A metal ore sample is analyzed by a back titration of EDTA f

A metal ore sample is analyzed by a back titration of EDTA for iron content. A 1.5936 g sample of the ore is fully dissolved to liberate the iron as Fe2+ ions. The resulting solution is treated with 100.0 mL of 0.0536 F Na2H2Y*2H2O. The excess EDTA was then titrated with a 0.110 F MgCl2 solution, needing 3.28 mL to reach equivalence. What is the mass percent of iron in the ore sample?

Solution

Determine the millimoles of EDTA, Na₂H₂Y.2H₂O added as millimoles = (volume of Na₂H₂Y.2H₂O in mL)*(formality of Na₂H₂Y.2H₂O) = (100.0 mL)*(0.0536 F) = 5.36 mmole.

Na₂H₂Y.2H₂O reacts with MgCl₂ on a 1:1 molar basis as

MgCl₂ (aq) + Na₂H₂Y.2H₂O (aq) --------> MgH₂Y (aq) + 2 NaCl (aq) + 2 H₂O (l)

Millimoles of MgCl₂ reacted with EDTA = millimoles of EDTA titrated by MgCl₂ = (3.28 mL)*(0.110 F) = 0.3608 mmole.

Millimoles of Na₂H₂Y.2H₂O reacted with Fe²⁺ = (5.36 – 0.3608) mmole = 4.9992 mmole.

Fe²⁺ reacts with Na₂H₂Y.2H₂O on a 1:1 molar basis as

Fe²⁺ (aq) + Na₂H₂Y.2H₂O (aq) ---------> FeH₂Y (aq) + 2 Na⁺ (aq) + 2 H₂O (l)

Therefore, millimoles of Fe²⁺ present in the sample = millimoles of Na₂H₂Y.2H₂O required to neutralize Fe²⁺ = 4.9992 mmole.

Atomic mass of Fe²⁺ = atomic mass of Fe (electron is supposed to be massless) = 55.845 g/mol.

Mass of Fe²⁺ present in the sample = (moles of Fe²⁺)*(atomic mass of Fe²⁺) = (4.9992 mmole)*(1 mole/1000 mmole)*(55.845 g/mol) = 0.279180 g 0.2792 g.

Percent Fe in the sample = (mass of Fe)/(mass of sample)*100 = (0.2792 g)/(1.5936 g)*100 = 17.52008% 17.5201% (ans).