All Ion Tests: Full Reference

Part of Tests for Ions · Section 6 of 15

ComparisonUnit: Chemical AnalysisGCSE
IonReagentPositive ResultKey Equation
Cu²⁺NaOH(aq)Blue precipitateCu²⁺ + 2OH⁻ → Cu(OH)₂(s)
Fe²⁺NaOH(aq)Green precipitateFe²⁺ + 2OH⁻ → Fe(OH)₂(s)
Fe³⁺NaOH(aq)Brown/orange precipitateFe³⁺ + 3OH⁻ → Fe(OH)₃(s)
Al³⁺NaOH(aq)White precipitate; redissolves in excessAl³⁺ + 3OH⁻ → Al(OH)₃(s)
Ca²⁺NaOH(aq)White precipitate (stays)Ca²⁺ + 2OH⁻ → Ca(OH)₂(s)
CO₃²⁻Dilute HClFizzing; gas turns limewater milkyCO₃²⁻ + 2H⁺ → CO₂ + H₂O
SO₄²⁻BaCl₂ + HClWhite precipitate of BaSO₄Ba²⁺ + SO₄²⁻ → BaSO₄(s)
Cl⁻AgNO₃ + HNO₃White precipitate of AgClAg⁺ + Cl⁻ → AgCl(s)
Br⁻AgNO₃ + HNO₃Cream precipitate of AgBrAg⁺ + Br⁻ → AgBr(s)
I⁻AgNO₃ + HNO₃Yellow precipitate of AgIAg⁺ + I⁻ → AgI(s)

This comparison covers All Ion Tests: Full Reference within Tests for Ions for GCSE Chemistry. Revise Tests for Ions in Chemical Analysis for GCSE Chemistry with 20 exam-style questions and 14 flashcards. This is a high-frequency topic, so it is worth revising until the explanation feels precise and repeatable. It is section 6 of 15 in this topic. Use this comparison to connect the idea to the wider topic before moving on to questions and flashcards.

Practice questions for Tests for Ions

Which reagents are used to test for carbonate ions in a solution?

  • A. Add barium chloride solution, then dilute HCl
  • B. Add dilute acid, then test the gas with limewater
  • C. Add silver nitrate solution, then dilute HNO3
  • D. Add sodium hydroxide solution and warm
1 markfoundation

Describe how sodium hydroxide solution can be used to distinguish between iron(II) ions and iron(III) ions in solution, including the expected observations.

3 marksstandard

Quick recall flashcards

How do you test for sulfate ions (SO₄²⁻)?
Add barium chloride solution + dilute HCl. White precipitate of BaSO₄ forms. Equation: Ba²⁺ + SO₄²⁻ → BaSO₄(s)
What is a precipitation reaction?
A reaction where two soluble ionic compounds react to form an insoluble precipitate. General form: A⁺(aq) + B⁻(aq) → AB(s)

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