Sulfite (SO3^2-) in wine was measured by the following procedure: 50.0ml of wine were added 5.00ml of solution containing (0.8043 g KIO3 + 5 g KI)/100ml. Acidification with 1.0ml of 6.0 M H2SO4 quantitatively converted IO3^- into I3^-. I3^- reacted with sulfite to generate sulfate, leaving excess I3^- in solution. Excess I3^- required 12.86ml of 0.04818 M Na2S2O3 to reach end point.

What is the concentration of sulfite in the wine? In milligrams of SO3^2- per liter?

Uh...I'm not quite sure how to start this problem. I know the answer just not the how to get there...

To determine the concentration of sulfite in the wine, we need to use the given information to calculate the number of moles of Na2S2O3 reacted with the excess I3^- and then convert it to the concentration of sulfite in the wine.

Let's break down the steps to solve the problem:

1. Calculate the number of moles of Na2S2O3 required to reach the endpoint:
- Given the volume and concentration of Na2S2O3 used (12.86 mL and 0.04818 M, respectively), we can calculate the number of moles of Na2S2O3 using the formula:
Moles of Na2S2O3 = Volume (L) x Concentration (M)

2. Calculate the number of moles of I3^- reacted with Na2S2O3:
- Since the reaction between I3^- and Na2S2O3 is 1:2, the number of moles of I3^- reacted will be half the number of moles of Na2S2O3.

3. Calculate the number of moles of I3^- initially present:
- Since the I3^- reacted with the sulfite, any remaining I3^- is due to the excess initially present. Hence, the number of moles of I3^- initially present is equal to the number of moles of I3^- reacted with Na2S2O3.

4. Calculate the number of moles of sulfite:
- Since the reaction between I3^- and sulfite is 1:1, the number of moles of sulfite will be the same as the number of moles of I3^-.

5. Calculate the concentration of sulfite in the wine:
- Concentration (mol/L) = Moles of sulfite / Volume (L)

6. Convert the concentration to milligrams of SO3^2- per liter:
- Since 1 mole of SO3^2- is equal to its molar mass (approximately 80 g), the concentration in milligrams per liter is obtained by multiplying the concentration (mol/L) by the molar mass (mg/g) and converting the units accordingly.

By following these steps, you should be able to determine the concentration of sulfite in the wine in milligrams of SO3^2- per liter.

IO3^- + 5I^- ==> 3I2 + .. you can finish balancing if you wish but this is all you need.

SO3^2- + I2 ==> 2I^- + SO4^2-
Again you can finish AND if you want to change the I2 to I3^- you may do so and make the 2I^- on the right side into a 3I^-.

moles IO3^- initially = 0.8043 x (5mL/100mL) x (1 mol/214.00 g) = ??
moles I2 generated = 3*moles S2O3^2-.

Then moles I2 react with sulfite according to the equation at the top of my post, which uses up part of the I2 there initially. How much is used? That is the difference between moles I2 initially and moles I2 at the end. The moles at the end were titrated with S2O3^- so
moles S2O3^2- = M x L and 1/2 that gives moles I2.
Convert moles I2 to moles sulfite and convert that to mg/L.

I omitted a step so disregard the first answer and just use this one.

IO3^- + 5I^- ==> 3I2 + .. you can finish balancing if you wish but this is all you need.
SO3^2- + I2 ==> 2I^- + SO4^2-
Again you can finish AND if you want to change the I2 to I3^- you may do so and make the 2I^- on the right side into a 3I^-.

moles IO3^- initially = 0.8043 x (5mL/100mL) x (1 mol/214.00 g) = ??
moles I2 generated = 3*moles S2O3^2-.

Then moles I2 react with sulfite according to the equation at the top of my post, which uses up part of the I2 there initially. How much is used? That is the difference between moles I2 initially and moles I2 at the end. The moles at the end were titrated with S2O3^- so
moles S2O3^2- = M x L and 1/2 that gives moles I2 at the finish. Subtract from moles there initially to find moles that reacted. Then
Convert moles I2 to moles sulfite and convert that to mg/L.