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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 ($\ce{I2}$)?

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  • $\begingroup$ @Nij States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different. $\endgroup$ – Martin - マーチン Jul 21 '16 at 3:23
  • $\begingroup$ @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. $\endgroup$ – Nij Jul 21 '16 at 3:26
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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.

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  • $\begingroup$ 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. $\endgroup$ – Martin - マーチン Jul 21 '16 at 3:25
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    $\begingroup$ I have down voted this because while it isn't completely wrong, it contains not explanation and should have probably better be left as a comment. $\endgroup$ – Martin - マーチン Jul 21 '16 at 3:30
  • $\begingroup$ @martin, Downvoting something that is not wrong (or that is honestly right) is an interesting practice. What do you disagree with? $\endgroup$ – sixtytrees Jul 21 '16 at 12:19

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