A 10.00mL sample of 0.02141mol/L potassium iodate was treated with excess iodide ion and titrated with sodium thiosulfate. If 32.33 mL of a Na2S2O3 solution was requried to reach the colorless endpoint (after addition of starch indicator) , what molarity of the Na2S2O3 solution should be reported?

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

I2 + 2S2O3^2- ==> SrO5^2- + 2I^-

Check those equations; I'm trying to remember it instead of working it out.

mol IO3^- = M x L = 0.0214 x 0.010 = ?

1 mol IO3^- = 3 mols I2
1 mol I2 = 2 mols S2O3^-

Therefore, 1 mol IO3^- = 6 mols S2O3^-
Convert mols IO3^- from above to mols S2O3^2- and substitute into
M S2O3^2- = mols S2O3^2-/L S2O3^2-

To determine the molarity of the Na2S2O3 solution, we first need to determine the number of moles of Na2S2O3 used in the titration.

The balanced chemical equation for the reaction between potassium iodate (KIO3) and iodide ion (I-) is:

KIO3 + 5I- + 6H+ -> 3I2 + 3H2O + K+

From the equation, we can see that 1 mole of KIO3 reacts with 5 moles of I-. Therefore, the number of moles of I- used in the titration is:

moles of I- = (0.02141 mol/L)(0.01000 L) = 0.0002141 moles

Now, let's determine the number of moles of Na2S2O3 used. From the balanced chemical equation for the reaction between I2 and Na2S2O3:

2Na2S2O3 + I2 -> Na2S4O6 + 2NaI

We can see that 1 mole of I2 reacts with 2 moles of Na2S2O3. Therefore, the number of moles of Na2S2O3 used in the titration is:

moles of Na2S2O3 = (0.0002141 moles I2) / 2 = 0.0001071 moles

Next, we calculate the molarity of the Na2S2O3 solution using the volume of the solution used in the titration.

molarity of Na2S2O3 = moles of Na2S2O3 / volume of Na2S2O3 solution (in L)

volume of Na2S2O3 solution = 32.33 mL = 32.33 / 1000 = 0.03233 L

molarity of Na2S2O3 = 0.0001071 moles / 0.03233 L = 0.003313 M

Therefore, the molarity of the Na2S2O3 solution should be reported as 0.003313 M.