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January 27, 2015

January 27, 2015

Posted by **Anonymous** on Monday, May 5, 2014 at 2:47pm.

- Chem -
**DrBob222**, Monday, May 5, 2014 at 3:17pmThis is an example of a back titration; i.e., the KI/KIO3 produces I2, the I2 reacts with some of the vit C, the excess I2 is titrated with thiosulfate.

6H^+ + I^- + IO3^- ==> 3I2 + 3H2O

Then vitC + I2 ==> 2I^- + vitCx

mols IO3 added initially = M x L = 0.025 x 0.01988 = approx and you must do this more carefully = 0.0005

mols I2 formed initially = 0.0005 x (3 mol I2/1 mol IO3^-) = 0.0015.

How much I2 was left; i.e., the amount left is the amount vit C did NOT react with.

2S2O3^2- + I2 ==> 2I^- + S4O6^2-

mols thiosulate = M x L = 0.08014 x 0.02385 = approx 0.002

Convert 0.002 mols S2O3^2- to mols I2. From the equation that is 0.002 x 1/2 = about 0.001 mols I2 not used by the vit C. So the difference between initial and end is amount used by vit C. That is 0.0015-0.001 = approx 0.0005 mols vit C analyzed.

g vit C = mols x molar mass in 19.8 mL sample. Convert to mg/100 mL.

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