LITo predict precipitation reactions using the solubility rules.
SC: I can:
Try these 2questions before today's new content. Click an answer for instant feedback — your teacher will walk through them with you.
On Wednesday (L3) you saw single-displacement — one element kicks another out of a compound. Today we meet its partner pattern, double displacement.
Watch this classic demo. Two clear, harmless-looking solutions are poured together — watch what happens the instant they touch.
Both starting solutions are clear. After mixing, a bright yellow SOLID appears inside the liquid. Where did the solid come from — and why is it yellow when neither starting solution was?
On Wednesday you saw single displacement: a free metal kicks a less-reactive metal out of an ionic compound (A + BC → AC + B). Here's the procedure as a flow chart — same shape as today's, just with the Activity Series (Figure 4.9) instead of Table 4.2.
Today's lesson is the partner pattern: instead of a free metal vs an ionic compound, we mix two ionic compounds and let them swap partners. The decision point uses Table 4.2 instead of the activity series. Same procedural shape; different signal.
An ion is an atom (or group of atoms) that has lost or gained electrons — so it carries an electric charge.
Lost electrons → positive charge. Usually a metal (or the ammonium ion NH₄⁺).
Gained electrons → negative charge. Usually a non-metal or polyatomic group.
Memory hook: CAT-ion = paws up = + · AN-ion = "A Negative" = −
A neutral ionic compound is just a cation + anion in the ratio that cancels the charges:
Drop an ionic salt into water and the water molecules pull the ions apart. Inside the beaker, every cation and anion is now floating around on its own:
That's why double displacement is even possible — once the ions are loose, they're free to swap partners with the next solution you pour in.
This is the loop you'll run for every prediction in §4.3 — and a version of it for §4.4 acid reactions later this week. Same shape, different "signal" at the end.
The precipitate is marked (s) in the balanced equation, even though it sits inside a liquid. The soluble salt stays (aq).
Use this at step 3 of today's flow chart to decide whether each new compound is soluble or insoluble:
| Ions | Solubility |
|---|---|
| Group 1 salts (Na⁺, K⁺) | All soluble |
| Ammonium salts (NH₄⁺) | All soluble |
| Nitrates (NO₃⁻) | All soluble |
| Chlorides (Cl⁻) | All soluble except for AgCl, PbCl₂ and Hg₂Cl₂ |
| Iodides (I⁻) | All soluble except for AgI, PbI₂ and HgI₂ |
| Sulfates (SO₄²⁻) | All soluble except for PbSO₄, CaSO₄ and BaSO₄ |
| Carbonates (CO₃²⁻) | All insoluble except for group 1 carbonates (e.g. Na₂CO₃, K₂CO₃) and (NH₄)₂CO₃ |
| Hydroxides (OH⁻) | All insoluble except for group 1 hydroxides (e.g. NaOH, KOH), Ca(OH)₂, Ba(OH)₂ and NH₄OH |
Mix potassium iodide solution and lead(II) nitrate solution. Predict what happens.
Step 1 — swap partners. Pb²⁺ pairs with I⁻; K⁺ pairs with NO₃⁻.
Step 2 — check each new partnership against Table 4.2.
Step 3 — write the equations (textbook order).
Word equation:
potassium iodide + lead(II) nitrate → lead(II) iodide + potassium nitrate
Balanced equation with state symbols:
That's the molecular equation — what you write for §4.3 prediction questions today. Next period (P3, straight after this one) goes one layer deeper: we'll split each aqueous compound into its separate ions (the complete ionic equation), cancel the spectator ions that don't actually react, and finish with the net ionic equation showing just the chemistry that happens.
We do · whole class · 3 prompts
Mix potassium sulfate solution K2SO4(aq) with lead(II) nitrate solution Pb(NO3)2(aq). Same Pb2+ as before, but a sulfate anion instead of iodide — what changes?
Prompt 1 — what are the cations and anions?
Prompt 2 — swap partners → check Table 4.2.
Prompt 3 — write the balanced equation with state symbols.
The precipitate is white lead(II) sulfate — different anion, different colour. (Real-world: PbSO₄ is what builds up on the plates of a lead–acid car battery.)
Work through all four questions in order. Then open Ch 4 Booklet pp.92–93 and complete the double-displacement section on paper.
Lead(II) iodide precipitating from solution — the moment two clear liquids are mixed. The yellow solid is PbI₂.
Source — Wikimedia Commons (CC BY-SA)
A student mixes silver nitrate solution with sodium chloride solution. A white cloudy solid forms instantly.
The white solid is silver chloride — check Table 4.2: chlorides are all soluble EXCEPT AgCl, PbCl₂, and Hg₂Cl₂. So AgCl drops out as the precipitate. The symbol equation with states:
A student mixes copper(II) sulfate solution with sodium hydroxide solution.
One diagnostic before you leave. Write the answer in your notebook AND type it in here — the paper copy gets collected.
Note — your printed handout shows a different reaction (AgNO₃ + KI → AgI). Use the website prompt below instead: cross out the printed one and write the new pair (Pb(NO₃)₂ + Na₂SO₄) on the worklines. Same skill, different row of Table 4.2.
One thing about double displacement or precipitation that I now understand, that I didn't understand before this lesson:
Next class (Fri 1 May, P3): application and consolidation of §4.3. We'll work through the textbook Learning Ladder questions covering classification, observations, net ionic equations, contaminant removal, and the Use & Influence of Science questions. Bring Table 4.2 with you.
