What volume of .10 M KIO3 required to reach the end point?

Question

Im stuck with this prelab problem and I dont know what to do any help? or suggestions of what should I do ? Thanks

a.400 g sample of ascorbic acid is titrated with .10 M KIO3. The solution contains 5 ml of 1 M hcl and 1 g of KI(s). the iodate ion reacts with iodite I- , and H + to produce iodine and water.

IO3- +5 i- + 6H+ ---> 3 i2 + 3H2O
the ascorbic acid reacts immediately with the i2

when all the ascorbic acid has reacted an increase in the concentration of I2 and the reaction of I2 with I- to produce I3- signals the blue end point. triiodine I3- complexes with starch to produce the dark navy blue color.

I2 + I - -----> I3-

what volume of .10 M KIO3 is required to reach the end point.?

Answer 1

First of all I- is iodiDe and NOT iodiTe.
I- acts catalytically here (since it is produced again by the reaction of ascorbic acid and I2)

Ascorbic acid is C6H8O6 with MW=176.13. I think you have written 0.400 g and not 400g, right?
So in your sample you have moles C6H8O6 =0.4/176.13 = 2.27*10^-3

The titration reaction is
C6H8O6 + I2 -> C6H6O6 + 2I- +2H+

So 1mole C6H8O6 reacts with 1 mole I2
2.27*10^-3 .. .. .. .. .. .. .. .. . .. . .x
=> x=2.27*10^-3 mole I2

But I2 was formed by the KIO3

IO3- +5I- +6H+ -> 3I2 + 3H2O

1 mole IO3- produces 3 moles I2
x .. .. .. .. .. .. .. .. .. .. 2.27*10^-3

3x=2.27*10^-3 => x =(2.27/3)*10^-3 = 7.57*10^-4 mole KIO3

but you know that M=mole/V => V=mole/M= (7.57*10^-4)/0.1 = 7.57*10^-3 L= 7.57 mL