Question

A 0.5242 g solid sample containing a mixture of LaCl3 (molar mass = 245.26 g/mol) and Ce(NO3)3 (molar mass = 326.13 g/mol) was dissolved in water. The solution was titrated with KIO3, producing the precipitates La(IO3)3(s) and Ce(IO3)3(s). For the complete titration of both La3 and Ce3 , 42.70 mL of 0.1237 M KIO3 was required. Calculate the mass fraction of La and Ce in the sample.

Answer 1

Moles of IO3^-1 added = 0.04396 x 0.1237 = 5.43785 x 10^-3
Mols of La^+3 + moles of Ce^+3 = 5.43785 x 10^-3 / 3 = 1.8126x10^-3

Let X = g of LaCl3
X / 245.26 = moles of LaCl3
0.5242 - X = g of CeCl3
(0.5242 - X) / 326.13 = moles of CeCl3

X / 245.26 + (0.5242 - X) / 326.13 = 1.8126x10^-3
Solve for X

X = 0.2030 g LaCl3
0.5242 - 0.2030 = 0.3212 g CeCl3
(138.9055 / 245.26) x 0.2030 = 0.1149 g La in 0.5242 g sample (0.2193 g La / 1 g sample)
(140.12 / 326.13) x 0.3212 = 0.1380 g Ce in 0.5242 g (0.2632 g Ce in 1 g sample)