Difficulty: Introductory to Intermediate | Prerequisites: Ionic compound formulas, strong electrolyte dissociation, balancing equations.
Precipitation reactions are the first major class of reactions in solution that you study in detail. They combine your knowledge of ionic compounds with solubility rules to predict whether mixing two solutions will produce a solid. The skill of writing net ionic equations strips away the spectator ions so you can see the chemistry that is actually happening. If you can write ionic formulas and balance simple equations, you are set.
When two aqueous ionic compounds are mixed, the cation of one may combine with the anion of the other to form an insoluble solid (a precipitate). You predict this using solubility rules, then write the balanced molecular equation, the complete ionic equation, and finally the net ionic equation, which shows only the ions that form the precipitate.
Precipitate
An insoluble solid that forms when two aqueous solutions are mixed. Think of it as the stuff that crashes out of solution because it cannot stay dissolved.
Solubility rules
A set of empirical guidelines that tell you which ionic compounds dissolve in water (soluble) and which do not (insoluble). These are the lookup table you apply to every precipitation problem.
Molecular equation (formula equation)
The balanced equation written with complete ionic formulas for every compound, showing state symbols (aq), (s), (l), (g). It looks like a standard balanced equation.
Complete ionic equation
The equation in which every strong electrolyte that is dissolved (aq) is written as its separated ions. Solids, liquids, gases and weak electrolytes stay as complete formulas.
Net ionic equation
The equation left after you cancel all spectator ions from the complete ionic equation. It shows only the species that undergo a chemical change.
Spectator ions
Ions present in solution that do not participate in the reaction. They appear on both sides of the complete ionic equation and cancel out. In simple terms, they are the bystanders.
Write out the ions present in each solution.
Consider the possible new pairings: the cation from solution A with the anion from solution B, and vice versa.
Check each pairing against the solubility rules. If one combination is insoluble, a precipitate forms.
Example: Pb(NO3)2(aq) + CaBr2(aq). Possible products: PbBr2 and Ca(NO3)2. Lead(II) bromide is insoluble (Pb2+ is an exception to the "most bromides are soluble" rule). Ca(NO3)2 is soluble. So PbBr2 precipitates.
Some problems ask you to precipitate one ion from a mixture while leaving another in solution. The key is to find a reagent whose anion forms an insoluble compound with one cation but a soluble compound with the other.
Example: to precipitate Ba2+ but not Fe3+, add Na2SO4. BaSO4 is insoluble, but Fe2(SO4)3 is soluble. Iron stays in solution.
Write complete formulas for all reactants and products with correct subscripts, charges balanced within each formula, and state symbols. Coefficients balance the atoms.
Example: Na2S(aq) + ZnCl2(aq) → ZnS(s) + 2 NaCl(aq).
Start from the balanced molecular equation.
Split every soluble strong electrolyte marked (aq) into its ions.
Cancel any ion that appears identically on both sides (spectator ions).
What remains is the net ionic equation.
Example: Ca2+(aq) + CO3 2-(aq) → CaCO3(s). The Na+ and CH3CO2- ions from calcium acetate and sodium carbonate are spectators.
Spectator ions are those that do not appear in the net ionic equation. For the reaction of HBr(aq) + NaOH(aq), both are strong electrolytes. The net ionic equation is H+(aq) + OH-(aq) → H2O(l). The spectator ions are Na+ and Br-.
Soluble by default: compounds of Na+, K+, NH4+; all nitrates (NO3-); all acetates (CH3CO2-); most chlorides, bromides and iodides (exceptions: Ag+, Pb2+, Hg2 2+); most sulfates (exceptions: Ba2+, Pb2+, Ca2+ is slightly soluble).
Insoluble by default: most hydroxides (exceptions: NaOH, KOH, Ba(OH)2); most carbonates, phosphates and sulfides (exceptions: those of Na+, K+, NH4+). AgF is soluble, but AgCl, AgBr and AgI are not.
Water treatment plants use precipitation reactions to remove heavy metals such as lead and mercury from wastewater. Qualitative analysis in the lab, a technique you may encounter in later chemistry courses, relies entirely on selectively precipitating ions to identify unknowns.
Students often forget the exceptions in the solubility rules. "All chlorides are soluble" is only true if you ignore AgCl, PbCl2 and Hg2Cl2.
Writing the net ionic equation does not mean removing all ions. Only spectator ions cancel. The ions that form the precipitate stay.
A weak electrolyte such as HCO2H (formic acid) is not split into ions in the complete ionic equation. Only strong acids, strong bases and soluble salts are fully dissociated.
Students sometimes write the precipitate as (aq) instead of (s). The precipitate is a solid; getting the state symbol wrong changes the meaning of the equation.
⚠️ You will need to have the solubility rules memorised. The exam will not provide them.
⚠️ Selective precipitation questions (precipitate ion A but not ion B) are a favourite. Know BaSO4 is insoluble while most iron salts of common anions are soluble.
⚠️ Expect to write both the balanced molecular equation and the net ionic equation for a given reaction. Getting the charges and subscripts right in the molecular equation is essential; everything else follows from it.
⚠️ Weak acids (HCO2H, CH3CO2H, HCN) stay as molecules in ionic equations. Confusing them with strong acids is a common exam trap.
True or false: all sodium salts are soluble. (True.)
Fill in the blank: ions that appear on both sides of a complete ionic equation and cancel are called ______ ions. (Spectator.)
True or false: PbCl2 is soluble in water. (False; Pb2+ is an exception to the chloride solubility rule.)
Fill in the blank: in a net ionic equation, the precipitate is written with the state symbol ______. ((s).)
True or false: HCO2H is split into H+ and HCO2- in a complete ionic equation. (False; it is a weak acid and stays as HCO2H.)
Q: A precipitate forms when aqueous Pb(NO3)2 is added to which of the following: Cu(NO3)2, CaBr2, NaCH3CO2, Ca(ClO4)2, NaNO3?
A: CaBr2. The product PbBr2 is insoluble (Pb2+ is an exception to bromide solubility). The other pairings all produce soluble compounds.
Q: What is the net ionic equation when aqueous calcium acetate reacts with aqueous sodium carbonate?
A: Ca2+(aq) + CO3 2-(aq) → CaCO3(s). Calcium carbonate is insoluble. Na+ and CH3CO2- are spectators.
Q: What is the net ionic equation for the reaction of NaOH(aq) and FeCl2(aq)?
A: Fe2+(aq) + 2 OH-(aq) → Fe(OH)2(s). Iron(II) hydroxide is insoluble. Na+ and Cl- are spectators.
Q: Write the balanced molecular equation for Na2S(aq) + ZnCl2(aq).
A: Na2S(aq) + ZnCl2(aq) → ZnS(s) + 2 NaCl(aq). Zinc sulfide is insoluble; NaCl is soluble.
Q: If you add Na2SO4 to a mixture of Ba2+ and Fe3+, which ion precipitates and which stays in solution?
A: Ba2+ precipitates as BaSO4(s). Fe3+ stays in solution because iron(III) sulfate is soluble.
Q: What are the spectator ions in the reaction between HBr(aq) and NaOH(aq)?
A: Na+ and Br-. Both HBr and NaOH are strong electrolytes. The net ionic equation is H+(aq) + OH-(aq) → H2O(l).
Q: Formic acid (HCO2H) is a weak acid. Write the net ionic equation for HCO2H(aq) + KOH(aq).
A: HCO2H(aq) + OH-(aq) → HCO2-(aq) + H2O(l). Because formic acid is weak, it stays as a molecule. K+ is the spectator.
Precipitation reactions connect directly to the solubility and intermolecular forces topics you will cover later: why certain ionic compounds dissolve while others do not comes down to the balance between lattice energy and hydration energy.
The net ionic equation framework reappears in acid-base and redox reactions. Once you can write a net ionic equation for a precipitation, the same logic carries over to neutralisation and electron-transfer reactions.
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