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Analysis of Sulphur Dioxide Content in Wine

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Introduction

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. ...read more.

Middle

4. 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. 5. Record the actual molarity of the iodine solution. 6. Record the titre required to produce the first faint but permanent blue colour. 7. 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 Titration 1 2 3 4 Final Burette Reading (ml) 39.8 21.4 18.2 16.2 Initial Burette Reading (ml) 17 7.6 2.9 2.2 Volume of titrant (ml) 22.8 13.8 15.3 14 Mean volume of titrant (ml) = ____(13.8 + 15.3 + 14) ...read more.

Conclusion

= 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. 2. 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. 3. 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. 4. What are the assumptions in arriving at your result ? Assuming that no any Sulphur dioxide evaporated. ...read more.

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