BACK TITRATION OF AS CORBIC ACID For the standardization, 5 ml of KIO 3 was adde

Question

BACK TITRATION OF ASCORBIC ACID

For the standardization, 5 ml of KIO3 was added, as well as 0.5087 g of solid KI and 2.5 mL of 0.5 M H2SO4. It was titrated with the Na2S2O3 solution previously prepared.

0.01 M KIO3 (+excess KI and acid)           0.07 M Na2S2O3

1- Calculate the ascorbic acid concentration from your manual titration. (HOW TO CALCULATE THE EXCESS IODINE)

IO3- + 8I- + 6H+ ?3I 3-+ 3H2 O                                

   I3- + 2S2O3-2 ? 3I- +S4 O6-2        

C6H8O6 + I3- ? C6H6O7 + 4H+ +3I-

Standardization of Thiosulfate solution

Trial

KI mass (g)

Thiosulfate added (ml)

KIO3 (ml) (0.01M)

1

0.5087

3.610

5 ml

analysis of ascorbic acid

trial

Ascorbic acid (g)

KI (g)

KIO3 (ml)

Thiosulfate added (ml)

Manual titration

0.1504

0.5002

15

2.360

C

Trial

KI mass (g)

Thiosulfate added (ml)

KIO3 (ml) (0.01M)

1

0.5087

3.610

5 ml

Explanation / Answer

Standardization of Thiosulfate solution

KI mass (g)=0.5087

Thiosulfate added (ml)=3.610

KIO3 (ml) (0.01M)= 5 ml

calculated concentration of S2O3-2= 0.0802



analysis of ascorbic acid

Ascorbic acid (g)=0.1504

KI (g)=0.5002

KIO3 (ml)=15

Thiosulfate added (ml)=2.360


so far, I was trying to find the excess Iodide
# moles I3- ( iodate reaction)- # moles I3- (thiosulfate reaction)= excess iodine

0.01M KIO3 X0.005L X( 3 mol I3-/ 1 mol KIO3)=4.5*10^-4 mol I3- excess

0.0802M S2O3-2 x 0.00236L x (1mol I3-/ 2mol S2O3^-2 )=9.5* 10^-5 mol I3-

(4.5*10^-4)-(9.5* 10^-5)= 3.55*10^-4 mol I3- and then I guess I'll multyply it by the ratio of 1:1 from the ascorbic acid which will give me the the moles and now I don't know what volume to use to find the concentration or if I am right. please help

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