In the standardization of na2s2o3 solution, if you did not titrate the liberated iodine immediately, how would this likely affect the value of the molarity you calculate?

The titration ought to be done fastly in light of the fact that if not titrated immediately the freed I2 will leave the system. So it won't respond with the Na2S2O3. then the esteem we are getting for the Na2S2O3 is not the actual value. The majority will littler than the real esteem.