Solubility of AgCl in Ammonia: Ksp, Kf and Complex-Ion Equilibria
Silver chloride is almost insoluble in water, yet it dissolves readily in aqueous ammonia. The reason is complex-ion formation: Ag+ is removed from solution as [Ag(NH3)2]+, so more AgCl must dissolve to restore the Ksp equilibrium. Problems of this type appear in IIT-JAM, GATE, CSIR-NET and CUET-PG physical and inorganic chemistry because they combine two equilibria in one calculation.
The two equilibria and their sum
Ag+ + 2 NH3 ⇌ [Ag(NH3)2]+ Kf
AgCl(s) + 2 NH3 ⇌ [Ag(NH3)2]+ + Cl− K = Ksp × Kf
Ksp is the solubility product. Kf is the formation (stability) constant of the complex. When two reactions are added, their equilibrium constants multiply. The overall K is the equilibrium constant for dissolution in the presence of ligand. For the calculations below I use Ksp(AgCl) = 1.8 × 10−10 and Kf = 1.6 × 107. Tables differ slightly (for example 1.7 × 107), so always use the value given in the question.
Because Kf in this form is for the overall formation from the free ion and NH3, the stepwise constants K1 and K2 multiply to give it: Kf = β2 = K1 K2.
Setting up the solubility expression
Let s be the solubility of AgCl in mol/L in an ammonia solution of initial concentration C. All the dissolved silver is either free Ag+ or complex. If Kf is large, free Ag+ is negligible, so [complex] = [Cl−] = s. Each complex uses 2 NH3, so [NH3] = C − 2s.
The square root makes the algebra linear, so no quadratic is needed.
Worked example 1: solubility in 1.0 M ammonia
Find the solubility of AgCl in 1.0 M NH3, and compare it with its solubility in pure water.
K = Ksp Kf = 1.8 × 10−10 × 1.6 × 107 = 2.88 × 10−3.
√K = 0.05367. Then s / (1.0 − 2s) = 0.05367, so s = 0.05367 − 0.10733 s, giving s (1 + 0.10733) = 0.05367 and s = 0.05367 / 1.10733 = 0.0485 mol/L.
In pure water, s = √Ksp = √(1.8 × 10−10) = 1.34 × 10−5 mol/L.
Ratio = 0.0485 / 1.34 × 10−5 = 3.6 × 103. Ammonia raises the solubility about 3600 times.
Check: [NH3] = 1.0 − 2(0.0485) = 0.903 M. K = (0.0485)2 / (0.903)2 = 2.352 × 10−3 / 0.8154 = 2.88 × 10−3. It matches.
Worked example 2: ammonia needed to dissolve a given amount
What minimum initial concentration of NH3 is needed to dissolve 0.010 mol of AgCl in 1.0 L of solution?
At the point of complete dissolution, s = 0.010 M. From s / [NH3]free = √K, [NH3]free = 0.010 / 0.05367 = 0.186 M.
The NH3 bound in the complex is 2 × 0.010 = 0.020 M. Total NH3 = 0.186 + 0.020 = 0.206 M.
Remember to add the bound ammonia, because it is the total that you must prepare.
Worked example 3: free silver ion concentration
In the 1.0 M NH3 solution of example 1, estimate [Ag+].
From Ksp = [Ag+][Cl−], with [Cl−] = 0.0485 M: [Ag+] = 1.8 × 10−10 / 0.0485 = 3.7 × 10−9 M.
Check using Kf: [Ag+] = [complex] / (Kf [NH3]2) = 0.0485 / (1.6 × 107 × 0.8154) = 0.0485 / 1.305 × 107 = 3.7 × 10−9 M.
The two routes agree. Free Ag+ is about ten million times smaller than the complex, so the assumption that all silver is in the complex was valid.
Worked example 4: will the precipitate return?
The clear solution of example 1 contains 0.0485 M Cl− and a very small free Ag+. What happens if the NH3 is removed by adding acid?
Acid converts NH3 to NH4+, so free ammonia falls, the complex breaks down (Le Chatelier), and [Ag+] rises. When Q = [Ag+][Cl−] exceeds Ksp, AgCl precipitates again. This is the classic confirmatory test in qualitative analysis: AgCl dissolves in ammonia and reappears on adding dilute nitric acid.
Related situations
- Other ligands. Thiosulfate and cyanide dissolve AgBr and AgI because their Kf values are larger. The same method applies with different numbers and a different stoichiometry.
- Amphoteric hydroxides. Al(OH)3 dissolves in excess OH− as [Al(OH)4]−, a similar combination of Ksp and Kf.
- If the complex has one ligand. The equation becomes s / (C − s) = K instead of a square root.
Common mistakes
- Forgetting the factor 2 in C − 2s because the complex binds two NH3.
- Using Ksp alone and ignoring the ligand, which gives the pure-water solubility.
- Adding K values instead of multiplying them when the reactions are added.
- Treating the initial NH3 as the equilibrium NH3. A noticeable part is tied up in the complex, as example 1 shows (0.903 M, not 1.0 M).
- Quoting [Ag+] as s in ammonia. Most of the silver is in the complex, not free.
Exam relevance
| Question type | Method |
|---|---|
| Solubility in a ligand solution | K = Ksp Kf, then s / (C − 2s) = √K |
| Ligand needed to dissolve a given amount | Free ligand from s / √K, then add bound ligand |
| Free metal ion concentration | Ksp / [anion] or [complex] / (Kf [L]n) |
| Qualitative analysis | Dissolution in NH3 and re-precipitation with acid |
Constants differ slightly between data tables and exam papers, so use the value given in the paper. For the topics and pattern of your exam in a given year, check the current official notification.
The solubility product tool helps you check the Ksp part of these calculations.
Open the Solubility Product (Ksp) Calculator →Trends and exceptions are asked every year: practise previous-year CSIR-NET questions and quizzes.
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