posted by Yulieth on .
in an experiment 5.00ml of 0.1440 mol L^1- aqueos KIO3 was reacted with an excess of aqueous iodide and the molecular iodine I2, Formed was titrated against 0.1000mol L^1- aqueous S2O3^2- the equation for the titration reaction is:
I2(aq)+ 2 S2O3^2-(aq)---->S4O6^2-(aq)+ 2I^-1(aq)
calculate the number of moles of potassium iodate (KIO) that were present in the 5.00mL aliquot of standar potassium iodate solution
can some1 please help me?
I've answered this for you earlier tonight. moles KIO3 = M x L = 0.144M x 0.005 mL = ?