Ionic equations and spectator ions
Retain the particles that change and check that both atoms and electrical charge are conserved.
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
Before you begin
- Balance symbol equations without changing formulas.
- Read ionic charges and aqueous state symbols.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What total charge do Ag⁺ and Cl⁻ have together?
What you will learn
- Deduce balanced equations from stated reaction information.
- Construct ionic equations with appropriate state symbols and charge balance.
Write the full reaction before removing spectators
Use supplied observations and substance identities to choose reactants and products before balancing. In a full ionic equation, represent soluble aqueous ionic substances as their ions when appropriate. Keep solids, liquids and gases as intact formulas. In introductory strong-acid and alkali examples, the aqueous acid and alkali supply their reacting ions.
An ion appearing unchanged on both sides is a spectator and cancels from the net ionic equation. The remaining equation describes the actual particle change. Cancellation is not permission to remove a particle that forms a solid, gas or new molecule.
Balance atoms and charge together
For a silver chloride precipitate, Ag⁺(aq) + Cl⁻(aq) → AgCl(s). One atom of each element appears on each side and total charge is zero on both sides. In acid–alkali neutralisation, H⁺(aq) + OH⁻(aq) → H₂O(l); the hydrogen and hydroxide ions form a new water molecule, while the other dissolved ions remain spectators.
For copper(II) hydroxide precipitation, one Cu²⁺ needs two OH⁻: Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s). Equal atom totals alone are insufficient; charge totals must also match. Include states so the equation shows dissolved ions becoming a solid rather than cancelling the precipitate.
Sources for this lesson
Pause and explain
Which particles cancel in the silver nitrate–sodium chloride example?
Worked example
Aqueous silver nitrate and sodium chloride form a white solid silver chloride. Deduce a balanced full equation and its net ionic equation.
Show the worked solution
- The remaining soluble product is sodium nitrate: AgNO₃(aq)+NaCl(aq) → AgCl(s)+NaNO₃(aq).
- Na⁺ and NO₃⁻ remain aqueous and unchanged, so they are spectators.
- Retain Ag⁺ and Cl⁻ forming AgCl(s); atoms and charge balance.
Answer Ag⁺(aq) + Cl⁻(aq) → AgCl(s).
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Write the full and net ionic equations for HCl(aq) reacting completely with NaOH(aq). Identify the spectators.
Give me a hint
Begin with salt plus water, then compare unchanged dissolved ions.
Compare my reasoning
- The full equation is HCl(aq)+NaOH(aq) → NaCl(aq)+H₂O(l).
- Na⁺ and Cl⁻ remain unchanged in solution.
- The net equation is H⁺(aq)+OH⁻(aq) → H₂O(l), with zero charge each side.
Spectators: Na⁺ and Cl⁻. Net: H⁺(aq) + OH⁻(aq) → H₂O(l).
Look for these in your work
- Spectators are unchanged aqueous ions.
- Atoms, charge and water's liquid state are correct.
Barium ions and sulfate ions form insoluble barium sulfate. Write and check the net ionic equation.
Give me a hint
The charges are Ba²⁺ and SO₄²⁻; the product is a solid.
Compare my reasoning
- Combine one Ba²⁺ with one SO₄²⁻.
- Write Ba²⁺(aq)+SO₄²⁻(aq) → BaSO₄(s).
- Ba, S and O counts match; total charge +2−2=0 matches the neutral product.
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).
Look for these in your work
- The solid is not split into aqueous ions.
- Both atoms and charge are checked.
A student writes Cu²⁺(aq)+OH⁻(aq) → Cu(OH)₂(s). Explain two conservation failures and repair the equation.
Give me a hint
Count hydroxide groups and then total charge.
Compare my reasoning
- One hydroxide supplies only one O and one H, but the product needs two of each.
- The left charge is +1 while the product is neutral.
- Use 2OH⁻ to match both atoms and charge: Cu²⁺+2OH⁻ → Cu(OH)₂, retaining the stated phases.
Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s).
Look for these in your work
- Both conservation failures are identified.
- One coefficient repairs both atom and charge totals.
Common mistakes
- Splitting an insoluble solid into aqueous ions and cancelling it.
- Checking atoms but forgetting total charge.
What can you explain now?
Write the net ionic equation for copper(II) ions forming copper(II) hydroxide with hydroxide ions.
Compare with the explanation
Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s)
Two hydroxides supply the needed O and H counts and balance the copper ion's +2 charge.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Take a full precipitation equation and show which ions stay dissolved. Explain every cancellation and verify the remaining total charge.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.