A solution of I2 was standardized with ascorbic acid. Using a 0.1000-g sample of pure ascorbic acid (C6H8C6), 25.32 ml of I2 were required to reach the starch end point.

Question

What is the molarity of the iodine solution?

Please check to see if I did this right:

(0.1000g Ascorbic Acid/25.32 mlI2)x(1 mole AA/176 g AA)x(1 mole I2/1 mole AA) = 2.244x10^(-5)

The text says that you need one mole of Ascorbic acid to react with 1 mole of I2) I have my doubt with my calculations because I am not sure if you can cancel out grams and ml as I did in the first part.

Answer 1

I will do the problem and let you check it. I'm glad you included the 1:1 ratio for I had forgotten what it was. I assume you meant the formula to be C6H8O6
mols ascorbic acid = grams/molar mass = 0.1000/176.13 = ??

Since it is a 1:1 titration, then mols I2 = the same number of ??

The definition of molarity = mols/L.
You know mols, ?? so divide by 0.02532 L. I find 0.02242 M. Check my work.

Answer 2

I see what I did wrong now, I forgot that Molarity= mols/L, I get the same number for the answer. Thank You!