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.
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...
2 answers
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.
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.
1 answer