08P3

Fri 1 May

§4.3

Week 2 · Lesson 5 of 17

Displacement reactions — application and consolidation

You've now met both single and double displacement, the activity series (Figure 4.9), and the solubility rules (Table 4.2). Today you put them to work — running through the textbook's Learning Ladder questions for §4.3 to make sure you can predict, explain, and write equations for any displacement scenario.

Learning Intentions + Success Criteria

LITo apply the activity series and solubility rules to a range of displacement scenarios.

SC: I can:

01I can identify a reaction as single- or double-displacement and predict its products.
02I can identify spectator ions and write the net ionic equation.
03I can construct an activity-series ranking from virtual-lab evidence and compare it to Figure 4.9.

Engage

5 min

Quick recap · from last class

L7 · §4.3 Displacement reactions — double displacement and precipitation

Try these 3questions before today's new content. Click an answer for instant feedback — your teacher will walk through them with you.

ClickView video · school login↗

Displacement Reactions

Replay if you want a quick recap before today's practice.

Then watch this short refresher on double-displacement reactions to lock in yesterday's L4 ideas before you tackle today's Learning Ladder questions.

YouTube · Double-displacement reactions — refresher · open in new tab

This is your application and consolidation lesson for §4.3. You'll work through the textbook Learning Ladder on pp.92–93. Have these two open in front of you for the whole period:

Figure 4.9 (the activity series — 16 metals, Li at top, Au at bottom).
Table 4.2 (the solubility rules — 8 ion groups).

Predict · your turn

Write before you watch

Before we start: which of the §4.3 ideas do you feel LEAST confident about — predicting from the activity series, identifying spectator ions, writing net ionic equations, or applying the solubility rules? Write down the one you most want to nail today.

Explicit

12 min

Two tools you'll use today

Both live in the Toolbox (bottom-left icon) — keep it open through the whole period.

Toolbox · Activity series

A free metal displaces another only if it sits above it in Figure 4.9.

Toolbox · Solubility rules

A precipitate forms whenever a new partnership matches an "except…" entry in Table 4.2.

Single vs double — what you actually see

Single displacement

One free metal + a salt solution.

Free metal grows a coating of the displaced metal.
Solution colour may fade or shift (the original cation is consumed).
Predict using Figure 4.9 (activity series).

e.g. Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

Double displacement

Two ionic solutions swap partners.

A solid suddenly appears in the mix and settles.
Both reactants stay liquid until they touch.
Predict using Table 4.2 (solubility rules).

e.g. AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

The four-step net ionic equation method

This is the textbook's signature derivation (Figures 4.10–4.12). Memorise the four steps — every Q5 stretch question below uses the same recipe.

Four-step net ionic equation

1. Balanced MOLECULAR equation

with state symbols (s) / (aq) / (l) / (g)

2. Split each (aq) compound into ions

= complete ionic equation. Solids/liquids/gases stay together — never split.

3. Cross out SPECTATOR IONS

identical on both sides — they take no part in the reaction

4. Rewrite what is left

= net ionic equation. The actual chemistry.

Same recipe, every reaction. Use it on every Rung-5 question.

I do — KI + Pb(NO₃)₂

Step 1. KI and KNO₃ are soluble (Group 1); Pb(NO₃)₂ is soluble (nitrate); PbI₂ is the iodide-row EXCEPTION → solid.

2KI(aq)+Pb(NO₃)₂(aq)⟶PbI₂(s)+2KNO₃(aq)

Step 2. Split every (aq) into ions:

2K⁺(aq)+2I⁻(aq)+Pb²⁺(aq)+2NO₃⁻(aq)⟶PbI₂(s)+2K⁺(aq)+2NO₃⁻(aq)

Step 3. K⁺ and NO₃⁻ appear unchanged on both sides → they are the spectator ions.

Step 4. Net ionic equation:

Pb²⁺(aq)+2I⁻(aq)⟶PbI₂(s)

Two ions colliding and locking together as a yellow solid — that's the actual chemistry.

We do — Na₂CO₃ + CaCl₂ (together)

We do · whole class · 4 prompts

Mix sodium carbonate solution Na₂CO₃(aq) with calcium chloride solution CaCl₂(aq). Calcium carbonate CaCO₃ is the white solid in chalk, limestone, and eggshells — let's see how it forms.

Prompt 1 — what are the cations and anions?

Prompt 2 — swap partners → check Table 4.2 → write the balanced molecular equation.

Prompt 3 — split every (aq) into ions (the complete ionic equation).

Prompt 4 — cross out the spectators → net ionic.

Same recipe, different precipitate. Now you do the rest below.

Apply

30 min

A 5-rung ladder, easiest to hardest. Climb in order — each rung uses what the previous one set up. If today's energy is low, a strong stop at rung 4 is fine; rung 5 is the stretch.

Interactive simulation · opens in new tab↗

Optional — Virtual Lab: Activity Series (Mr Palermo)

If time permits at the end of class, or as optional extension: run a few metal-in-solution combinations and compare against Figure 4.9.

Question 1Rung 1 — warmup: identify the ions

When an ionic compound dissolves in water, it splits into its cation (positive) and anion (negative). Pick the right pair for each.

Question 2Rung 2 — classify single vs double displacement

For each reaction, decide whether it is single-displacement (one free element kicks out another) or double-displacement (two compounds swap partners).

Question 3Rung 3 — predict single-displacement

For each, open the Toolbox → Activity Series. Compare the free metal to the metal in the compound — does a reaction happen?

Question 4Rung 4 — predict double-displacement (and name the precipitate)

For each pair of solutions, open the Toolbox → Solubility Rules (Table 4.2). Decide whether a precipitate forms — and if it does, name it.

Question 5Rung 5 (stretch) — write the net ionic equation

For each reaction, follow the 4-step recipe from the worked example above: balanced molecular → complete ionic → cancel spectators → net ionic.

Your turnShort answer · Have a go first

(a) NaCl + AgNO₃ → NaNO₃ + AgCl

Have a go first:

Your turnShort answer · Have a go first

(b) Pb(NO₃)₂ + 2 NaCl → PbCl₂ + 2 NaNO₃

Have a go first:

Extension — only if you've climbed all 5 rungs

External video · opens in new tab↗

Khan Academy — Double Replacement Reactions

Sal Khan walks through the same chemistry from a different angle. Note: Khan calls these 'double replacement'; we call them 'double displacement' — same reactions, two names. Worth a watch if you've already finished rungs 1–5.

The full solubility chart (13 cations × 10 anions)

Table 4.2 has the 8 ion-group rules of thumb you need for §4.3. The chart below is the full reference — every common cation × every common anion in one grid — and it's what you'll meet again in Year 11 and 12 VCE Chemistry. Each cell tells you whether the salt is soluble (SOL · green), insoluble (INSOL · red), or slightly soluble (SL SOL · amber). Blank cells are compounds rare enough that you're not expected to know them.

SOL = soluble

INSOL = insoluble (precipitates)

SL SOL = slightly soluble

	Li⁺	Na⁺	K⁺	NH₄⁺	Mg²⁺	Ca²⁺	Ba²⁺	Al³⁺	Fe³⁺	Cu²⁺	Zn²⁺	Ag⁺	Pb²⁺
F⁻	SOL	SOL	SOL	SOL	INSOL	INSOL	SL SOL	SOL	SL SOL	SOL	SOL	SOL	INSOL
Cl⁻	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	INSOL	SL SOL
Br⁻	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	INSOL	SL SOL
I⁻	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	—	—	SOL	INSOL	INSOL
OH⁻	SOL	SOL	SOL	SOL	INSOL	SL SOL	SOL	INSOL	INSOL	INSOL	INSOL	—	INSOL
NO₃⁻	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL
C₂H₃O₂⁻	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL	SOL
SO₄²⁻	SOL	SOL	SOL	SOL	SOL	SL SOL	INSOL	SOL	SOL	SOL	SOL	SL SOL	INSOL
CO₃²⁻	SOL	SOL	SOL	SOL	INSOL	INSOL	INSOL	—	INSOL	SL SOL	INSOL	INSOL	INSOL
PO₄³⁻	INSOL	SOL	SOL	SOL	INSOL	INSOL	INSOL	INSOL	INSOL	INSOL	INSOL	INSOL	INSOL

Spot the patterns: every nitrate (NO₃⁻) and acetate (C₂H₃O₂⁻) row is all green — they're always soluble. Every Group 1 / NH₄⁺ column is also all green. Most carbonates and phosphates are all red — almost always precipitate. The interesting chemistry is in the boundary cases — Cl⁻ with Ag⁺, SO₄²⁻ with Ba²⁺, OH⁻ with most transition metals.

Catch

5 min

Bonus (stretch): write the balanced symbol equation for the reaction. Hint: Li forms a +1 ion (Li⁺), so the product is Li₂SO₄.

Reflect

5 min

Your turnReflect · One thing you learned

Look back at the question you flagged at the start as 'least confident'. After today's practice, do you feel solid on it? If not — what specifically still doesn't make sense?

I've finished this lesson

Success criteria — where are you right now?

Next class (Tue 5 May, P4): §4.2–§4.3 recap — synthesis, decomposition, and displacement consolidated before §4.4 begins.