Question

A 0.5206 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 , 43.38 mL of 0.1236 M KIO3 was required. Calculate the mass fraction of La and Ce in the sample.​

Answer 1

Let mass of LaCl₃ = x

Therefore, mass Ce(NO₃)₃ = (total mass) - (mass of LaCl₃)

mass Ce(NO₃)₃ = 0.5206g - x

moles LaCl₃ = (mass of LaCl₃) / (molar mass of LaCl₃)

moles LaCl₃ = (x) / (245.26 g/mol)

moles LaCl₃ = (x / 245.26) mol

moles La(IO₃)₃ formed = moles LaCl₃

moles La(IO₃)₃ formed = (x / 245.26) mol

moles KIO₃ consumed = 3 * (moles La(IO₃)₃ formed)

moles KIO₃ consumed = (3x / 245.26) mol

moles Ce(NO₃)₃ = (mass Ce(NO₃)₃) / (molar mass Ce(NO₃)₃)

moles Ce(NO₃)₃ = (0.5206g - x) / (326.13 g/mol)

moles Ce(NO₃)₃ = (0.5206 - x) / 326.13 mol

moles Ce(IO₃)₃ formed = moles Ce(NO₃)₃

moles Ce(IO₃)₃ formed = (0.5206 - x) / 326.13 mol

moles KIO₃ consumed = 3 * (moles Ce(IO₃)₃ formed)

moles KIO₃ consumed = (1.5618 - 3x) / 326.13 mol

Total moles KIO₃ consumed = (3x / 245.26) mol + (1.5618 - 3x) / 326.13 mol

Total moles KIO₃ required = (molarity KIO₃) * (volume KIO₃ in Liter)

Total moles KIO₃ required = (0.1236 M) * (43.38 x 10^-3 L)

Total moles KIO₃ required = 5.3618 x 10^-3 mol

Total moles KIO₃ required = Total moles KIO₃ consumed

5.3618 x 10^-3 mol = (3x / 245.26) mol + (1.5618 - 3x) / 326.13 mol

Solving for x, x = 0.1889 g

mass LaCl₃ = 0.1889 g

mass fraction LaCl₃ = (mass LaCl₃ / mass sample) * 100

mass fraction LaCl₃ = (0.1889 g / 0.5206 g) * 100

mass fraction LaCl₃ = 0.3628

mass fraction Ce(NO₃)₃ = 1 - mass fraction LaCl₃

mass fraction Ce(NO₃)₃ = 1 - 0.3628

mass fraction Ce(NO₃)₃ = 0.6372