In order to standardise a solution of sodium thiosulphate, a solution potassium iodate was made by dissolving 0.980g in water to give exactly 250cm^3 of solution. when 20.00cm^3 of this solution was added to an access of acidified potassium iodide solution, 21.35cm^3 of the sodium thiosulphate solution was requires to react with the liberated iodine. what was the concentration of the original sodium thiosulphate solution?
edit: ok equation is 2S2O3 +I2 -->2I- +S406 + I2
moles of iodate is ...don't know
dont know, confused