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.

Type your answer to 4 significant figures (no units as it will not be able to be marked by Blackboard).

I - ¡æ ¨öI2 + e -

IO3- + 6H+ + 5e - ¡æ ¨öI2 + 3H2O

The liberated iodine reacts as follows:

I2 + 2e - ¡æ 2I-

2S2O3 2- ¡æ S4O62- + 2e-

Hint: 6 equivalents of S2O32- are equivalent to IO3-.

mols S2O3^2- = M x L = ?

mols IO3^- = 1/6 * mols S2O3^-
grams KIO3= mols KIO3 x molar mass KIO3
Post your work if you get stuck.

How am I going to find the mols? Actualy I have corrected the units in my way of finding the answer. So far I have done the following:

30.65cm3/1000*0.1005= 3.080325x10-3(on the top, I don't know how to write it here)
What about next?

My first line shows you how to find mols S2O3^2- and the second line shows how to convert to mols IO3^-.

To calculate the mass of KIO3 in the sample, we need to determine the number of moles of thiosulfate solution used in the titration.

First, let's find the number of moles of thiosulfate solution used. We can use the equation:

2S2O3 2- → S4O62- + 2e-

From this equation, we see that every mole of thiosulfate solution (S2O3 2-) reacts with 2 moles of electrons (e-). Therefore, the number of moles of thiosulfate solution used is half the number of moles of electrons.

The 30.65 cm3 of thiosulfate solution used corresponds to 30.65 / 1000 L = 0.03065 L.

Now, we need to find the number of moles of electrons used. We can use the balanced chemical equation for the oxidation of I2:

2I- → I2 + 2e-

From this equation, we see that every mole of iodine (I2) produced corresponds to 2 moles of electrons (e-). Therefore, the number of moles of electrons used is half the number of moles of iodine produced.

Since the iodine is produced from the reaction between IO3- and I-, we need to determine the number of moles of IO3-. We can use the equation:

IO3- + 6H+ + 5e- → I2 + 3H2O

From this equation, we see that every mole of IO3- reacts with 5 moles of electrons (e-). Therefore, the number of moles of electrons used is 5 times the number of moles of IO3-.

To find the number of moles of IO3-, we can use the molarity of the thiosulfate solution and the volume used:

Molarity of thiosulfate solution = 0.1005 M
Volume of thiosulfate solution used = 0.03065 L

Number of moles of thiosulfate solution used = (Molarity of thiosulfate solution) * (Volume of thiosulfate solution used)
= 0.1005 M * 0.03065 L

Now, we can find the number of moles of electrons used:

Number of moles of electrons used = 0.1005 M * 0.03065 L * 5 moles of electrons per mole of IO3-
= 0.015325 moles of electrons

Finally, the number of moles of thiosulfate solution used is half the number of moles of electrons:

Number of moles of thiosulfate solution used = 0.015325 moles of electrons / 2
= 0.0076625 moles of thiosulfate solution

Now we can calculate the number of moles of KIO3 in the impure sample. According to the balanced equation:

6 equivalents of S2O3 2- are equivalent to IO3-

Therefore, the number of moles of KIO3 is:

Number of moles of KIO3 = (Number of moles of thiosulfate solution used) / 6
= 0.0076625 moles of thiosulfate solution / 6
= 0.0012771 moles of KIO3

To calculate the mass of KIO3, we need to use its molar mass:

Molar mass of KIO3 = (atomic mass of K) + (atomic mass of I) + 3 * (atomic mass of O)
= 39.10 g/mol + 126.90 g/mol + 3 * 16.00 g/mol
= 214.10 g/mol

Now we can determine the mass of KIO3:

Mass of KIO3 = (Number of moles of KIO3) * (Molar mass of KIO3)
= 0.0012771 moles of KIO3 * 214.10 g/mol
= 0.2738 g

Therefore, the mass of KIO3 in the sample is approximately 0.2738 grams.