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Derivation of SO₂ Calculation formula - SO₂ ppm = 16 x Titre ml (0.01M NaOH)

The formula and calculations are based on the use of a 20.00 ml juice/wine sample and the use of exactly 0.010 M NaOH solution.

Calculating the reaction ratio between NaOH and SO₂

SO₂ + H₂O₂ < == > H₂SO₄ (see full equation)

i.e. 1 molecule of SO₂ forms 1 molecule of H₂SO₄

H₂SO₄ + 2NaOH < == > 2Na⁺ + SO₄ ⁼ + 2H₂O (see full equation)

i.e. 1 molecule of SO₂ forms 1 molecule of H₂SO₄

These two equations indicate that for one (1) molecule of SO₂ present, two (2) molecules of NaOH are involved.

or

It takes two (2) molecules of NaOH to react with or neutralise one (1) molecule of H₂SO₄, forming one (1) neutral molecule of water (H₂O).

The molecular weight (MW) of SO₂ is 64
The molecular weight (MW) of NaOH is 40

This is to say that for every 64 g of SO₂, 80g (2 x 40) of NaOH is involved in the titration



			64 g SO2

i.e. a ratio of		--------		= 	0.80 SO2/NaOH                     

			80 g NaOH

 

The reaction ratio between NaOH and SO₂ = 0.80 SO2/NaOH

• Calculating the quantity of SO₂ in 20 ml sample at 1 ppm SO₂
  
  

  
                  1mg SO2         1 L
  
  20ml     X      -------   X   -------    =    0.020 mg SO2
  
                     L          1000 ml
  
  

  20ml of juice or wine with 1ppm (mg/L) of SO₂ contains 0.020 mg of SO₂
  
  
  
• Calculating the quantity of NaOH required to react with 0.020 mg SO₂
  
  
  Applying the 0.8 SO₂/NaOH reaction ratio
  
  0.020 mg SO₂ x NaOH/0.80 SO₂ = 0.025 mg NaOH.
  
  ie. 0.020 mg of SO₂ will react with 0.025 mg of NaOH
  
  
  

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

  
            40 g NaOH       0.01 mole      1000 mg        1L
  
  1 ml  X   ---------   X   ---------  X  -------   X   -------   =  0.40 mg NaOH/ml
  
              mole             L             g          1000 ml
  
  

  
  Therefore 1ml of 0.01 M NaOH contains 0.40mg of NaOH
  
  
  
• Calculating the number of mls of a 0.01M NaOH solution contains 0.40 mg NaOH
  
  

  
   
  
              0.40 mg NaOH                        
  
              -------------     =          16 ml
  
              0.025 mg NaOH/ml
  
  

  
  Therefore it requires 16 ml of 0.010 M NaOH to indirectly react with 0.020 mg SO₂ (20ml @ 1ppm)
  
  
  
  Therefore using a 20 ml sample and exactly 0.01M NaOH
  
  SO₂ ppm = 16 x Titre ml (0.01M NaOH)