Practical investigations

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For my wine investigation I'm trying to find the content of sulphur dioxide in wine and its effect on sweetness. I carried out the aspiration method where I aspirated the mixture (wine and phosphoric acid) with air. I had the mixture in one Dreschel bottle and H₂O₂ in another. I’ve learnt that the SO₂ (g) produced reacts with excess H₂O₂ (aq) solution to produce sulphuric acid: H₂O₂ (aq) + SO₂ (g) ---> H₂SO₄ (aq)
What I don’t understand is why I then titrated the sulphuric acid with 0.01 M NaOH - is it to neutralise? What is the equation for this?
I attempted an equation and the best I could come with is 2NaOH + 2H₂SO₄ --> Na₂S₂O₃ + 3H₂O + 2O₂ but this doesn’t look right.
Can you please tell me how to work out the concentration of sulphur dioxide from this? Thank you. I really appreciate it.

Igloo writes ...
Yes, the 0.01M sodium hydroxide is used to find out the number of moles of sulphuric acid in your oxidised mixture. The neutralisation equation is far simpler than the one you have given.
 
It is: 2NaOH(aq) + H₂SO₄(aq) ----> Na₂SO₄(aq) + 2H₂O(l)
 
[1] You know the concentration of the sodium hydroxide solution (0.01M). So use this, together with your titration reading to find the number of moles of sodium hydroxide used in each titration.
 
[2] Use the above equation to deduce the number of moles of sulphuric acid being neutralised.
 
[3] With your first equation (H₂O₂ + SO₂ --> H₂SO₄) you can now deduce the number of moles of sulphur dioxide originally present in the sample of wine used for your titration.
 
[4] Scale this up to the number of moles of sulphur dioxide in 1 dm³. This is the concentration of SO₂ in mol dm^-3.
 
[5] If necessary, this value can be converted into ppm (parts per million) or perhaps mg per dm³ in order to compare your results with official limits data.

Risk assessment
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updated: 17 December 2006