Chrome Green pigment is obtained by mixing lead(II) chromate and iron(II) hexacyanoferrate(III). A t — Physical Chemistry — Thermodynamics Chemistry Question
Analysis of Chrome Green
Chrome Green pigment is obtained by mixing lead(II) chromate and iron(II) hexacyanoferrate(III). A titrimetric method of Chrome Green analysis involves the following steps: an accurate weight of the pigment sample is treated with sodium carbonate solution while heating and then filtered.
Write down the reactions occurring on treatment of Chrome Green with carbonate. What is left on the filter?
Model Answer
PbCrO4 + Na2CO3 → Na2Pb(OH)4 + Na2CrO4 + NaHCO3
Fe4[Fe(CN)6]3 + 12 Na2CO3 + 12 H2O → 4 Fe(OH)3↓ + 3 Na4[Fe(CN)6] + 12 NaHCO3
Fe(III) hydroxide is left on the filter.
To determine chromate, the iodometric method is used. An excess of KI is added to the acidified solution, and the released iodine is titrated with the standardized Na2S2O3 solution in the presence of starch.
Write down the reactions occurring when chromate is determined by this method. Why is it not recommended to titrate dichromate directly with thiosulfate?
Model Answer
Direct oxidation of thiosulfate with dichromate is not stoichiometric. The reactions normally used are:
Cr2O7 2– + 6 I– + 14 H+ → 2 Cr3+ + 3 I2 + 7 H2O
I2 + 2 S2O3 2– → 2 I– + S4O6 2–
Na2S2O3 solution should be standardized before using it as the titrant. The standardization is carried out against a standard K2Cr2O7 solution in the same way as described above for the determination of chromate. If the acidity of the solution significantly exceeds 0.4 mol dm-3, the reaction between dichromate and iodide induces the oxidation of iodide with atmospheric oxygen.
Propose a scheme for such an induced process. How would it affect the results of thiosulfate determination?
Model Answer
If reaction B is induced by reaction A, it implies that reaction A produces some intermediates active with the components of reaction B. In our case, the reduction of Cr(VI) occurs via the formation of intermediate oxidation states of chromium, predominantly Cr(V) species. (At the same time, the oxidation of I– to I0 may not require any iodine-containing intermediates.) A reasonable reaction scheme is as follows:
H2Cr2O7 + I– → Cr(V) + I;
Cr(V) + O2 → Cr2O7 2–, etc.
As a result of oxygen involvement, a higher amount of free iodine is obtained, which results in a greater amount of Na2S2O3 titrant spent and lower apparent concentration determined.
One aliquot of the filtered sample of Chrome Green solution (10.00 cm3 out of the total volume of 50.0 cm3) was used for the iodometric determination of chromate following the procedure described above (5.01 cm3 of Na2S2O3 solution (c = 0.0485 mol dm-3) were spent).
Calculate the amount of lead chromate in the sample (mg PbCrO4).
Model Answer
The amount of chromium is found as follows: 3 nCr = nthios = 0.0485 mol dm-3 × 5.01 cm3 = 0.2430 mmol (nCr = 0.0810 mmol). This corresponds to 26.2 mg of PbCrO4 (M = 323.2 g mol–1 ) in the aliquot, or 262 mg totally.
A reaction of chromium(VI) with [Fe(CN)6] 4– might occur upon adding the acid.
Estimate whether any analytical errors might be caused by this side reaction.
Model Answer
The side reaction
CrO4 2– + 3 [Fe(CN)6] 4– + 8 H+ → Cr3+ + 3 [Fe(CN)6] 3– + 4 H2O
produces an amount of [Fe(CN)6] 3– equivalent to CrO4 2– reacted. At the titration stage that hexacyanoferrate(III) would also liberate free iodine; hence, the side process can be neglected.
Another aliquot of the filtered solution (10.00 cm3 out of the total volume of 50.0 cm3) were mixed with 10.00 cm3 of K4Fe(CN)6 solution (c = 0.0300 mol dm-3) acidified with H2SO4 to obtain [H+] ≅ 1 mol dm-3 and titrated with KMnO4 solution (c = 0.00500 mol dm-3 ) (2.85 cm3 were spent).
What reaction did occur upon acidification of the sample? Write down the reaction of titration with permanganate.
Model Answer
Acidification of the sample:
CrO4 2– + 3 [Fe(CN)6] 4– + 8 H+ → Cr3+ + 3 [Fe(CN)6] 3– + 4 H2O
Titration:
MnO4 – + 5 [Fe(CN)6] 4– + 8 H+ → Mn2+ + 5 [Fe(CN)6] 3– + 4 H2O
Calculate the amount of Turnbull's Blue in the sample (mg Fe3[Fe(CN)6]2).
Model Answer
On acidification of the 2nd aliquot, chromium is reduced by [Fe(CN)6] 4–. Then permanganate is spent for the oxidation of [Fe(CN)6] 4–, namely, the amount of [Fe(CN)6] 4– added plus the amount contained initially in the sample less the amount spent for the reduction of Cr(VI):
5 n(MnO4 -) = n(Fe added) + n(Fe from sample) – 3 n(Cr).
From this equation one can find n(Fe from sample):
n(Fe from sample) = 5 n(MnO4 -) – n(Fe added) + 3 n(Cr)
= 5 × 0.00500 × 2.85 – 10 × 0.0300 + 0.2430 = 0.07125 – 0.3000 + 0.2430 = 0.0155 mmol.
This corresponds to 4.44 mg of Fe4/3[Fe(CN)6] (M = 286.3 g mol–1) in the aliquot, or 22.2 mg totally.