If 7.50 x 10-4 mol of IO3- (the limiting reagent) were required to reach the endpoint of the titration of a vitamin C sample, how many moles of vitamin C were present in the sample?
+37084 The titration steps in question are : (believe me...I had to look this one up)
- the iodate ions are reduced to form iodine
IO3- + 6 H+ + 5 e- --> ½ I2 + 3 H2O
- while the iodide ions are oxidised to form iodine.
2 I- --> I2 + 2 e-
Combining these half-equations demonstrates the reaction between iodate and iodide
IO3- + 5 I- + 6 H+ --> 3 I2 + 3 H2O
It is the iodine formed by this reaction that oxidises the ascorbic acid to dehydroascorbic acid as the iodine is reduced to iodide ions.
ascorbic acid + I2 --> 2 I- + dehydroascorbic acid
SO each mole of IO3- used results in 3 I2 's Each I2 reacts with one molecule of ascorbic acid (vit C).....so you will use THREE times the vit C as moles of IO3-
7.5 x 10^-4 mole x 3= = 22.5 x 10^-4 moles of vit C
(I think !)
