Ethylenediamine tetraacetic acid (EDTA) is used as a reagent to titrate the metal ions in the comple — Organic Chemistry Chemistry Question
Complex compounds
Ethylenediamine tetraacetic acid (EDTA) is used as a reagent to titrate the metal ions in the complexometric titration. EDTA is a tetraprotic acid, abbreviated as H4Y, with the structure:
As EDTA is sparingly soluble in water, a more soluble sodium form, Na2H2Y, is usually used and H2Y 2− is commonly known as EDTA. EDTA forms strong 1:1 complexes with most metal ions Mn+.
I.
8.1 How many atoms of an EDTA molecule are capable of binding with the metal ion upon complexation?
Check in the appropriate box.
[ ] 2
[ ] 4
[ ] 6
[ ] 8
Model Answer
6
8.2 Draw the structure of the complex of a metal ion M2+ with EDTA.
II. Complexation reaction between Y4− form of EDTA and metal ion Mn+ has a large formation constant (stability constant) β:
Mn+ + Y4− ⇌ MY(4−n)−
β = [MY(4-n-] / ([Mn+][Y4-])
Besides complexation reaction between Y4− form of EDTA and metal ion Mn+, other processes in the solution also develop such as formation of hydroxo complexes of the metal ion, acid-base equilibrium of H2Y 2−… To account for such processes conditional formation constant β’ is used for the calculations. β’ is determined from β as the following expression:
β’ = β × α(Y4-) × α(Mn+)
where: α(Y4-) and α(Mn+) are fractions of Y4– (α(Y4-) = [Y4-]/[Y]') and free metal ion Mn+ (α(Mn+) = [Mn+]/[M]'), with [Y]’ and [M]’ being the total concentrations of all forms of Y4− and Mn+, excluding MY(4−n)−. Given that: H4Y has pKa1 = 2.00; pKa2 = 2.67; pKa3 = 6.16; and pKa4 = 10.26 (pKa values for H5Y+ and H6Y2+ are ignored).
pKs(Mg(OH)2) = 10.95; log β(MgY2-) = 8.69
Mg2+ + H2O ⇌ MgOH+ + H+ β = 1.58×10-13; (pKa = – logKa; pKs = – logKs)
In a typical experiment, 1.00 cm3 of MgCl2 solution (c = 0.10 mol dm-3) and 1.00 cm3 of Na2H2Y solution (c = 0.10 mol dm-3) are mixed together. pH of the resulting solution is adjusted to 10.26 by an NH3/NH4+ buffered solution.
8.3 Calculate conditional formation constant (β’) of the MgY2− complex at pH = 10.26 given that acid-base equilibrium of H2Y 2− and formation of mononuclear hydroxo complex of Mg2+ occur in the solution.
Model Answer
Let [H+] be h and at pH = 10.26:
α(Y4-) = K_a4 / (h + K_a4) = 10^-10.26 / (10^-10.26 + 10^-10.26) = 0.5
α(Mg2+) = 1 / (1 + β_MgOH h^-1) = 1 / (1 + 1.58×10^-13 × 10^10.26) ≈ 1
β'(MgY2-) = β × α(Mg2+) × α(Y4-) = 10^8.69 × 1 × 0.5 = 2.45 × 10^8
8.4 Does the Mg(OH)2 precipitate in this experiment? Check in the appropriate box.
[ ] Precipitation
[ ] No precipitation
Model Answer
At pH = 10.26:
[Mg2+]' = 0.05 M, [Y4-]' = 0.05 M.
[Mg2+] = 1.43×10^-5 M.
[Mg2+][OH-]^2 = 1.43×10^-5 (10^-3.74)^2 = 10^-12.32 < Ks(Mg(OH)2) = 10^-10.95. Hence no Mg(OH)2 precipitate appears.
Correct box: No precipitation
III. In order to titrate metal ions by EDTA, the conditional formation constant (β’) of the complex metal – EDTA (MY(4−n)−) must be large enough, usually β’ ≥ 1.00 ⋅ 10^8 – 1.00 ⋅ 10^9. To determine the concentrations of Mn2+ and Hg2+ in an analytical sample, two experiments are carried out.
Experiment 1: Add 25.00 cm3 of 0.040 mol dm-3 EDTA solution to 20.00 cm3 of the analytical solution. Adjust the pH of the resulting solution to 10.50. Titrate the excess EDTA with a suitable indicator; 12.00 cm3 of 0.025 mol dm-3 Mg2+ solution is consumed.
Experiment 2: Dissolve 1.400 g of KCN in 20.00 cm3 of the analytical solution (assuming that the volume is unchanged upon dissolution) and then add 25.00 cm3 of 0.040 mol dm-3 EDTA solution. Titrate the excess EDTA in the resulting mixture at the pH of 10.50; 20.00 cm3 of 0.025 mol dm-3 Mg2+ solution is consumed.
8.5 Prove that: in the experiment 2, Hg2+ cannot be determined by titration with EDTA in the presence of KCN in solution (or Hg2+ is masked in the complex form of Hg(CN)4 2-).
Model Answer
[CN-] = (1.400 / 65) × (1000 / 20.00) = 1.00 M.
α(Hg2+) = 1 / (1 + β_Hg(CN)4 [CN-]^4) = 1 / (1 + 10^38.97 × 1^4) ≈ 10^-38.97
α(Y4-) at pH 10.50 = K_a4 / (h + K_a4) = 10^-10.26 / (10^-10.50 + 10^-10.26) = 0.635
β'(HgY2-) = β_HgY2- × α(Hg2+) × α(Y4-) = 10^21.80 × 10^-38.97 × 0.635 = 4.29 × 10^-18.
Since β'(HgY2-) is very small, Hg2+ cannot be titrated in experiment 2.
8.6 Write down chemical equations for the reactions in the two experiments and calculate molar concentrations of Mn2+ and Hg2+ in the analytical solution. Given that:
log β(Hg(CN)4 2-) = 38.97; log β(HgY 2-) = 21.80; pKa(HCN) = 9.35
(Other processes of Hg2+ are ignored; the pKa values of H4Y are provided in question 2).
Model Answer
Chemical equations:
Experiment 1: Hg2+ + Y4- → HgY2-
Mn2+ + Y4- → MnY2-
Mg2+ + Y4-(excess) → MgY2-
(c(Mn2+) + c(Hg2+)) × 20.00 = 25.00 × 0.040 - 12.00 × 0.025 (1)
Experiment 2: Hg2+ + 4 CN- → Hg(CN)4 2-
Mn2+ + Y4- → MnY2-
Mg2+ + Y4-(excess) → MgY2-
c(Mn2+) × 20.00 = 25.00 × 0.040 - 20.00 × 0.025 (2)
According to (1) and (2): c(Mn2+) = 0.025 mol dm-3; c(Hg2+) = 0.010 mol dm-3.
IV. In the titration of polyprotic acids or bases, if the ratios of consecutive dissociation constants exceed 1.00 ⋅ 10^4, multiple titrations are possible with an error less than 1%. To ensure the allowed error, only acids or bases with equilibrium constants larger than 1.00 ⋅ 10^-9 can be titrated. To find the end-point, pH range of the indicator must be close to that of the equivalence point (pHEP); the point at which the stoichiometric amounts of analyte and titrant has reacted. Titrate 10.00 cm3 of 0.25 mol dm-3 Na2H2Y solution by 0.20 mol dm-3 NaOH solution in a typical experiment.
8.7 Write down the chemical equation for the titration reaction.
Model Answer
As Ka3 / Ka4 > 10^4 and Ka4 < 10^-9 only one endpoint can be determined for the titration of H2Y2−:
Titration reaction: H2Y2− + OH− → HY3− + H2O
8.8 Determine the value of pHEP.
Model Answer
pHEP = pH(HY3−) = (pKa3 + pKa4) / 2 = 8.21
8.9 Choose the most suitable indicator (check in the appropriate box) for the above titration from the following: bromothymol blue (pH = 7.60); phenol red (pH = 8.20); phenolphtalein (pH = 9.00).
[ ] Bromothymol blue
[ ] Phenol red
[ ] Phenolphtalein
Model Answer
pHEP = pH(phenol red), hence the most suitable indicator is phenol red.
8.10 Titration error q defined as the difference between the titrant amount added and the titrant amount needed to reach the equivalence point is expressed as:
q = ((c_NaOH V_1 - c_NaOH V_2) / (c_NaOH V_2)) × 100% = ((V_1 - V_2) / V_2) × 100%
where c_NaOH is the NaOH concentration; V1: end-point volume of NaOH; V2: equivalence point volume of NaOH.
Calculate the consumed volume of NaOH solution and the titration error if the final pH is 7.60.
Model Answer
If the final pH is 7.60 the percentage of H2Y2− that is titrated:
% = [HY3-] / ([H2Y2-] + [HY3-]) × 100 = 100 × K_a3 / ([H+] + K_a3) = 100 × 10^-6.16 / (10^-7.60 + 10^-6.16) = 96.5%
The volume of NaOH solution needed to reach pH of 7.60 is:
V_NaOH = V_1 = (0.25 × 10 × 0.965) / 0.2 = 12.06 (cm3)
V_EP = V_2 = (0.25 × 10) / 0.20 = 12.50 (cm3)
q = (12.06 - 12.50) / 12.50 × 100% ≈ -3.5%
(As 96.5% of H2Y2- is titrated, 3.5% of the analyte has not been titrated, or the error q = - 3.5%.)