Question

A 0.5072 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 , 44.01 mL of 0.1289 M KIO3 was required. Calculate the mass fraction of La and Ce in the sample.

Answer 1

Answer – Given, mass of sample = 0.5072 g , [KIO₃] = 0.1289 M , volume = 44.01 mL

For the complete titration of both La³⁺ and Ce³⁺ there required 44.01 mL of 0.1289 M KIO₃

So first we need to calculate the moles of KIO₃

Moles of KIO₃ = 0.1289 M * 0.044 L

                          = 0.00567 moles

When KIO₃ reacts with La³⁺ and Ce³⁺ then there is formed La(IO₃)₃(s) and Ce(IO₃)₃(s).

So each ions need 3 moles of KIO₃ for reacts completely.

So total moles of La³⁺ and Ce³⁺ = 0.00567 moles / 3

                                                    = 0.00189 moles

Now we need to assume the mass of LaCl₃ = x g, so

Mass of Ce(NO₃)₃ = 0.5072 – x

We know, moles of LaCl₃ = x / 245.26 g.mol^-1

Moles of Ce(NO₃)₃ = (0.5072 – x) g / 326.13 g.mol^-1

We know

Total moles of La³⁺ and Ce³⁺ = 0.00189 moles

So, x / 245.26 g.mol^-1 + (0.5072 – x) g / 326.13 g.mol^-1 = 0.00189 moles

For solving the x

0.003066 (0.5072 – x) + 0.004077x = 0.00189

0.001555 -0.003066x+0.004077x = 0.00189

0.001011x = 0.00189 - 0.001555

               x   = 0.331

so mass of LaCl₃ = 0.331 g

mass of Ce(NO₃)₃ = (0.5072 – x)

                               = 0.5072 -0.331

                               = 0.176 g