The mass of potassium iodate (KIO3) contained in an impure sample was determined by titration with sodium thiosulfate (0.1005 M). The impure KIO3 was dissolved in 50 cm 3 of water, an excess of potassium iodide (KI) and 5 cm^3 of dilute sulphuric acid was then added. At the starch endpoint, 30.65 cm^3 of thiosulfate solution had been added. Calculate the mass of KIO3 in the sample.
I - → ½I2 + e -
IO3- + 6H+ + 5e - → ½I2 + 3H2O
The liberated iodine reacts as follows:
I2 + 2e - → 2I-
2S2O3 2- → S4O62- + 2e-
Hint: 6 equivalents of S2O32- are equivalent to IO3-.
If you want this done with equivalents, then
mL x M x (molar mass KIO3/6000) = grams KIO3
How do I find mL?
The problem tells you.
30.65 cm^3 = 30.65 cc = 30.65 mL.
1 cc = 1 mL = 1 cubic centimeter = 1 cm^3.
I got 0.1099g
Is that correry?
To calculate the mass of KIO3 in the sample, we need to use the information provided in the titration.
First, let's identify the stoichiometry of the reaction between KIO3 and Na2S2O3. From the given equations, we can see that for every 1 mole of IO3- ions, 6 moles of S2O32- ions are required for complete reaction.
Given that the concentration of the Na2S2O3 solution is 0.1005 M, and 30.65 cm^3 of it was required for the titration, we can calculate the number of moles of S2O32- used:
Moles of S2O32- = (0.1005 mol/L) x (0.03065 L) = 0.003079 moles
Since 6 moles of S2O32- are required per mole of IO3-, the number of moles of IO3- can be calculated:
Moles of IO3- = 0.003079 moles / 6 = 0.0005132 moles
Now, we can calculate the mass of KIO3 using the molar mass of KIO3 (214.00 g/mol):
Mass of KIO3 = Moles of IO3- x Molar mass of KIO3
= 0.0005132 moles x 214.00 g/mol
≈ 0.1099 g
Therefore, the mass of KIO3 in the sample is approximately 0.1099 grams.