# Iodine Titration with Free SO2 in wine

To determine the free $\ce{SO2}$ in wine, a common method used is called the Ripper Titration Method.

The equation is as follows: $$\ce{SO2 (aq) + I2 (aq) + 2H2O (l) -> 4H+ (aq) + SO4^2- (aq) + 2I- (aq)}$$

This method involves titrating an Iodine solution into a sample of wine. I have found that triiodide is made from reacting $\ce{I-}$ with $\ce{I2}$. However, if it is the triiodide we are titrating, how is the iodine reacting with the wine?

Is it triiodide ($\ce{I3-}$) that is being titrated with the wine sample or iodine ($\ce{I2}$)?

• @Nij States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different. – Martin - マーチン Jul 21 '16 at 3:23
• @Martin-マーチン just fitting form based on how I would write for P&P. I'll check those recommendations for other details later, thanks. – Nij Jul 21 '16 at 3:26

Reaction goes along the lines $\ce{3 I2 -> 2 I3- -> 6 I-}$. The beginning of titration is $\ce{I2}$. The end of titration is $\ce{I-}$. $\ce{I3-}$ forms as an intermediate, but it can be ignored in titration.
• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. – Martin - マーチン Jul 21 '16 at 3:25