To study in vitro gastric digestion of iron, the following procedure is carried out. Dried and homog — Physical Chemistry — Kinetics Chemistry Question
Spectrophotometric determination of iron
To study in vitro gastric digestion of iron, the following procedure is carried out. Dried and homogeneously ground supplement tablet with a mass of 0.4215 g is accurately weighed, mixed with 10 cm3 of water and adjusted to pH 2.0 with a solution of HCl (6 mol dm-3). Then 0.375 cm3 of pepsin solution (16% w/v) is added and the mixture is adjusted to 12.50 cm3 with a HCl solution (0.01 mol dm-3). This mixture was quantitatively transferred into a dialysis bag of a fixed volume, which is further immersed for 2 hours in a 20.00 cm3 solution of HCl (0.01 mol dm-3). Iron released by gastric digestion is dialyzed until the concentrations of iron inside and outside the dialysis bag are equal.
[VISUAL]
To determine the gastric digestible iron from the supplement tablet, colorimetric measurement after complex formation between ferrous ion (M) and complexing agent (L) is carried out at pH 5.0. The resulting ML3 complex exhibits light absorption at 520 nm, whereas M and L do not absorb light at this particular wavelength.
Under a certain condition that the complexed iron is in the form of ML3, consider the absorbance values obtained from the total concentration of metal (cM) and the total concentration of ligand (cL) in the following table:
[VISUAL]
In an excess of L, all of iron is in the form of ML3.
a) Calculate the molar absorptivity (Ɛ) of ML3 complex
b) Calculate the overall formation constant (Kf) of ML3 complex
Model Answer
a) At cL = 2.20×10^-2 mol dm-3, [ML3] = 6.25×10^-5 mol dm-3.
Using Beer's law: A = Ɛ * b * c
Ɛ = A / (b * [ML3]) = 0.750 / (1 cm * 6.25×10^-5 mol dm-3) = 12,000 L mol-1 cm-1
b) The reaction is: M + 3 L ⇌ ML3
Kf = [ML3] / ([M] * [L]^3)
At cL = 9.25×10^-5 mol dm-3 and cM = 3.25×10^-5 mol dm-3:
[ML3] = A / Ɛ = 0.360 / 12,000 = 3.0×10^-5 mol dm-3
[M] = cM - [ML3] = 3.25×10^-5 - 3.0×10^-5 = 0.25×10^-5 mol dm-3
[L] = cL - 3 * [ML3] = 9.25×10^-5 - 3 * (3.0×10^-5) = 0.25×10^-5 mol dm-3
Kf = (3.0×10^-5) / ((0.25×10^-5) * (0.25×10^-5)^3) = 7.68×10^17
The CHN analysis shows that the complexing agent (L) contains 80 % of C, 4.44 % of H, and 15.56 % of N. The molar mass of this compound is 180 g mol-1. Determine the molecular formula of L.
Model Answer
Calculate ratio of moles of atoms:
n(C) : n(H) : n(N) = 80 / 12.011 : 4.44 / 1.008 : 15.56 / 14.007 = 6.66 : 4.40 : 1.11 = 6 : 4 : 1
The empirical formula of L is C6H4N.
Empirical formula mass = 6 * 12.011 + 4 * 1.008 + 14.007 = 90.11 g mol-1.
Molar mass / Empirical mass = 180 / 90.11 ≈ 2.
Therefore, the molecular formula of L is C12H8N2.
The Fe2+ complexes (ML, ML2 and ML3) adopt the octahedral structure. (Assume a perfect octahedral geometry for each isomer of these three complexes). Sketch the d-orbital splitting diagram for ML3. Draw all possible isomers of Fe2+ complexes. Order the magnitudes of Δo (crystal field splitting) of these three complexes and explain.
Spectrochemical series:
I- < Br- < Cl- ≈ SCN- < F- ≈ urea < ONO- ≈ OH- < H2O < NCS- < pyridine ≈ NH3 < en < bipy < o-phen < NO2- < CN- ≈ CO
Model Answer
[VISUAL]
- d-orbital splitting diagram for ML3 shows octahedral splitting: a triply degenerate set of t2g orbitals (d_xy, d_xz, d_yz) at lower energy and a doubly degenerate set of eg orbitals (d_x^2-y^2, d_z^2) at higher energy.
- Possible isomers:
* For ML (coordinating with 1 bidentate L and 4 monodentate H2O ligands in octahedral geometry): only 1 isomer exists.
* For ML2 (coordinating with 2 bidentate L and 2 monodentate H2O ligands in octahedral geometry): cis and trans isomers can form (the cis isomer also has an optically active enantiomeric pair).
* For ML3 (coordinating with 3 bidentate L ligands in octahedral geometry): Delta (Δ) and Lambda (Λ) enantiomers.
- Magnitude of crystal field splitting Δo order: ML < ML2 < ML3.
- Explanation: According to the spectrochemical series, H2O is a weaker field ligand compared to bipyridine (or similar nitrogen-donor bidentate ligand L, like bipy or o-phen). As the number of stronger-field L ligands replacing weaker-field H2O ligands increases from ML to ML3, the overall crystal field strength increases, thereby increasing the value of Δo.
To determine the dialyzable iron concentration (iron outside the dialysis bag), 5.00 cm3 of the solution outside the dialysis bag is added with a reducing agent to ensure that all of dissolved iron is in the ferrous ion form. Then, the solution is adjusted to the suitable pH, followed by addition of excess amount of complexing agent (L) and deionized water added to make up the volume to 50.00 cm3 in a volumetric flask. The absorbance measured at 520 nm is 0.550. Calculate the concentration of dialyzable iron (in unit of mg dm-3).
Model Answer
Use Beer's law: A = Ɛ * b * c_meas
c_meas = A / (Ɛ * b) = 0.550 / (12,000 L mol-1 cm-1 * 1 cm) = 4.58×10^-5 mol dm-3
Since the sample was diluted from 5.00 cm3 to 50.00 cm3, the concentration of dialyzable iron outside the bag is:
c_dialyzable = c_meas * (50.00 / 5.00) = 4.58×10^-4 mol dm-3
To convert this to mass concentration (mg dm-3):
Mass concentration = c_dialyzable * M(Fe) * 1000 mg/g
Mass concentration = 4.58×10^-4 mol dm-3 * 55.845 g mol-1 * 1000 mg/g = 25.58 mg dm-3
Presume all of the iron in the supplement tablet is completely digestible in the gastric condition. Determine in mg the iron in 1.0000 g of supplement tablet.
Model Answer
At equilibrium, the concentration of iron inside the dialysis bag is equal to the concentration of dialyzable iron outside the bag, which is 25.58 mg dm-3.
The total volume of the digestion mixture is 12.50 cm3 (inside the bag) + 20.00 cm3 (outside the bag) = 32.50 cm3 (or 0.03250 dm3).
The total mass of digestible iron released from the supplement tablet is:
Total mass of Fe = 25.58 mg dm-3 * 0.03250 dm3 = 0.8314 mg
This amount of iron was released from 0.4215 g of the supplement tablet.
For 1.0000 g of the supplement tablet, the amount of iron is:
Iron content per 1.0000 g = 0.8314 mg * (1.0000 g / 0.4215 g) = 1.972 mg