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Precipitation and Net Ionic Equations: Quiz

Seven questions on solubility rules, precipitates, spectator ions and net ionic equations, with explanations.

7 questionsIntermediate

7 questions

  1. Question 1EasyWhich compound is insoluble in water?
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    Answer: BaSO₄

    Sulfates are usually soluble, but barium sulfate is an exception. Sodium, potassium and ammonium compounds, and all nitrates, are soluble.

  2. Question 2EasyWhich mixture of solutions gives a precipitate?
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    Answer: AgNO₃ + KBr

    Silver bromide (AgBr) is insoluble and precipitates (cream colour). In the other mixtures all possible products are soluble.

  3. Question 3MediumIn the reaction of BaCl₂(aq) with Na₂SO₄(aq), which ions are spectators?
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    Answer: Na⁺ and Cl⁻

    Ba²⁺ and SO₄²⁻ form the BaSO₄ precipitate. Na⁺ and Cl⁻ stay dissolved, unchanged: they are spectator ions.

  4. Question 4MediumWhat is the net ionic equation for Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)?
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    Answer: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)

    K⁺ and NO₃⁻ are spectators. The net reaction is lead(II) ions combining with two iodide ions; atoms and charge both balance.

  5. Question 5MediumWhich substance should NOT be split into ions in a complete ionic equation?
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    Answer: AgCl(s)

    AgCl is a solid precipitate, so it is written as a formula. Soluble salts and strong acids in solution are written as ions.

  6. Question 6MediumWhich hydroxide is soluble in water?
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    Answer: KOH

    Hydroxides are mostly insoluble, except those of Group 1 metals (and Ba(OH)₂). Potassium hydroxide is soluble.

  7. Question 7HardExcess AgNO₃ is added to 25.0 mL of 0.100 mol/L NaCl. What mass of AgCl (143.32 g/mol) precipitates?
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    Answer: 0.358 g

    n(Cl⁻) = 0.100 mol/L × 0.0250 L = 2.50 × 10⁻³ mol; mass = 2.50 × 10⁻³ mol × 143.32 g/mol = 0.358 g. 358 g comes from using 25.0 L.