Two important factors that affect the solubility of a sparingly soluble salt are pH and the presence — Analytical Chemistry Chemistry Question
Solubility of sparingly soluble salts
Two important factors that affect the solubility of a sparingly soluble salt are pH and the presence of a complexing agent. Silver oxalate is one such salt, which has low solubility in water (2.06×10 –4 at pH = 7.0). Its solubility is affected by pH as the anion oxalate reacts with hydronium ions, and also by a complexing agent such as ammonia as the cation silver forms complexes with ammonia.
Calculate the solubility of silver oxalate in acidified water with pH = 5.0. The first and second dissociation constants for oxalic acid are 5.6×10 –2 and 6.2×10 –5 , respectively.
Model Answer
Ag2C2O4(s) = 2 Ag+ + C2O4 2-
The solubility product Ksp is given by
Ksp = [Ag+] 2 [C2O4 2-]
If S is the solubility of Ag2C2O4
[Ag+] = 2S (1)
The total oxalate concentration, denoted by cox, is
cox = S = [C2O4 2–] + [H2C2O4] (2)
The dissociation reactions are:
H2C2O4 H+ + HC2O4 – K1 = 5.6×10–2 (3)
HC2O4 – H+ + C2O4 2– K2 = 6.2×10–5 (4)
Eqs. (2), (3) and (4) give
cox = S = [C2O4 2–] + [H+][C2O4 2-]/K2 + [H+]2[C2O4 2-]/K1K2
where 1/ = 1 + [H+]/K2 + [H+]2/K1K2 (5)
At pH = 7, [H+] =10^-7 and 1
Ksp = 4 S3 = 3.5×10^-11
At pH = 5.0, [H+] = 1×10^-5
From the values of K1, K2 and [H+] we get
= 0.861 (6)
Ksp = [2S]2 [S]
S = 2.17 × 10^-4
In the presence of ammonia in aqueous solution, silver ion forms two complexes Ag(NH3)+ and Ag(NH3)2+. The values of the stepwise stability constants for the formation of these complexes are 1.59×103 and 6.76×103. What is the solubility of silver oxalate in an aqueous solution in which the concentration of NH3 is 0.02 mol dm–3 and its pH = 10.8 ?
Model Answer
Eq. (5) implies = 1
i. e. cox = S = [C2O4 -2] (7)
The total silver ion in the solution is given by
cAg = 2 S = [Ag+] + [AgNH3+] + [Ag(NH3)2+] (8)
The complex formation reactions are:
Ag+ + NH3 = AgNH3+ K3 = 1.59×103 (9)
AgNH3+ + NH3 = Ag(NH3)2+ K4 = 6.76×103 (10)
From eqs. (8), (9) and (10)
cAg = 2 S = [Ag+] {1 + K3[NH3] + K3K4[NH3]2}
[Ag+] = × cAg = × 2S
where = 1 / (1 + K3[NH3] + K3K4[NH3]2)
Using the values of K3, K4 and [NH3],
= 2.31×10-4
Ksp = [Ag+]2 [C2O4 2-] = [× 2 S]2 [S]
S = 5.47×10-2