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 .

Question

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- = ?