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Topic: Titrations dealing with redox reactions  (Read 3937 times)

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Offline psychkub

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Titrations dealing with redox reactions
« on: July 20, 2008, 08:17:59 PM »
Can someone just confirm I did this right?

Ascorbic Acid (C6H8O6, MM=176.126) acts as a reducing agent as follows
C6H8O6 -> C6H6O6 + 2H+ +2e-
It can be oxidized by I2 +2e- -> 2I-
A 200.0mL sample of a citrus fruit drink is acidified, and 100.0mL of .0500M I2 is added. All of the ascorbic acid is oxidized and some iodine remains in excess. After the reaction is complete this excess iodine is titrated with 38.62mL of .0120M Na2S2O3. Compute the number of milligrams of ascorbic acid per mL of fruit drink.

So the first thing I did was find the net result of the redox reaction and got:
C6H8O6 + I2 -> C6H6O6 + 2 I- + 2H+

Next I found the total amount of iodine that reacted in both the redox part and in the titration and got:
.0500mol I2/L * .01000L = 5.00e-4 mol I2

Next, I found the amount of Na2S2O3 that would react with the iodine and  used that amount to stoichiometrically find the amount of iodine that was used for the titration and got:
.0120 mol Na2S2O3/L * .03862L = 4.6344e-4 mol Na2S2O3
which is also the amount of S2O3 available due to the one to one relationship.

Since I2+ 2S2O3 -> 2I-+ S4O62-
4.6344e-4 mol S2O3 * mol I2/2 mol S2O3 = 2.3172e-4 mol I2

To find the amount of iodine that initially reacts:
5.00e-4 - 2.317e-4 = 2.6828e-4 mol I2

To find the mg of ascorbic acid:
2.6828e-4 mol 2 * mol C6H8O6/ mol I2 = 2.6828e-4 mol C6H8O6

2.6828e-4 mol C6H8O6 * 176.126g/mol C6H8O6 = .047251g

47.251mg/200.0mL = .2363mg/mL

is that right?

Offline Astrokel

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Re: Titrations dealing with redox reactions
« Reply #1 on: July 21, 2008, 08:35:18 AM »
hey psychkub,

Brief through it, your steps are all correct, just remember to put the charge of thiosulphate in your equations.
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

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