Question

Iron(II) can be oxidized by an acidic K2Cr2O7 solution according to the net ionic equation.

Cr2O72- + 6 Fe2+ + 14 H+ 2 Cr3+ + 6 Fe3+ + 7 H2O

If it takes 30.0 mL of 0.0250 M K2Cr2O7 to titrate 20.0 mL of a solution containing Fe2+, what is the molar concentration of Fe2+?

I have seen this question answered before, and I understand the mechanics of taking the ratio of CrO7 ion to Fe ion (1:6) and using MV=MV the answer (.225M). Could someone please try to clear up for me precisely why the moles of K2Cr2O7 corresponds directly with the moles of the Cr2O7 ion? Perhaps while walking through the solution again. And is it possible to work backwards into the non-net ionic equation so we can see the whole titration?

Answer 1

Fortitrations, it is always better to use concentration in terms of normality rather than molority. Using molarity calculations will be confusing.

We know the relation Normlity = n x molarity

Where n = n-factor.

For redox titration, n = change in oxidation state for 1 molecule

Hence for Fe²⁺ Fe³⁺, n = 3 - 2 = 1

For K₂Cr₂O₇, 2Cr⁶⁺ 2Cr³⁺, n = 2x6 - 2x3 = 6

Since K₂Cr₂O₇ dissociates 100% to give Cr₂O₇^2- ions, we can consider

K₂Cr₂O₇ Cr₂O₇^2- + 2K⁺

conc. of K₂Cr₂O₇ = conc. of Cr₂O₇^2-

Given molarity of K₂Cr₂O₇ M₁ = 0.0250 M

Hence normality of Cr₂O₇^2- N₁ =6 x M₁ = 6 x 0.0250 N = 0.15 N

Volume of K₂Cr₂O₇ V₁ = 30 ml

Volume of Fe²⁺ V₂ = 20.0 ml

let normality of Fe²⁺ = N₂

We know for titration equivalence point

V₁N₁ = V₂N₂

hence N₂ = V₁N₁/V₂

therefore normality of Fe2+ N₂ = 30 x 0.15 / 20 = 0.225 N

For Fe²⁺ Normality = n x molarity

Hence molarity = normality/ n

therefore for Fe²⁺ molarity M = 0.225 / 1 = 0.225 M

As per your other question regarding the non-net ionic equation, it shooldn't matter whether you are using the net ionic or non-ionic reaction. Because, we only consider those ions which change their oxidation states and the other counter ions does not influnce the concentration.