Question

The question: If the purity of thiosulfate exceeds 100% what could the explanation for that be?

More information if needed:

I was doing experiment and was supposed to find out the purity of thiosulfate.

I measured 1,3430 g of sodium thiosulfate and put that into 100mL volumetric glassware. I added de-ionized water, so that the solution would be 100mL together(the water and sodium thiosulfate). I put this solution into burette.

Then I calculated the molarity of that solution by assuming 100% purity.

So:

mass Na₂S₂O₃ * 5H₂O = 1,3430g

Molar mass Na₂S₂O₃ * 5H₂O = 248,18g/mole

Volume  Na₂S₂O₃ solution = 0,100L

Molarity  Na₂S₂O₃ solution = (1,3430g/248,18g/mole)/0,100L = 0,05411 moles/L

After that I made another solution by mixing 0,2377 g of KIO₃ with de-ionized water so that the solution would be 100mL together. I took 10,00mL of this solution with pipette and put it into 250 mL erlenmeyer flask. I diluted that amount with 50 mL of de-ionized water and added to that about 0,8 g of KI. When this was soluted I acidified the solution with 10,00mL of 1M HCl. Then I quickly titrated the solution with the thiosulfate solution until the color was light brown and then I added 1mL of starch solution and then the solution became dark blue(kind of light black like coca cola though). After that I titrated more until the solution became clear again. I did this 3 times and the average volume of Na₂S₂O₃ used was 12,25 mL

Then I calculated the molarity of the KIO₃ solution.

So:

Mass KIO₃ = 0,2377 g

Molar mass KIO₃ = 214,001 g/mole

Volume KIO₃ solution = 0,100 L

Molarity KIO₃ solution = (0,2377g / 214,001g/mole)/0,100L = 0,01111 moles/L

Then I calculated the molarity of Na₂S₂O₃ solution.

So:

Moles of KIO₃ in each titration = 1,11074 * 10^-4 moles.

Moles of Na₂S₂O₃ in each titration = 3*2* 1,11074 * 10^-4 = 6,66444 * 10^-4 moles.

Volume Na₂S₂O₃ (on average) = 12,25 * 10^-3 L

Molarity Na₂S₂O₃ solution = 6,66444 * 10^-4 moles / 12,25 * 10^-3 L = 0,05440 moles/L.

Purity in % = (Molarity Na₂S₂O₃, measured/ Molarity Na₂S₂O₃, calculated) * 100% = ((0,05440 moles/L) /( 0,05411 moles/L)) * 100% = 100,5%

I got that the purity was more than 100% how can that be?

Answer 1

Hello I have see your calculation part which is corrected. Now you are getting purity 100.5% that mean error is 0.5% which can be explained below

1. There are chance of permanent adsorption of iodine on starch. Starch should be added only near the equivalence point to reduce permanent adsorption of iodine on starch.

2. Loss of liberated iodine due to it's high volatility. To avoid this loss, KI is added which reacts with iodine to form KI3 which is soluble in water.