Lead II Nitrate and Potassium Iodide — Reaction, Equation & Explanation

The balanced chemical equation for the reaction between lead(II) nitrate and potassium iodide is:

Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) ↓ + 2KNO₃(aq)

Word equation: Lead(II) nitrate + Potassium iodide → Lead(II) iodide + Potassium nitrate

Balancing steps:

1. Unbalanced: Pb(NO₃)₂ + KI → PbI₂ + KNO₃
2. Count atoms: Pb has 2 iodine atoms on the product side but only 1 KI on the reactant side
3. Place coefficient 2 before KI: Pb(NO₃)₂ + 2KI → PbI₂ + KNO₃
4. Now 2K on left, so place 2 before KNO₃: Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃
5. Verify: Pb = 1:1, N = 2:2, O = 6:6, K = 2:2, I = 2:2 ✓

State symbols: • (aq) = aqueous (dissolved in water) • (s) = solid (precipitate) • ↓ indicates a precipitate forming

This reaction is classified as:

1. Double Displacement Reaction (Metathesis) The cations and anions of the two reactants exchange partners: • Pb²⁺ was with NO₃⁻, now pairs with I⁻ → PbI₂ • K⁺ was with I⁻, now pairs with NO₃⁻ → KNO₃

General pattern: AB + CD → AD + CB Here: Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃

2. Precipitation Reaction One of the products (PbI₂) is insoluble in water and forms a solid precipitate. This is what drives the reaction forward.

3. Non-Redox Reaction No change in oxidation states occurs: • Pb remains +2 • N remains +5 • O remains −2 • K remains +1 • I remains −1

This is NOT a redox reaction — it is purely an exchange of ions.

Full Ionic Equation (all soluble compounds split into ions): Pb²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq)

Spectator ions (appear unchanged on both sides): • K⁺ — present on both sides • NO₃⁻ — present on both sides

Net Ionic Equation (remove spectator ions): Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)

The net ionic equation shows the actual chemical change: lead ions combine with iodide ions to form the insoluble yellow precipitate of lead(II) iodide.

This net ionic equation represents the driving force of the reaction — the formation of an insoluble product that is removed from solution.

When lead(II) nitrate solution is mixed with potassium iodide solution, the following observations are made:

Before mixing: • Lead(II) nitrate solution: colourless • Potassium iodide solution: colourless

After mixing: • A bright yellow precipitate forms immediately • The yellow solid is lead(II) iodide (PbI₂) • The remaining solution contains dissolved potassium nitrate (colourless) • The reaction occurs instantly at room temperature

Appearance of PbI₂: • Bright yellow, sometimes described as 'golden yellow' • Heavy, crystalline precipitate that settles to the bottom • When heated and cooled slowly, PbI₂ forms beautiful golden flakes known as 'golden rain' or 'gold rain' experiment

The 'Golden Rain' experiment:

1. Mix the two solutions to form yellow PbI₂ precipitate
2. Heat the mixture until PbI₂ dissolves (PbI₂ is more soluble in hot water)
3. Cool slowly — golden, glittering flakes of PbI₂ crystallise out
4. This demonstration is popular in chemistry classes

Lead(II) Iodide (PbI₂): • Chemical formula: PbI₂ • Molar mass: 461.01 g/mol • Colour: bright yellow • State: solid at room temperature • Solubility: slightly soluble in cold water (0.076 g/100 mL at 20°C), more soluble in hot water (4.1 g/100 mL at 100°C) • Crystal structure: hexagonal • Melting point: 402°C • Used in: photography, printing, and as a semiconductor material

Potassium Nitrate (KNO₃): • Chemical formula: KNO₃ • Molar mass: 101.10 g/mol • Colour: colourless/white • State: solid (dissolves readily in water) • Solubility: highly soluble in water (31.6 g/100 mL at 20°C) • Common name: saltpetre or nitre • Used in: fertilisers, fireworks, food preservation, and gunpowder

Understanding why PbI₂ precipitates requires knowledge of solubility rules:

Key solubility rules:

1. All nitrates (NO₃⁻) are soluble → Pb(NO₃)₂ dissolves, KNO₃ dissolves ✓
2. All potassium (K⁺) compounds are soluble → KI dissolves, KNO₃ dissolves ✓
3. Most iodides (I⁻) are soluble, EXCEPT PbI₂, AgI, and HgI₂ → PbI₂ is insoluble ✗

So when Pb²⁺ and I⁻ ions meet in solution, they combine to form insoluble PbI₂, which precipitates out as a bright yellow solid.

Predicting precipitation: • Check if any combination of cations and anions from the two reactants forms an insoluble compound • Pb²⁺ + 2I⁻ → PbI₂ (insoluble) → precipitate forms • K⁺ + NO₃⁻ → KNO₃ (soluble) → stays in solution

Ksp (Solubility Product) of PbI₂: • Ksp = 9.8 × 10⁻⁹ at 25°C • This very small value confirms that PbI₂ is poorly soluble

From the balanced equation: Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃

Mole ratio: • 1 mol Pb(NO₃)₂ reacts with 2 mol KI • Produces 1 mol PbI₂ and 2 mol KNO₃

Molar masses: • Pb(NO₃)₂ = 331.2 g/mol • KI = 166.0 g/mol • PbI₂ = 461.0 g/mol • KNO₃ = 101.1 g/mol

Sample calculation: Q: If 33.12 g of Pb(NO₃)₂ reacts with excess KI, how much PbI₂ is formed?

Solution: • Moles of Pb(NO₃)₂ = 33.12 ÷ 331.2 = 0.1 mol • Mole ratio: 1 mol Pb(NO₃)₂ → 1 mol PbI₂ • Moles of PbI₂ = 0.1 mol • Mass of PbI₂ = 0.1 × 461.0 = 46.1 g

Answer: 46.1 g of bright yellow PbI₂ precipitate is formed.