Question

How many milliliters of 0.450 M Na2S2O3 solution are needed to titrate 2.431 g of I2 to the equivalence point?
I2(aq) + 2 S2O32-(aq) S4O62-(aq) + 2 I -(aq)


mL

Answer 1

I2 + 2 S2O32- S4O62- + 2I-
M(I2) = 127g/mol
n(I2) = 2.431 / 127 = 0.019(mol)
2C(I2)*V(I2)= C(S2O32-)*V(S2O32-)
C(I2)*V(I2)= n(I2)
2n(I2)= C(S2O32-)*V(S2O32-)
V(S2O32-) = 2n(I2) / C(S2O32-)
V(S2O32-) = 0.019 mol * 2 / 0.45 mol/l = 0.084 l = 84 ml