A0.5222 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 and ce(io3)3 . for the complete titration of both la3+ and ce3+ , 42.81 ml of 0.1250 m kio3 was required. calculate the mass fraction of la and ce in the sample.

Mole fraction of La is 19,58 wt% and of Ce is 28,11 wt%

Explanation:

The reactions of LaCl₃ and Ce(NO₃)₃ with KIO₃ are:

LaCl₃ + 3KIO₃ → La(IO₃)₃ + 3KCl

Ce(NO₃)₃ + 3KIO₃ → Ce(IO₃)₃ + 3KNO₃

The moles required for a complete titration of the sample are:

0,04281L× 0,1250M = 5,351x10⁻³ moles of KIO₃

By the reactions, and defining X are grams of LaCl₃ and Y as grams of Ce(NO₃)₃, these moles are:

5,351x10⁻³ = + (1)

As total mass of the solid sample are 0,5222g:

0,5222g = X + Y (2)

Replacing (2) in (1):

Y = 0,3417g

Thus:

X = 0,1805g

Grams of La are:

0,1805gLaCl₃×× = 0,1022g of La

Mass fraction of La is:

0,1022g of La÷ 0,5222g × 100 = 19,58 wt%

Grams of Ce are:

0,3417gCe(NO₃)₃×× = 0,1468g of Ce

Mass fraction of Ce is:

0,1468g of Ce÷ 0,5222g × 100 = 28,11 wt%

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