Hi, i'm doing an experiment that uses iodine and sodium thiosulphate and also sodium sulphite. First, i added 50.0 cm3 of 0.05 mol dm-3 aqueous iodine into a titration flask. Then added 25.0 cm3 of Solution that contain 24.0 g of anhydrous sodium sulphite, Na2SO3 per dm3 into it. Lastly, i added 2g of solid hydrogen carbonate into it. Then the solution is titrated with 0.1 mol dm-3 aqueous sodium thiosulphate. From the equation given, which is SO32- + I2 + H2O will gives you SO42- + 2HI And 2HI + 2HCO3- will gives 2I- + 2H2O + 2CO2 So from what i know is that the aqueous iodine solution reacts with the anhydrous sodium sulphite, Na2SO3 and water will gives a SO42- (is it pronounced as sulphide ions?) and 2 Hydrogen iodide. The hydrogen iodide then reacts with 2HCO3- (which should be the hydrogen carbonate ions) and gives out 2I- + 2H2O + 2CO2 Am i right so far? Please correct me if i'm wrong. If i am right, here's the question. Calculate the volume of I2, that did not react with the sulphite ions (which i think should be SO32-). In my experiment, what i did is exactly what i mentioned above, the color was first reddish, as i added my aqueous sodium thiosulphate in, the color slowly turns pale yellow, then i added starch as an indicator when it is pale yellow, and titrate until the blue color is gone. By that, it should mean that no more Iodine/Iodide is present in the soultion, so all Iodine should have completely been reacted, how do i solve the problem above? Thanks, sorry for the long post.