The solubility product of silver chloride is 2.10·10–11 at 9.7 °C while 1.56·10–10 at a room tempera — Physical Chemistry — Kinetics Chemistry Question
Theoretical Problem 13
The solubility product of silver chloride is 2.10·10–11 at 9.7 °C while 1.56·10–10 at a room temperature (25 °C).
Although AgCl is practically insoluble in water, it dissolves in solutions containing complexing agents. For example, in the presence of a high excess of Cl– ions, a part of the AgCl precipitate dissolves forming [AgCl2]– ions.
The equilibrium constant of the reaction Ag+(aq) + 2 Cl–(aq) ⇌ AgCl2–(aq) is β = 2.50·105 at 25 °C.
If a substance is present in a solution in various oxidation states, it cannot be determined directly by a redox titration. In this case, the sample has to be first reduced. For this purpose, so-called reductors are used. A reductor is a column, containing a strong reducing agent in the solid phase. An acidified sample is passed through the reductor, collected, and titrated with a strong oxidizing titrant of known concentration (for example KMnO4). The most common version is the so-called Jones-reductor that contains amalgamated zinc granules.
When a milder reducing agent is required, sometimes the Ag/HCl-reductor (containing porous silver granules and aqueous HCl) is used. This might seem surprising, since Ag metal is not a good reducing agent. Considering only the standard potentials, the reduction of Fe3+ to Fe2+ by Ag is not a spontaneous reaction.
Now let us suppose that the reduction of Fe3+ with Ag is carried out in a solution that also contains 1.00 mol dm–3 HCl.
Estimate the solubility product and the solubility (in mg dm–3) of AgCl at 50 °C.
Model Answer
T1 = 282.85 K and Ksp(AgCl) = 2.10·10–11.
Therefore ∆rGº1 = –RT1 lnKsp 1 = 57.8 kJ mol-1.
T2 = 298.15 K and Ksp 2 = 1.56·10–10.
Therefore ∆rGº2 = –RT2 lnKsp 2 = 56.0 kJ mol–1.
Using ∆G = ∆H – T∆S gives ∆rSº = 119 J mol–1 K–1 and ∆rHº = 91.3 kJ mol–1, if we assume that ∆rHº and ∆rSº are independent of temperature in this limited range.
Extrapolating to 50 °C ∆rGº3 is 53.0 kJ mol-1, thus Ksp 3 = exp(–∆rGº/RT) = 2.71·10–9.
The solubility is c = √Ksp 3 = 5.2·10–5 mol dm–3, that is 7.5 mg dm–3.
Calculate the concentration of a KCl solution (at room temperature), in which the solubility of AgCl is equal to its solubility in water at 50 °C.
Model Answer
Let us suppose that the concentration of Cl– ions is relatively high at equilibrium. This means that [Ag+] is relatively low and it can be neglected in comparison with [AgCl2–].
[AgCl]total = 5.2·10–5 mol dm–3 = [AgCl2–] + [Ag+] ≈ [AgCl2–]
[AgCl]total / [Cl–]^2 ≈ [AgCl2–] / [Cl–]^2 = β * Ksp
Therefore [Cl–] = sqrt([AgCl]total / (β * Ksp)) = 1.34 mol dm–3.
[KCl]total = [Cl–] + 2 [AgCl2–] ≈ [Cl–] = 1.34 mol dm–3.
What reaction would take place if the zinc was not amalgamated?
Model Answer
Zn + 2 H+ → Zn2+ + H2
Give the reactions that take place when the following solutions are passed through a Jones-reductor:
0.01 mol dm–3 CuCl2
0.01 mol dm–3 CrCl3
0.01 mol dm–3 NH4VO3 (pH = 1)
Model Answer
Cu2+ ions:
Since Eº(Cu2+/Cu) > Eº(Cu2+/Cu+) >> Eº(Zn2+/Zn), the preferred reaction is:
Cu2+(aq) + Zn(Hg) → Cu(s) + Zn2+(aq)
Cr3+ ions:
Since Eº(Cr3+/Cr2+) > Eº(Cr3+/Cr) >> Eº(Zn2+/Zn) > Eº(Cr2+/Cr), the preferred reaction is:
2 Cr3+(aq) + Zn(Hg) → 2 Cr2+(aq) + Zn2+(aq)
VO2+ ions:
VO2+ + 2 H+ + e– → VO2+ + H2O
At pH = 1 Eº‘(VO2+/VO2+) = 1.00 V + 0.059 V lg 0.1^2 = 0.88 V.
VO2+ + 2 H+ + e– → V3+ + H2O
At pH = 1, Eº‘(VO2+/V3+) = 0.36 V + 0.059 V lg 0.1^2 = 0.24 V.
Eº(V3+/V2+) = – 0.255 V
Since all three half-reactions have a standard potential higher than the Zn2+/Zn system, vanadium reaches an oxidation number of +II. The standard potential for further reduction is lower; therefore the preferred reaction is:
2 VO2+(aq) + 3 Zn(Hg) + 8 H+(aq) → 2 V2+(aq) + 3 Zn2+(aq) + 4 H2O(l)
Estimate the equilibrium constants of these reactions using the redox potentials in the table.
Model Answer
Amalgamation supposedly does not change the zinc potential.
Cu2+(aq) + Zn(Hg) → Cu(s) + Zn2+(aq)
The number of electrons is n = 2.
Eºcell = 0.34 V – (–0.76 V) = 1.10 V.
K = e^(n F Eºcell / RT) = 1.6·10^37
2 Cr3+(aq) + Zn(Hg) → 2 Cr2+(aq) + Zn2+(aq)
The number of electrons is n = 2.
Eºcell = –0.40 V – (–0.76 V) = 0.36 V.
K = e^(n F Eºcell / RT) = 1.5·10^12
2 VO2+ + 3 Zn + 8 H+ → 2 V2+ + 3 Zn2+ + 4 H2O
The number of electrons is n = 6.
For the half reaction VO2+ + 4 H+ + 3 e– → V2+ + 2 H2O:
Eº = (1.00 V + 0.36 V - 0.255) / 3 = 0.368 V
At pH=1 Eº = 0.368 V + (4 * 0.059 V / 3) * lg 0.1 = 0.290 V
Eºcell = 0.290 V – (–0.76 V) = 1.05 V.
K = e^(n F Eºcell / RT) = 2.9·10^106
Consider a silver rod that is immersed in a 0.05 mol dm–3 Fe(NO3)3 solution. Calculate the equilibrium concentration of the various metal ions. What percentage of Fe3+ ions has been reduced?
Model Answer
The reaction that takes place is: Fe3+(aq) + Ag(s) ⇌ Fe2+(aq) + Ag+(aq)
Eºcell = 0.77 V – 0.80 V = – 0.03 V
K = e^(n F Eºcell / RT) = 0.31
If [Ag+] = [Fe2+] = x, [Fe3+] = 0.05 – x, thus:
x^2 / (0.05 - x) = 0.31
From here x = [Ag+] = [Fe2+] = 4.4·10–2 and [Fe3+] = 6·10–3. Thus 88 % of the Fe3+ ions are reduced.
What reaction takes place in this case? Calculate the equilibrium constant of the reaction.
Model Answer
The reaction taking place is:
Fe3+(aq) + Ag(s) + Cl–(aq) → Fe2+(aq) + AgCl(s)
The potential of the half reaction AgCl(s) + e– → Ag(s) + Cl–(aq) is:
Eº' = 0.80 V + 0.059 V lg Ksp = 0.22 V
Eºcell = 0.77 V – 0.22 V = 0.55 V
K = e^(n F Eºcell / RT) = 1.99·10^9
Calculate [Fe3+] at equilibrium if the initial concentration of Fe3+ was 0.05 mol dm–3.
Model Answer
If [Fe3+] = y, [Fe2+] = 0.05 – y ≈ 0.05 mol dm–3, [Cl–] = 1 – (0.5 – y) ≈ 0.95 mol dm–3 (since the equilibrium constant is relatively high).
K = [Fe2+] / ([Fe3+][Cl–]) = 0.05 / (y * 0.95) = 1.99·10^9
From here, y = [Fe3+] = 2.65·10–11
Which of the following substances are reduced in an Ag/HCl reductor?
0.01 mol dm–3 CrCl3
0.01 mol dm–3 TiOSO4 (cHCl = 1 mol dm–3)
[VISUAL]
Model Answer
Both reactions have a standard potential under 0.22 V, so the cations are not reduced.