A student studied the chemical reactions between cations, A2+, B2+, C2+, D2+, E2+ in nitrate aqueous — Organic Chemistry Chemistry Question
Separation and Identification of Ions
A student studied the chemical reactions between cations, A2+, B2+, C2+, D2+, E2+ in nitrate aqueous solutions and anions X–, Y–, Z–, Cl–, OH– in sodium aqueous solutions as well as an organic ligand L. Some precipitation (ppt) products and colored complexes were found as shown in Table 1:
Design a flow chart for the separation of A2+, B2+, C2+, D2+, E2+ in a nitrate aqueous solution by using various aqueous solutions containing anions X–, Y–, Z–, Cl–, OH–, respectively, as testing reagents. Write down the product of the chemical reaction for each step in the flow chart.
Model Answer
1.1 For example
Design a flow chart for the separation of anions X–, Y–, Z–, Cl–, OH– in a sodium aqueous solution by using various aqueous solutions containing cations A2+, B2+, C2+, D2+, E2+, respectively, as testing reagents. Write down the product of the chemical reaction for each step in the flow chart.
Model Answer
1.2 For example
The white ppt BY2 and brown ppt CY2 have low solubilities in water with solubility products Ksp at 25 °C of 3.20×10–8 and 2.56×10–13, respectively.
a) Calculate the solubility of BY2.
b) Calculate the solubility of CY2.
Model Answer
a) BY2 = B2+ + 2 Y– Ksp = (S1)(2 S1)2 = 3.20×10–8
S1 2 S
4 S1^3 = 3.20×10–8, S1(solubility of BY2) = 2.0×10–3 mol dm–3
b) CY2 = C2+ + 2 Y– Ksp = (S2)(2S2)2 = 2.56×10–13
S2 2 S2
4 S2^3 = 2.56×10–13, S2(solubility of CY2) = 4.0×10–5 mol dm–3
A series of solutions containing B2+ and L were prepared in 50 cm3 volumetric flasks by adding 2 cm3 of solution of B2+ (8.2×10–3 mol dm–3) to each flask. Varying amounts of a solution of the ligand L (c = 1.0×10–2 mol dm–3) are added to each flask. The solution in each volumetric flask was diluted with water to the mark (50 cm3). The absorbance (A) of complex BLn was measured at 540 nm for each solution in a 1.0 cm cell. The data are summarized in Table 2. (Both B2+ and ligand L show no absorption (A = 0) at 540 nm.) [Mole Ratio Method]
a) Calculate the value of n (coordination number) in the complex BLn2+.
b) Calculate the formation constant (Kf) of complex BLn2+.
Table 2
Model Answer
a) Plot of absorbance (A) vs volume (VL) of L added as follows:
From the volume of L at break point B (all B2+ ions form complex with L) in the plot, n can be calculated:
n/1 = n(L) / n(B2+) = (0.0051 dm3 × 1.0×10–2 mol dm–3) / (0.0020 dm3 × 8.2×10–3 mol dm–3) ≅ 3
It means that B2+ forms BL3^2+ complex with L.
b) (1) Calculation of molar absorption coefficient ε.
At break point, A = 0.66 = ε × 1 × c(BL3^2+)
And ε = 0.66 / (2.0 cm3 × 8.2×10–3 / 50 cm3) = 2.01×103
(2) Choose a point in the curve of the plot, for example:
At point P (2.0 cm3 of L added): A = 0.26
A = 0.26 = ε × 1 × c(BL3^2+)
c(BL3^2+) = 0.26 / ε = 0.26 / (2.01×103) = 1.29×10–4 mol dm–3
c(B2+) = (2.0 cm3 × 8.2×10–3 – 50 cm3 × 1.29×10–4 mol dm–3) / 50 cm3
c(B2+) = 1.99×10–4 mol dm–3
c(L) = (2.0 cm3 × 1.0×10–2 – 3 × 50 cm3 ×1.29 ×10–4 mol dm–3) / 50 cm3
c(L) = 1.3 × 10–5 mol dm–3
[Calculation of formation constant]
So Kf = [BL3^2+] / ([B2+][ L ]3) = (1.29×10–4) / ((1.99×10–4)(1.3×10–5)3)
Kf = 2.95×1014
Solid NaY (soluble) was added very slowly to an aqueous solution with a concentration of B2+ and C2+ equal to 0.10 mol dm–3 and 0.05 mol dm–3, respectively, prepared from their respective nitrate aqueous salts.
a) Which cation (B2+ or C2+) precipitates first? What is the [Y–] when this happens? (Ksp = 3.20×10–8 for BY2 and Ksp = 2.56×10–13 for CY2, at 25 oC.) [Separation by Precipitation]
b) What are the concentrations of Y– and the remaining cation when complete precipitation of the first precipitating cation has occurred (assume that the concentration of the first cation in solution after complete precipitation is ≤ 10–6 mol dm–3)? Is it possible to separate B2+ and C2+ by the precipitation method with Y– ion as a precipitating agent?
Model Answer
a) For CY2: Ksp = [C2+] [Y–]2 = 2.56×10–13
[Y–] = ((2.56×10–13) / 0.05)1/2 = 7.16×10–6 when CY2 begins to form
For BY2: Ksp = [B2+] [Y–]2 = 3.20×10–8
[Y–] = ((3.20×10–8) / 0.05)1/2 = 5.66×10–4 when BY2 begins to form
CY2 forms first
b) The precipitation of C2+ as CY2 considered to be completed at [C2+] = 1×10–6
Thus Ksp = [C2+] [Y–]2 = 2.56 × 10–13 and [Y–] = ((2.56 × 10–13) / 10–6)1/2 = 5.06×10–4
It means that [Y–] = 5.06×10–4 when CY2 precipitates completely.
When [Y–] = 5.06×10–4 for BY2 :
[B2+] [Y–]2 = (0.1) (5.06×10–4)2 = 2.56×10–8 < Ksp of BY2 (3.20×10–8)
BY2 (ppt) can not be formed at [Y–] = 5.06×10–4 and [B2+] = 0.1 when CY2 precipitates completely.
It means that it is possible to separate B2+ and C2+ ions by precipitation method with Y– as a precipitating agent.