In Iodometry, I already know that I- will be oxidized to I2 if titration of Cu(II) is not done immediately, but what effect does it have to the molarity of S2O3^2-?

Are you standardizing thiosulfate with Cu metal? Then air oxidizes I^- to free I2 and that takes more thiosulfate in the titration; therefore, volume of thiosulfate is too large and molarity is too ???

low.. thanks!

To determine the effect on the molarity of S2O3^2-, we need to understand the reaction that takes place in iodometry. In iodometry, I- ions are oxidized to I2 by a strong oxidizing agent. The iodine formed will then react with S2O3^2- ions in a 1:2 ratio to produce I3^- ions.

The overall reaction can be represented as follows:

2S2O3^2- + I2 → S4O6^2- + 2I^-
(Iodine combines with the thiosulfate ion to form tetrathionate ion and iodide ion)

Initially, the molarity of S2O3^2- can be determined through the titration of a standard solution of iodine using a known concentration of thiosulfate solution (Na2S2O3). However, if the Cu(II) ions are not titrated immediately, the I2 formed from the oxidation of I- will react with S2O3^2- in solution.

This reaction will decrease the available concentration of S2O3^2- in the solution, leading to a decrease in the molarity of S2O3^2- over time. Therefore, if the titration is delayed, the molarity of S2O3^2- will decrease, resulting in inaccurate or incorrect titration results.

It is crucial to carry out the titration of Cu(II) immediately after the addition of I-, ensuring accurate determination of the concentration of S2O3^2-.