The solubility of metals and their salts played an important role in Earth's history changing the sh — Physical Chemistry — Thermodynamics Chemistry Question
Solubility of salts
The solubility of metals and their salts played an important role in Earth's history changing the shape of the Earth's surface. Furthermore, solubility was instrumental in changing the Earth's atmosphere. The atmosphere of the primitive Earth was rich in carbon dioxide. Surface temperature of the early Earth was maintained above the boiling point of water due to continued bombardment by asteroids. When the Earth cooled, it rained and a primitive ocean was formed. As metals and their salts dissolved the ocean became alkaline and a large amount of carbon dioxide from the air dissolved in the ocean. The CO2 part of most carbonate minerals is derived from this primitive atmosphere.
As life arose about 3.8 billion years ago and photosynthetic bacteria evolved about 3 billion years ago, molecular oxygen was produced as a by–product of photosynthesis. As oxygen reacted with the metal ions in the ocean, metal oxides with low solubility were deposited on the ocean floor which later became dry land through plate tectonic motion. Iron and aluminum ores were, and still are, of particular importance as raw materials in human civilization.
Let's consider a solubility problem using silver halides. Ksp values for AgCl and AgBr are 1.8×10–10 and 3.3×10–13, respectively.
Excess AgCl was added to deionized water. Calculate the concentration of Cl– in equilibrium with solid AgCl. Repeat the calculation for Br– assuming that AgBr was added instead of AgCl.
Model Answer
AgCl(s) → Ag+(aq) + Cl–(aq)
Ksp = [Ag+][Cl–] = x^2 = 1.8×10^–10 ⇒ [Ag+] = [Cl–] = 1.34×10^–5 mol dm^-3
AgBr(s) → Ag+(aq) + Br–(aq)
Ksp = [Ag+][Br–] = x^2 = 3.3×10^–13 ⇒ [Ag+] = [Br–] = 5.74×10^–7 mol dm^-3
Assume that 0.100 dm3 of Ag+ solution (c = 1.00×10–3 mol dm–3) is added to a Cl– solution of the same volume and concentration. What is the concentration of Cl– in the solution once equilibrium has been established? What is percentage of the total chloride in solution?
Model Answer
In this hypothetical case, [Ag+] = [Cl–] = 1.34×10^–5 mol dm^-3 just as in 7.1.
[VISUAL]
Cl–(aq) / Cl(total) = Cl–(aq) / (Cl–(aq) + AgCl(s)) = (1.34×10^–5 mol dm^–3 × 0.200 dm^3) / 1.00×10^–4 mol = 0.027 = 2.7 %
Assume that 0.100 dm3 of Ag+ solution (c = 1.00×10–3 mol dm–3) is added to a Br– solution of the same volume and concentration. What is the concentration of Br– in the solution once equilibrium has been established? What is percentage of the total bromide solution?
Model Answer
Similarly, [Ag+] = [Br–] = 5.7×10^–7 mol dm^-3 just as in 7.1.
[VISUAL]
Br–(aq) / Br(total) = Br–(aq) / (Br–(aq) + AgBr(s)) = (5.7×10^–7 mol dm^–3 × 0.200 dm^3) / 1.00×10^–4 mol = 1.1×10^–3 = 0.11 %
Experimental verification of the answers in 7.2 and 7.3 is difficult, because the exact volume and concentration of the solutions are unknown. Repeat the calculations in 7.2 and 7.3 assuming that the concentration of the Ag+ solution is 1.01×10–3 mol dm–3.
Model Answer
Assume that 1.00×10–4 mol of AgCl is precipitated, and 1.00×10–6 mol of Ag+ ions remains in solution. Then a portion of AgCl dissolves.
[Ag+] = 5.0×10–6 + x , [Cl – ] = x
Ksp = [Ag+][Cl – ] = (5.0×10–6 + x) x = 1.8×10–10
⇒ [Cl – ] = 1.1×10–5 (slightly decreased)
[Ag+] = 1.6×10–5 (slightly increased)
[VISUAL]
Cl–(aq) / Cl(total) = Cl–(aq) / (Cl–(aq) + AgCl(s)) = (1.1×10^–5 mol dm^–3 × 0.200 dm^3) / 1.00×10^–4 mol = 0.022 = 2.2 %
Similarly,
[Ag+] = 5.0×10–6 + x , [Br–] = x
Ksp = [Ag+][Br–] = (5.0×10–6 + x) x = 3.3×10–13
x < 5.0×10–6; therefore, (5.0×10–6) x = 3.3×10–13
⇒ [Br–] = 6.6×10–8 significant decrease from 5.7×10–7
[Ag+] = 5.1×10–6 significant increase from 5.7×10–7
[VISUAL]
Br–(aq) / Br(total) = Br–(aq) / (Br–(aq) + AgBr(s)) = (6.5×10^–8 mol dm^–3 × 0.200 dm^3) / 1.00×10^–4 mol = 1.3×10^–4 = 0.013 %
Now let's assume that Ag+ solution (c = 1.00×10–3 mol dm–3) is slowly added with constant stirring to a 0.100 dm3 solution containing both Cl– and Br– at a concentration of 1.00×10–3 mol dm–3.
Which silver halide will precipitate first? Describe the situation when the first precipitate appears.
Model Answer
AgBr will precipitate first. Theoretically, AgBr will begin to precipitate when the Ag+ concentration reaches 3.3×10–10 mol dm –3. At this concentration of Ag+, AgCl will not precipitate.
AgBr: [Ag+] = Ksp / [Br-] = 3.3×10^-13 / 1.00×10^-3 = 3.3×10^–10 mol dm^-3
This corresponds to 3.3×10–8 dm3 of the Ag+ solution, which is much less than the smallest volume one can deliver with a micropipet.
Determine the percentage of Cl–, Br– and Ag+ ions in the solution and in the precipitate after addition of 100, 200, and 300 cm3 of Ag+ solution.
[VISUAL]
Model Answer
This problem can be solved using the mass conservation relations. However, the solution can be simplified as shown below.
A = total amount of Ag = [Ag+]0 Vadd = (1.00×10–3 mol dm–3) Vadd
B = total amount of Br– = [Br–]0 Vorignal = (1.00×10–3 mol dm–3) (0.100 dm3) = 1.00×10–4 mol
C = total amount of Cl = [Cl–]0 Vorignal = 1.00×10–3 mol dm–3 × 0.100 dm3 = 1.00×10–4 mol
A = [Ag+] Vtot + nAgCl(s) + nAgBr(s) (1)
B = [Br–] Vtot + nAgBr(s) (2)
C = [Cl–] Vtot + nAgCl(s) (3)
Ksp(AgBr) = [Ag+][Br–] (4)
Ksp(AgCl) = [Ag+][Cl–] (5)
- Vadd = 100 cm3,
Vtot = 200 cm3 (total n(Ag) = 1.00×10–4 mol). Assume that all Ag+ are used to precipitate Br– as AgBr(s).
[Ag+] = [Br–] = 0, [Cl–] = 5.0×10–4 n(AgBr) = 1.00×10–4 mol, n(AgCl) = 0
At equilibrium:
[Ag+] = Ksp(AgCl) / [Cl–] = 3.6×10–7
[Br–] = Ksp(AgBr) / [Ag+] = 9.2×10–7
Total Ag = Ag+(aq) + AgBr + AgCl, total Br = Br– (aq) + AgBr
Since total Ag = total Br, Ag+(aq) + AgCl = Br–(aq)
n(AgCl) = ([Br–] – [Ag+])Vtot = [(9.2 – 3.6)×10–7 mol dm–3] × 0.200 dm3 = 1.1×10–7 mol (0.11 % of the total Cl)
[Cl–] = 5.0×10–4 (still valid, because very little AgCl is formed) n(AgBr) = 1.00×10–4 mol (still valid, because [Br–] is small)
- Vadd = 200 cm3,
Vtot = 300 cm3 (total n(Ag) = 2.00×10–4 mol)
Assume complete precipitation of Br– and Cl– with Ag+
[Ag+] = [Br–] = [Cl–] = 0, n(AgBr) = 1.0×10–4 mol, n(AgCl) = 1.0×10–4 mol
At equilibrium:
[Ag+] = [Br–] + [Cl–] = Ksp(AgBr)/[Ag+] + Ksp(AgCl)/[Ag+]
[Ag+] = 1.3×10–5
[Br–] = Ksp(AgBr) / [Ag+] = 2.5×10–8
[Cl–] = Ksp(AgCl) / [Ag+] = 1.3×10–5
n(AgBr) = 1.00×10–4 mol – c(Br–) Vtot = 1.00×10–4 mol
n(AgCl) = 1.00×10–4 mol – c(Cl–) Vtot = 9.6×10–5 mol
- Vadd = 300 cm3
Vtot = 400 cm3 (total n(Ag) = 3.00×10–4 mol)
Assume complete precipitation of Br– and Cl– with Ag+.
[Ag+] = 2.5×10–4 , [Br–] = [Cl–] = 0, n(AgBr) = 1.0×10–4 mol, n(AgCl) = 1.0×10–4 mol
[Br–] = Ksp(AgBr) / [Ag+] = 1.3×10–9
[Cl–] = Ksp(AgCl) / [Ag+] = 7.2×10–7
n(AgBr) = 1.00×10–4 mol – c(Br–) Vtot = 1.00×10–4 mol
n(AgCl) = 1.00×10–4 mol – c(Cl–) Vtot = 9.97×10–5 mol
Summary Table:
[VISUAL]
| Vadd | % Br in soln | % Br in ppt | % Cl in soln | % Cl in ppt | % Ag in soln | % Ag in ppt |
|------|--------------|-------------|--------------|-------------|----------- ---|-------------|
| 100 cm3 | 0.18 | 99.8 | 99.9 | 0.11 | 0.07 | 99.9 |
| 200 cm3 | 0.007 | 100 | 4.0 | 96.0 | 2.0 | 98.0 |
| 300 cm3 | 0.0005 | 100 | 0.3 | 99.7 | 33.3 | 66.7 |