼ + 2H₂O (see full equation)

These two equations indicate that one (1) molecule of H₂Ta requires two (2) molecules of NaOH to neutralize it by forming water (H₂O).

The molecular weight (MW) of H₂Ta (C₄H₆O₆) is 150
The molecular weight (MW) of NaOH is 40

This is to say that for every 150 g of H₂Ta, 80g (2 x 40) of NaOH is required to neutralize through titration



                     80 g NaOH

i.e. a ratio of     ----------   =    0.53 NaOH/ H2Ta                   

                    150 g H2Ta

 

The reaction ratio between NaOH and H₂Ta = 0.53 NaOH /H₂Ta

• Calculating the quantity (g) of H₂Ta in 10 ml sample at a concentration of 1 g/L
  
  

  
              1g H2Ta        1 L
  
  10 ml   X   ------   X   ----------  = 0.01g H2Ta 
  
                L           1000 ml
  
  

  10ml of juice or wine with 1g/L of H₂Ta contains 0.01 g of H₂Ta
  
  
  
• Calculating the quantity of NaOH required to react with 0.01 g H₂Ta
  
  
  Applying the NaOH/H₂Ta reaction ratio of 0.53
  
  0.01 g H₂Ta x 0.53 NaOH/H₂Ta = 0.0053 g NaOH.
  
  ie. 0.01 g of H₂Ta will react with 0.0053 g of NaOH
  
  or
  
  ie. 10 mg of H₂Ta will react with 5.3 mg of NaOH
  
  
  

• Calculating the quantity of NaOH in 1ml of 0.01 M NaOH solution
  
  

  
           40 g NaOH     0.10 mole        1000 mg        1L
  
  1 ml  X ---------  X  ---------   X    -------   X  -------  =  4.0 mg /ml of 0.1M NaOH
  
            mole            L              g           1000 ml
  
  

  
  Therefore 1ml of 0.10 M NaOH contains 4.0mg of NaOH
  
  
  
• Calculating the number of mls of a 0.10M NaOH solution contains 6.7 mg NaOH
  
  (remember 5.3 mg of NaOH is the amount required to react with quantity of H₂Ta in a 10 ml of sample with a concentration of 1g/L)
  
  

  
   
  
              5.3 mg NaOH              
  
              -------------      =    1.325 ml of 0.1 M NaOH
  
              4.0 mg NaOH/ ml 
              of 0.1 M NaOH
  
  

  
  Therefore it requires 1.325 ml of 0.10M NaOH to react with 6.6 mg H₂Ta (10ml @ 1 g/L)
  
  
  i.e. titre (ml)/1.325 ml = H₂Ta (g/L) or equals 0.75 * titre (ml)
  
  
  Therefore using a 10 ml sample and exactly 0.10M NaOH
  
  Titratable acidity (as H₂Ta g/L) = 0.75 x Titre ml (0.10M NaOH)