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?

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- = ?

To determine the molar concentration of thiosulfate in the solution, we need to use the given information about the reaction between KIO3 and excess KI. The reaction can be represented as follows:

2 KIO3 + 6 KI + 6 HCl → 6 KCl + 2 I3 + 3 H2O

From the balanced equation, we can see that 2 moles of KIO3 react with 2 moles of I3 (triiodide ion). Therefore, the ratio of moles of KIO3 to I3 is 1:1.

Given that 25 ml of 0.01 M KIO3 solution reacted, we can calculate the moles of KIO3 used:

Moles of KIO3 = Volume (in L) × Molarity
= 0.025 L × 0.01 mol/L
= 0.00025 moles

Since KIO3 reacts with I3 in a 1:1 ratio, the moles of I3 formed is also 0.00025 moles.

Now, we can use the volume of sodium thiosulfate solution required to titrate the I3 (32.04 ml) to determine its molar concentration.

Moles of thiosulfate = Moles of I3
Volume of thiosulfate solution (in L) × Molarity of thiosulfate = Moles of thiosulfate

Molarity of thiosulfate = Moles of thiosulfate / Volume of thiosulfate solution (in L)
= 0.00025 moles / 0.03204 L

Therefore, the molar concentration of thiosulfate in the solution is 0.0078 M (rounded to four decimal places).