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-
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?
1 answer