Step by step procedure on how to determine the value of n.

Question

The following solutions are needed to be prepared for the titration procedure:

1.) 20 mL of 2 M HCl

2.) 20 mL of 0.05 M KOH

After the approval of the procedure, you went through the titration process wherein 1.0867 grams of Fe(OH)_n was mixed with the 20 mL of 2 M HCl that you initially prepared. This solution was diluted to 200 mL and was labeled as solution A. 20.0 mL aliquot of solution A was taken and titrated with 20.0 mL of 0.05 M KOH. From this data, determine for the value of n.

Answer 1

n is the number of moles of (OH) per mole of Fe. For example, it could be 2, as in Fe(OH)₂ or 3 as in Fe(OH)₃. These would be the two most common possibilities. So, how are we to determine which it is?

This is somewhat involved, but I'll do my best to explain. This involves a back titration where the original sample is placed in EXCESS HCl, some of which will react with the iron hydroxide. But some HCl will be left over, and we can determine that amount by back titrating the excess with KOH. The difference will be the amount of HCl that actually reacted with the original sample of iron hydroxide.

Reaction 1: Fe(OH)_n + nHCl == > nH₂O + FeCl_n

mols HCl used = 20 ml x 1 L / 1000 ml x 2 mol/L = 0.04 mols HCl used

But not all of this HCl reacted with the Fe(OH)_n, because we can then back titrate the excess HCl with KOH

Reaction 2: HCl + KOH ==> H₂O + KCl

mols KOH used = 20.0 ml x 1 L / 1000 ml x 0.05 mol/L = 0.001 mols KOH = 0.001 mols HCl in excess

Correcting for the aliquot of 20 ml out of 200 mls, this equates to 0.01 mols HCl in excess in original sample

Moles HCl used to react with original Fe(OH)_n = 0.04 mols - 0.01 mols = 0.03 mols HCl

This is equal to mols OH- present in original sample (see reaction 2 above)

molar mass OH^- = 17 g/mol

0.03 mols OH^- x 17 g/mol = 0.51 g OH^- in original sample

mass Fe in original sample = 1.0867 g - 0.51 g = 0.5767 g Fe

mols Fe in original sample = 0.5767 g x 1mol Fe / 55.85 g = 0.01 mols Fe

NOTE the mole ratio of Fe to OH^- is 1:3

This makes the compound Fe(OH)₃

n = 3