Analysis of Sulphur Dioxide Content in Wine

EXPERIMENT ( 3 )

Topic : Analysis of Sulphur Dioxide Content in Wine

Introduction : In the presence of atmospheric oxygen, the alcohol content of wine can be converted to ethanoic acid making the wines sour & unpalatable. Even small amount of air, over a period of time, can adversely affect wines. The problem can be minimized by the introduction of a suitable reductant which will preferentially react with oxygen. One such reductant is sulphur dioxide but because it is toxic & pungent in odour, limits are set on the amount of “free” sulphur dioxide allowed in wine. Most of the preservative present in wine is “fixed” in the form of NaHSO3. Although this can act as a source of sulphur dioxide, the actual amount of free sulphur dioxide is quite low.

In this experiment, the amount of total available SO2 in the wine, irrespective of its actual form in the sample is determined. The method involves first the conversion of all SO2 into SO32-.

Acidification of the solution then liberates all SO2 : SO32-(aq) + 2 H+(aq) → SO2(aq) + H2O(l)

SO2 is then titrated with iodine solution : SO2(aq) + I2(aq) + 2 H2O(l) → 2 HI(aq) + H2SO4(aq)

Chemicals : White wine (non-sparkling or non-carbonated), 1M NaOH , 2M H2SO4 , 0.005M I2 , KIO3 ,

Starch solution (freshly prepared)

Procedures : 1. Determine, from the label, the volume of wine in a bottle.

Using a pipette, transfer 25 cm3 of white wine into a 250 cm3 conical flask.
Add about 12 cm3 of 1M NaOH(aq) & allow to stand for about 15 minutes.
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Acidification of the solution then liberates all SO2 : SO32-(aq) + 2 H+(aq) → SO2(aq) + H2O(l)

SO2 is then titrated with iodine solution : SO2(aq) + I2(aq) + 2 H2O(l) → 2 HI(aq) + H2SO4(aq)

Chemicals : White wine (non-sparkling or non-carbonated), 1M NaOH , 2M H2SO4 , 0.005M I2 , KIO3 ,

Starch solution (freshly prepared)

Procedures : 1. Determine, from the label, the volume of wine in a bottle.

Using a pipette, transfer 25 cm3 of white wine into a 250 cm3 conical flask.
Add about 12 cm3 of 1M NaOH(aq) & allow to stand for about 15 minutes.
Add about 10 cm3 of 2M H2SO4(aq) to the mixture & then a few drops of starch indicator.

Quickly titrate the mixture with 0.005M iodine solution.

Record the actual molarity of the iodine solution.
Record the titre required to produce the first faint but permanent blue colour.
Record the procedure to get 2 additional concordant titres/titrants.

Results : Brand name of wine = ____________Senorio de Vadel__________________

Volume of wine = __________________750_____________________ ml

Actual molarity of iodine solution = ___________0.005_____________ M

Table 1

Mean volume of titrant (ml) = ____(13.8 + 15.3 + 14) / 3 = 14.37 ml____

Number of moles of I2 in the titrant = ______7.19 x 10^-5 mol________

Number of moles of SO2 in the original aliquot = ______7.19 x 10^-5 mol_______

Number of moles of SO2 in the bottle of wine = _______7.19 x 10^-5 mol______

Mass of SO2 in the bottle of wine = __________6.92 x 10^-3____________ mg

Calculation:

Number of moles of I2 titrated with the SO2 :

0.005 x (14.37 / 1000)

= 7.19 x 10^-5 mol

Number of moles of SO2 in the original aliquot= Number of moles of I2 titrated with the SOⁿ

= 7.19 x 10^-5 mol

By Equation: NaHSO3 + NaOH → H2O + Na2SO3

2NaOH + SO2 → NaHSO3 + H2O

Number of moles of SO2 in the bottle of wine= Number of moles of SO2 in the original aliquot

= 7.19 x 10^-5 mol

Mass of SO2 in the bottle of wine:

7.19 x 10^-5 x (32 + 64.2)

= 6.92 x 10^-3 mg

Questions :

1. Compare your results with the limit of 450 mg dm-3 stated in the Hong Kong Preservatives in Food

Regulations. Comment on the comparison.

By comparing with the 450mg dm-3 ,I having the result ,which is 121mg dm-3 only.It have the safty result,it will not cause poisoning.

Why should the titration in step 4 be done quickly ?

It is because sulphur dioxide is volatile.It is easily to escape out of the aqueous solution and become gas state which cause the result being not accurate.

By using oxidation numbers, show that the reaction during the titration was a redox reaction.

The oxidation number of I charge from 0 to –1,which is undergo the reduction. The oxidation number of S charge form +4 to +6,which is undergo the oxidation. So,the titration was a redox reaction.