← Equilibrium principles in aqueous systems
ℹ
Common ion (adding an ion already in the equilibrium, e.g. Cl⁻ to AgCl): increases that ion, so less dissolves — solubility drops.
Complex ion (adding a ligand that grabs the metal, e.g. NH₃ to AgCl): decreases [Ag⁺], so more dissolves — solubility rises.
2022 asked about both in one question. Read which ion is being added and whether it is in the equilibrium or reacts with something in it.
8 Dissolving a precipitate with a complex ion
A precipitate that will not dissolve in water often dissolves when a ligand is added, because the metal ion is pulled out of the solubility equilibrium into a soluble complex ion. This came up in 2022, 2024 and 2025, and it is always worth Excellence.The chain of reasoning (this is the mark scheme)
- The ligand reacts with the metal ion to form a soluble complex ion — you must write the balanced equation for this.
- That removes Mn+ from the solubility equilibrium, so [Mn+] decreases.
- The rate of the forward (dissolving) reaction increases to replace the Mn+.
- So more solid dissolves. With excess ligand, all of it can dissolve.
Write it as a RATE, not as “the equilibrium shifts”
Across five years of marking schedules the phrase “Le Châtelier” never appears, and neither does “shifts”. What is marked is a rate sentence, almost word for word: “This decreases [Ni²⁺] in the system, so the rate of the dissolving reaction increases to replace the Ni²⁺ ions.” Write it that way and you cannot be marked down; “the equilibrium shifts right” on its own may not earn the mark.| Ligand added | Complex ion formed |
|---|---|
| NH₃ with Ag⁺ | [Ag(NH₃)₂]⁺ |
| NH₃ with Cu²⁺ | [Cu(NH₃)₄]²⁺ |
| NH₃ with Zn²⁺ | [Zn(NH₃)₄]²⁺ |
| NH₃ with Ni²⁺ | [Ni(NH₃)₆]²⁺ |
| CN⁻ with Ag⁺ | [Ag(CN)₂]⁻ |
| CN⁻ with Ni²⁺ | [Ni(CN)₄]²⁻ |
| OH⁻ (excess) with Al³⁺ | [Al(OH)₄]⁻ |
| OH⁻ (excess) with Zn²⁺ | [Zn(OH)₄]²⁻ |
| OH⁻ (excess) with Pb²⁺ | [Pb(OH)₄]²⁻ |
You are given these
The Resource Booklet supplies this list with formulae, so you do not have to memorise them. What you do need is to recognise that adding NH₃, CN⁻ or excess OH⁻ to a metal ion is a complex-ion question, and to get the coefficients right in the equation. Worked example Ni(OH)₂ dissolving in excess cyanide (2025) Solid Ni(OH)₂ is in equilibrium with its saturated solution. NaCN solution is added and the precipitate dissolves. Explain, using equilibrium principles. The solubility equilibrium is: Ni(OH)₂(s) ⇌ Ni²⁺(aq) + 2OH⁻(aq). The added CN⁻ forms a complex ion with the nickel: Ni²⁺(aq) + 4CN⁻(aq) → [Ni(CN)₄]²⁻(aq). This decreases [Ni²⁺] in the system. So the rate of the dissolving (forward) reaction increases to replace the Ni²⁺ ions. Because excess cyanide is added and the complex ion is water-soluble, eventually all of the Ni(OH)₂ dissolves. ✓ Complex-ion formation removes Ni²⁺ from the equilibrium, so the dissolving reaction speeds up to replace it and the precipitate fully dissolves. The balanced complex-ion equation is required for the mark. ✕Common-ion effect vs complex-ion formation — opposite directions
Both are “add something to a saturated solution”, and they do opposite things.Common ion (adding an ion already in the equilibrium, e.g. Cl⁻ to AgCl): increases that ion, so less dissolves — solubility drops.
Complex ion (adding a ligand that grabs the metal, e.g. NH₃ to AgCl): decreases [Ag⁺], so more dissolves — solubility rises.
2022 asked about both in one question. Read which ion is being added and whether it is in the equilibrium or reacts with something in it.