Iron is one of the most important elements necessary for the support of the vital functions of human — Physical Chemistry — Kinetics Chemistry Question
Iron determination
Iron is one of the most important elements necessary for the support of the vital functions of human organism. Its deficiency may cause anemia for treatment of which Fe(II) supplementation is usually employed. The therapeutic effect of Fe(III) compounds is much less pronounced.
Fe(II) is a fairly strong reducing agent which can be readily oxidized to Fe(III). Therefore methods for separate determination of Fe(II) and Fe(III) as well as for the determination of the total iron content are needed for quality control of pharmaceuticals. Here we will see how this problem can be solved.
Prior to determination of the total iron content it is usually transformed quantitatively either to Fe(II) or to Fe(III). Using standard redox potentials given below establish which of the oxidizing agents listed can oxidize Fe(II) to Fe(III) under standard conditions. Write down the balanced net ionic equations of corresponding reactions.
Model Answer
An oxidizing agent can convert Fe(II) to Fe(III) only if the corresponding redox potential is higher than that of the Fe(III) / Fe(II) couple. Therefore, all the oxidizing agents listed in Table with the exception of I2 could be used:
3 Fe2+ + NO3 – + 4 H+ → 3 Fe3+ + NO + 2 H2O
2 Fe2+ + H2O2 + 2 H+ → 2 Fe3+ + 2 H2O
2 Fe2+ + Br2 → 2 Fe3+ + 2 Br–
After oxidation of all the iron to Fe(III) its total amount can be determined by precipitation of iron in the form of Fe(OH)3 followed by annealing of the precipitate to Fe2O3 and weighing.
a) Estimate the pH of aqueous FeCl3 solution (c = 0.010 mol dm-3). Assume that Fe(OH2)6 3+ cation is a monoprotic acid with the dissociation constant Ka = 6.3×10–3.
b) Calculate the pH necessary to start precipitation of Fe(OH)3 from the solution above. Solubility product of Fe(OH)3 is Ksp = 6.3×10–38.
c) At what pH value precipitation of Fe(OH)3 from 100.0 cm3 of FeCl3 aqueous solution (c = 0.010 mol dm-3) will be complete? Consider the precipitation as complete if no more than 0.2 mg Fe remains in solution.
Note. All the pH values should be estimated with accuracy of 0.1 units pH. Neglect the effect of ionic strength.
Model Answer
a) Fe(H2O)6 3+ ⇌ Fe(H2O)5(OH2+ + H+,
Ka = [Fe(H2O)5(OH2+][H+] / [Fe(H2O)6 3+] = 6.3×10–3
[Fe(H2O)6] 3+ (further referred to as [Fe3+]) + [Fe(H2O)5(OH)]2+ (further referred to as [Fe(OH) 2+]) = c(Fe) = 0.010 mol dm-3, [Fe(OH)] 2+ = [H+] = x.
Therefore 6.3×10–3 = x^2 / (0.01–x) ⇒ x = 5.4×10–3 = [H+] ⇒ pH = 2.3
Note. In this case a simplified approach to calculate [H+] as √(Ka c) leading to the pH value of 2.1 is not acceptable since the dissociation constant of [Fe(OH2)6] 3+ is large and x in the denominator of the expression above should not be neglected compared to c.
b) Ksp = [Fe3+][OH–]3 = 6.3×10–38 ;
[Fe3+] + [Fe(OH2+] = c(Fe) = 0.010 ;
Ka = [Fe(OH2+][H+] / [Fe3+] ⇒ [Fe(OH2+] = [Fe3+] Ka / [H+] = [Fe3+][OH–] β, where β = Ka/Kw = 6.3×1011 and Kw = [H+][OH–] = 1.0×10–14.
A cubic equation relative to [OH–] can be obtained from the equations above, which may be solved iteratively as follows.
Denote [Fe3+] = x, [OH–] = y, then x(1+βy) = c ⇒ x = c / (1+βy)
Ksp = x y^3 ⇒ y = ∛(Ksp/x) ⇒ pH = – log Kw + log y.
Zeroth approximation: y = 0 ⇒ x = c / (1+βy) = 0.010 ⇒ y = ∛(Ksp/x) = 1.85×10–12 ⇒ pH = 2.27;
1st iteration: y = 1.85×10–12 ⇒ x = c / (1+βy) = 0.00462 ⇒ y = ∛(Ksp/x) = 2.39×10–12 ⇒ pH = 2.38;
2nd iteration: y = 2.39×10–12 ⇒ x = c / (1+βy) = 0.00399 ⇒ y = ∛(Ksp/x) = 2.51×10–12 ⇒ pH = 2.40 ≈ 2.4. Accuracy required obtained.
c) To be solved in a similar way with c(Fe) = 1.10–6 mol dm–3. pH = 4.3 (after 4 iterations).
Fe(II) can be determined in the presence of Fe(III) by titration with KMnO4 solution in acidic media. Since aqueous solutions of KMnO4 tends to decompose slowly over time, the exact concentration of KMnO4 has to be found immediately before determination of Fe(II). This is usually done by titration with KMnO4 of a solution of a primary standard, a pure substance of known composition. Such standard solution can be prepared by dissolving an exact amount of the primary standard in water in a volumetric flask of an exactly known volume.
12.3 For the titration of 10.00 cm3 of a primary standard solution containing 0.2483 g of As2O3 in 100.0 cm3 of water 12.79 cm3 of KMnO4 solution were used, whereas for titration of 15.00 cm3 of the solution containing 2.505 g Fe per liter 11.80 cm3 of the same solution of KMnO4 were used. What fraction of iron in the sample was present in the form of Fe(II)?
Model Answer
Determination of KMnO4 concentration:
5 As2O3 + 4 MnO4 – + 12 H+ + 9 H2O → 10 H3AsO4 + 4 Mn2+
M(As2O3) = 197.8 g mol–1
c(As2O3) = 0.2483 / (0.1000 × 197.8) = 0.01255 mol dm–3
c(KMnO4) = (0.01255 × 10.00 × 4) / (5 × 12.79) = 7.850×10–3 mol dm–3
Determination of Fe(II):
5 Fe2+ + MnO4 – + 8 H+ → 5 Fe3+ + Mn2+ + 4 H2O
Ar(Fe) = 55.85
c(Fe(II)) = (7.850×10–3 × 11.80 × 5 × 55.85) / 15.00 = 1.724 mg cm–3 = 1.724 g dm–3
w(Fe(II)) = (1.724 / 2.505) × 100 = 68.8 %
Tartaric acid was added to a solution containing Fe(II) and Fe(III). The solution was neutralized with aqueous ammonia and then excess KCN was added. The potential of the platinum electrode immersed in that solution was found to be +0.132 V against saturated calomel electrode.
12.4 Assuming that all iron in the last solution was present in the form of Fe(CN)6 n–, calculate the fraction of iron present in the form of Fe(II) in the original sample. Standard redox potential of Fe(CN)6 3– / Fe(CN)6 4– is +0.364 V. Potential of saturated calomel electrode is +0.241 V. The temperature of the sample solution is 25 °C.
Model Answer
From Nernst equation (at 25 °C)
E = E° + 0.059 log ([Fe(CN)6 3-] / [Fe(CN)6 4-])
E = 0.132 + 0.241 = 0.373 V; E° = 0.364 V ⇒
log ([Fe(CN)6 3-] / [Fe(CN)6 4-]) = (0.373 - 0.364) / 0.059 = 0.153 ⇒
[Fe(CN)6 3-] / [Fe(CN)6 4-] = 1.42; w(Fe(II)) = 1 / (1+1.42) × 100 % = 41.3 %
12.5 What concurrent reactions were prevented by the addition of tartaric acid and ammonia to the sample solution? Write down the net ionic equations of those reactions.
Model Answer
Adding ammonia prevents formation of HCN in acidic medium:
CN– + H+ → HCN
Adding tartaric acid leads to formation of stable Fe(III) and Fe(II) tartrate complexes and prevents:
(i) precipitation of Fe(OH)3 and, possibly, Fe(OH)2 with NH3:
Fe3+ + 3 H2O + 3 NH3 → Fe(OH)3 + 3 NH4 +
Fe2+ + 2 H2O + 2 NH3 → Fe(OH)2 + 2 NH4 +
(ii) formation of insoluble mixed Fe(II) – Fe(III) cyanide (Berlin blue, Prussian blue, Turnbull's blue):
Fe3+ + Fe2+ + K+ + 6 CN– → KFeIIFeIII(CN)6