Chemical equation for a modified Ripper titration for determining free and total sulfur dioxide

[Loulou74]

I have worked in a laboratory of a winery where I analyzed the free and total sulfur dioxide in wine with a manual titrator by the Ripper method.

This method is usually a simple iodine titration where iodine reacts with sulfur dioxide according to the following equation:

SO2+ I2+ 2H2O⟶ H2SO4+ 2HI

The reagents used in the method I followed are slightly modified. Instead of titrating the wine sample acidified with sulfuric acid against an iodine solution, I titrated it against a solution of potassium iodide (a starch indicator is used to see the color change and distilled water is added to the sample). 

I have thought of the following equations to represent this reaction:

1. SO2+ 2H2O+ H2SO4+ 4KI⟶ 2K2SO4+ 4HI
2. SO2+ 2KI+ 2H2O⟶ H2SO4+ 2HI+ 2K⁺
3. HSO3−+ KI⟶ KSO3⁻

However, in equation 1 the hydrogen is not balanced, and in equation 2 the charges are not balanced. Equation 3 looks fine at first, but the bisulfite ion would imply that this reaction only occurs when analyzing the free fraction of sulfur dioxide, which is not true. 

How does potassium iodide in solution react differently to iodine, supporting this explanation with a chemical equation preferably?

If you are interested, here you can find the user's manual of the apparatus I used. I contacted the company but their response was not very helpful. They only said that, in their method, the common iodine solution is replaced by potassium iodide solution because as a product it is cheaper and more stable.

Edited August 25, 2018 by Loulou74

[chenbeier]

The problem is it is a redox reaction. KI is already reduced so your first equation can not work. SO₂   is oxidised to sulfuric acid, so you need a oxidiser.  Only iodine can do it. 

In the discription of the method they mentionen the yuse KIO3

 

Set the rod in the corresponding hole from the back and turn it clockwise, later set the nut clip. Place a 25 mL burette in the burette support. with . Place the glass tubes in the rack and put one stirrer magnet inside them. Connect the power voltage feeder and press . The leds will shine on indicating current in the instrument. will remain . Fill it Titration Liquid (Potassium Iodate N/128

Edited August 25, 2018 by chenbeier

[Loulou74]

@chenbeier

Thank you very much for pointing out that potassium iodate is used in the method, not KI.

I have thought of the following reactions that could be theoretically possible with KIO₃:

1.  KIO₃ + 3SO₂ + 4H₂O ⟶ KOH + 3H₂SO₄ + HI
2. KIO₃ + 3SO₂ + 3H₂O ⟶ KHSO₄ + 2H₂SO₄ + HI
3. 2KIO₃ + 6SO₂ + 6H₂O ⟶ K₂SO₄ + 5H₂SO₄ + 2HI

Do you know which of those three products is more likely to form?

Edited August 26, 2018 by Loulou74

[chenbeier]

The first one is not possible, because sulfuric acid, hydogeniodide,  and hydoxide cannot  existing side by side.  It will be neutralized.

Equation 2 and 3 are the same, because in aqueous solution  we have ions and not specific salts. K₂SO₄ + H₂SO₄ => 2 KHSO₄ => 2 K⁺ + 2 H⁺  + SO₄^2-

But also this will not happen, because in the description they use starch indicater, what means iodine will be developed.

2 IO₃^-  + 6 H⁺ + 6 e^- => I₂ + 3 H₂O and  SO₂ + 2 H₂O => SO₄^2- + 4 H⁺ + 4 e^-

Edited August 27, 2018 by chenbeier

[Loulou74]

@chenbeier

Can iodate ions theoretically produce iodine directly as you said according to the following equation:

18 hours ago, chenbeier said:

2 IO₃^-  + 6 H⁺ + 6 e^- => I₂ + 3 H₂O

 

or should there be potassium iodide present in solution to react with KIO₃ and form iodine as follows:

KIO₃ + 5KI + 3H₂O ⟶ 3I₂ + 6KOH ?

[chenbeier]

You need a reducer to do so, Iodate can be reduced by sulfurdioxide, of course also by iodide. In your equations you create iodide and this would react according your suggestion. The end is iodine in both cases.  By the way spectator ions like K⁺ can be taken out.

Edited August 28, 2018 by chenbeier

[Loulou74]

@chenbeier

In that case, can the reduction of iodate to iodine by sulfur dioxide be represented with the following equation (including spectator ions for completion):

2KIO₃ + 5SO₂ + 4H₂O  ⟶ K₂SO₄ + I₂ + 4H₂SO₄  ?

[chenbeier]

I would think so.