Question

A 0.5065 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 , 44.81 mL of 0.1252 M KIO3 was required. Calculate the mass fraction of La and Ce in the sample.

Answer 1

m = 0.5065 g

LaCl3 + Ce(NO3)3

La+3 + 3IO3- = La(IO3)3(s)

Ce+3 + 3IO3- =Ce(IO3)3(s).

total mmol of KIO3 = MV = 44.81*0.1252 = 5.6102 mmol of IO3- used

3 mol of La+3 + 3 mol of Ce+3 = 5.6102 mmol (EQN1)

for EQN2:

MW of LaCl3 * mol of LaCl3 + MW of Ce(NO3)3 *mol of Ce(NO3)3 = 0.5065

let

mol of LaCl3 =mol of La+3 = x

mol of CeCl3 = mol of Ce+3 = y

then

EQN2

MW of LaCl3 * mol of LaCl3 + MW of Ce(NO3)3 *mol of Ce(NO3)3 = 0.5065

x * 245.26 + y * 326.13 = 0.5065

and EQN 2:

3 mol of La+3 + 3mol of Ce+3 = 5.6102 mmol (EQN1) =  (5.6102 *10^-3) = 0.0056102

3x + 3y = 0.0056102

y = (0.0056102/3 - x) = 0.001870-x

solve:

x * 245.26  + y * 326.13  = 0.5065

x * 245.26  + (0.001870-x)*326.13   = 0.5065

x(245.26 -326.13 ) + (0.001870*326.13 ) = 0.5065

-80.87x = 0.5065 -0.001870*326.13

x = (-0.1033631) / (-80.87) = 0.001278

y = (0.0056102/3 - x) = 0.001870-0.001278 = 0.000592

then

mol of LaCl3 = 0.001278

mol of CeCl3 = 0.000592

mass of LaCl3 = 0.001278*245.26 = 0.313 g

mass of CeCl3 = 0.000592*326.13 = 0.1931 g

total mass = 0.193068+0.313

which prooves the mass balance