Question

A 3.826 g sample containing Fe, Ti, and other inert material was dissolved and diluted to form 250.0 mL of solution. The dissolution conditions were such that Fe was converted to Fe(III) and Ti was converted to Ti(IV) in solution. A 25.00-mL aliquot of the solution was then passed through a Walden reductor and titrated with 0.2032 M Ce4 . The endpoint was reached following the addition of 22.42 mL of the Ce4 solution. A second 50.00-mL aliquot of the solution was passed through a Jones reductor and titrated with the same Ce4 solution. The endpoint was reached following the addition of 61.36 mL of the Ce4 solution. Calculate the mass percent of Fe and Ti in the sample.

Answer 1

Amount of Fe (III) in 25mL aliquote : reaction in walden reductor

25 mL * M = 0.2032 M * 22.42 mL

or, M = (0.2032 M * 22.42 mL) / 25mL = 0.182 M

The molarity of Fe = 0.182 M

= 0.182 mol/L

The moles of Fe in 250 mL solution = (0.182 *250mL)/1000mL

= 0.045 moles

The mass of Fe = moles/molar mass

= 0.045 moles * 56 g/mol. ( as, molar mass of Fe = 56 g/mol)

= 2.52 g

Amount of Ti + Fe : Reaction in Jones reductor

Molarity of Ti + Fe in 50mL aliquote = (61.36 * 0.2032) / 50

= 0.249 moles / L

The amount of Ti + Fe in 250mL = 0.249 * 250mL / 1000 mL

= 0.0622 moles

The moles of Ti present in 250mL = 0.0622 - moles of Fe

= 0.0622 - 0.045

= 0.0172 moles

The amount of Ti = 0.0172 * 47.87 (the molar mass of Ti = 47.87g/mol)

= 0.823 g

The mass % of Fe = (2.52 / 3.826) * 100

= 65.8 %

The mass % of Ti = (0.823 / 3.826) * 100

= 21.51 %