4.3.5Chemical Reactions
Precipitation Reactions — AP CHEMISTRY Revision Notes & Topical Questions
5 min read 5 parts Lesson 18 of 18 Unit quiz
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Section 1 of 5Recognising a Precipitation Reaction
A precipitation reaction is a double replacement in which an insoluble ionic solid forms when solutions are mixed.
Double replacement
- Double replacement The cations swap partners, as in AB + CD → AD + CB, so two new compounds are possible.
Precipitate
- Precipitate The insoluble solid that separates from solution, shown with the state symbol (s).
Soluble compounds
- Soluble compounds Ionic compounds that dissociate fully in water remain as separate ions and are written (aq).
Evidence
- Evidence A cloudy suspension or solid appearing on mixing two clear solutions suggests a precipitate has formed.
Section 2 of 5Solubility Guidelines
These general rules let you decide which product, if any, is a solid.
Solubility Guidelines
Always soluble
- Always soluble Compounds of Group 1 cations, ammonium ions and nitrate ions dissolve in water.
Halides
- Halides Chlorides, bromides and iodides are soluble, except those of silver and lead(II).
Sulfates
- Sulfates Most sulfates are soluble, except those of barium, lead(II) and silver;
- calcium sulfate is only slightly soluble.
Mostly insoluble
- Mostly insoluble Carbonates, phosphates, sulfides and hydroxides are generally insoluble unless the cation is from Group 1 or is ammonium.
| Mixture | Possible product | Result |
|---|---|---|
| AgNO₃ + NaCl | AgCl | AgCl(s) forms |
| Pb(NO₃)₂ + KI | PbI₂ | PbI₂(s) forms |
| NaNO₃ + KCl | NaCl, KNO₃ | Both soluble, no reaction |
Section 3 of 5Writing the Equations
Work in three stages, moving from formulae to the ions that actually change.
1
Swap the cations to write the possible products, balancing the charges so each formula is neutral.
2
Use the solubility rules to label each product (aq) or (s) and balance the molecular equation.
3
Split every aqueous ionic compound into its ions to give the complete ionic equation, leaving solids as formulae.
4
Cancel the spectator ions that appear unchanged on both sides to give the net ionic equation.
5
Check that atoms and total charge are balanced; if no solid forms, then all ions cancel and there is no reaction.
Section 4 of 5Worked Example
Question
- 50.0 mL of 0.100 M Pb(NO₃)₂ is mixed with excess KI solution.
- Write the net ionic equation and find the mass of precipitate formed.
- Molecular equation
- Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq);
- PbI₂ is the precipitate.
- Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq);
- Net ionic equationPb²⁺(aq) + 2I⁻(aq) → PbI₂(s).
- n(Pb²⁺) = 0.100 mol L⁻¹ × 0.0500 L = 5.00 × 10⁻³ mol, and the ratio Pb²⁺ : PbI₂ is 1 : 1.
- M(PbI₂) = 207.2 + 2(126.90) = 461.0 g mol⁻¹, so mass = 5.00 × 10⁻³ × 461.0 = 2.31 g.
Answer 2.31 g of PbI₂
Section 5 of 5Exam Checklist & Tips
4.3.5 – Key Points for the Exam
- Precipitate An insoluble solid that forms when two solutions are mixed.
- Spectator ion An ion that is present but does not take part in the net change.
- Net ionic equation An equation showing only the species that actually react or form.
- Double replacement A reaction in which the ions of two compounds exchange partners.
- Soluble Able to dissolve in water to a significant extent.
Common Mistakes
- Writing the precipitate as (aq) or leaving a soluble product as (s).
- Splitting a solid precipitate into ions in the complete ionic equation.
- Cancelling ions that are not identical, including their charge and state.
- Forgetting to balance charges when swapping cations, such as writing PbI instead of PbI₂.
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