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?
1 answer
I've answered this for you earlier tonight. moles KIO3 = M x L = 0.144M x 0.005 mL = ?
3 answers