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Annette [7]
3 years ago
11

When standardizing an iodine solution, 30.37 ml of the solution was required to completely oxidize 10.00 ml of a 5 mg/ml solutio

n of ascorbic acid. what is the molarity of the titrant??
Chemistry
1 answer:
Svetach [21]3 years ago
4 0
First, we need to determine the reaction that happens when standardizing. The reactions would be:

 KIO3(aq) + 6 H+ (aq) + 5 I- (aq) ------> 3 I2(aq) + 3 H2O(l) + K+ (aq)
C6H8O6(aq) + I2(aq) ------> C6H6O6(aq) + 2 I- (aq) + 2 H+ (aq)

5g/L C6H8O6 (0.01 L) ( 1 mol / 176.12 g) ( 1 mol I2 / 1 mol C6H8O6 ) ( 1 mol KIO3 / 3 mol I2 ) = 0.00009 mol KIO3

Molarity = 0.00009 mol KIO3 / 0.03037 L = 0.003 M

Hope this answers the question. Have a nice day.
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