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Standardization of the Sodium Thiosulfate Solution

Make up a standard solution of potassium iodate by accurately “weighing by difference” about 0.1 g of KIO3 and placing it in your 100 mL volumetric flask. Fill the flask to mark with distilled water. Potassium iodate reacts with excess KI in acid solution according to the following reaction:

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

RECALL:

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

To perform then standardized, first prepare the buret with the given solution of Na2S2O3. Then dispense 25 mL of KIO3 solution by pipet into a 250 mL Erlenmeyer flask, add 25 mL of 2% (v/v) KI solution and then add 10 mL of 1M HCL, and mix swirling. A deep brown colour should appear, indicating the presence of iodine. Titrate immediately with the Na2S2O3 solution in the buret until the brown fades to a pale yellow and then add 5 mL of starch indicator solution and continue the titration until the deep blue colour of the starch indicator disappears. Record the volume of the sodium thiosulafte solution used in the titration. Determine the stoichiometric ratio between the iodate and thiosulafte and use this to calculate the accurate concentration if the Na2S2O3 solution.

A.Standardization of Na2S2O3 ( Sodium thiosulfate)

Mass of KIO3 in 100 mL - 0.1255g

Find Molarity of potassium Iodate: (PLEASE CHECK IF CORRECT)

n= 0.1255/ 214 = 5.86*10^-4
100ml/1000= 0.1L

Molarity = 5.86*10^-4 /0.1
= 5.86*10^-3

Therefore, the concentration of KIO3 is: 5.86*10^-3

Titration:

Average Volume of Na2S2O3 = 27.15 mL

Concentration of Na2S2O3 _______________M.

I am not sure how to calculate this

Your calculation for the molarity of KIO3 is ok except that you didn't carry it out far enough. You have four places in the mass of 0.1255 g KIO3 you weighed, (and the molar mass is 214.00 and not 214) so you need at least 4 places in the molarity.
The procedure calls for you to pipet 25 mL of this KIO3 solution; therefore, the mols will be M x L = ??
Now use the equations to convert mols KIO3 to mols S2O3^-2.
mols KIO3 x (3 mols I2/1 mol KIO3) x (2 mols S2O3^-2/1 mol I2) = xx mols S2O3.
Then M S2O3^-2 = xxmols/0.02715 L.

Hey DrBobb i have a question...

I am sure what you mean by 4 places... since they are asking for the mass of KIO3 in 100 mL do we not assme its before we add it together with the other solutions?

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