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Question

You have prepared an iron(III) solution by diluting 13.50mL of 7.73M iron(III) nitrate to a total volume of 270.0mL. What is the iron(III) ion concentration in this new solution? (show steps)

Answer 1

Iron(III) nitrate = Fe(NO₃)₃

molar mass Fe(NO₃)₃ = 241.9 g / mol

When in water, we have Fe(NO₃)₃ ==> Fe³⁺(aq) + 3NO₃^-(aq) ... note 1 Fe³⁺ per mol of Fe(NO₃)₃

For the dilution, we can use V1M1 = V2M2

V1 = initial volume = 13.50 ml

M1 = initial molarity = 7.73 M

V2 = final volume = 270.0 ml

M2 = final molarity = ?

(13.50 ml)(7.73 M) = (270.0 ml)(x M)

x = 0.3865 M Fe(NO₃)₃

Since each mol Fe(NO₃)₃ produces 1 mol Fe³⁺ the concentration of Fe³⁺ = 0.387 M (3 sig. figs.)