Question

25ml of 0.01M KIO3 solution were put to react with an excess of KI. 32.04ml of the sodium thiosulfate solution were needed to titrate the triiodide ions that were present .
What is the molar concentration of thiosulfate in the solution?

Answer 1

I have not used the triiodide ion (I3^-) below but you may convert to that if you wish.
KIO3 + 5KI + 3H2SO4 ==> 3H2O + I2 + 3K2SO4

2S2O3^2- + I2 ==> S4O6^2 + 2I^-

mols KIO3 = M x L = ?
mols I2 produced = 3 x that = ?
mols S2O3^2- needed = 2 x mols I2 = ?
Then M S2O3^2- = mols S2O3^2-/L S2O3^2- = ?