Chemistry • Year 12 • Module 5 • Lesson 17

Solubility Product K_sp

Lock in the core vocabulary, write K_sp expressions from dissolution equations, and practise converting between K_sp and molar solubility.

Build • Band 3–4

1. Fill in the blanks

Complete the paragraph using the word bank below. Each word is used once. 10 marks

Word bank: equilibrium • solid • concentration • heterogeneous • exponents • saturation • ions • molar solubility • denominator • Ksp

The solubility product, _____________ (1), is the equilibrium constant for the dissolution of a sparingly soluble salt. It is a _____________ (2) equilibrium because the system involves a _____________ (3) phase (the undissolved salt) and a liquid phase. The K_sp expression contains only the _____________ (4) of the dissolved _____________ (5) raised to powers equal to their stoichiometric _____________ (6). Because the solid is a pure phase with constant activity, it is excluded from the expression — there is no _____________ (7). The K_sp expression applies only at _____________ (8), when the solution is saturated. The number of moles of salt that dissolve per litre to reach saturation is called the _____________ (9), symbol s. A smaller K_sp generally indicates lower _____________ (10) for compounds of the same formula type.

Stuck? Revisit lesson Key Terms panel and Card 1 (Writing K_sp Expressions).

2. Term–definition match

Match each term (column A) to the correct definition (column B) by writing the letter in the right-hand column. 8 marks

#	Definition (column B)	Term (write letter)
2.1	The equilibrium constant for the dissolution of a sparingly soluble ionic compound in water.
2.2	The number of moles of a sparingly soluble salt that dissolves per litre of water to form a saturated solution.
2.3	An equilibrium system that involves more than one phase (e.g. solid and aqueous).
2.4	A solution in which the dissolved ion concentrations are at their maximum value at a given temperature — no more salt can dissolve.
2.5	The reason the solid is omitted from the K_sp expression: a pure solid has a constant value of this quantity equal to 1.
2.6	The formula type in which each formula unit of salt dissolves to give one cation and two anions (e.g. BaF₂).
2.7	The algebraic relationship between K_sp and s for a 1:1 salt (e.g. AgCl): K_sp = s².
2.8	The algebraic relationship between K_sp and s for a 1:2 salt (e.g. CaF₂): K_sp = 4s³.

Column A — terms:
A. Saturated solution • B. Molar solubility (s) • C. 1:2 formula type • D. Activity • E. K_sp = s² • F. K_sp • G. Heterogeneous equilibrium • H. K_sp = 4s³

Stuck? Revisit lesson Key Terms and formula panel.

3. True or false — with correction

Circle T or F. If false, write the correct version on the line below. 8 marks (1 T/F + 1 correction where false)

3.1 The solid is included in the K_sp expression in the denominator because it is a reactant in the dissolution equilibrium. T / F

3.2 For AgCl dissolving in water, the correct K_sp expression is: K_sp = [Ag⁺][Cl⁻]. T / F

3.3 For any pair of sparingly soluble salts, the one with the smaller K_sp is always less soluble, regardless of formula type. T / F

3.4 For Mg(OH)₂, which dissolves as Mg(OH)₂ ⇌ Mg²⁺ + 2OH⁻, the molar solubility is found from K_sp = 4s³. T / F

Stuck? Revisit lesson Cards 1 and 4.

4. Write K_sp expressions

For each salt, write the balanced dissolution equation and then the K_sp expression. The first row is completed as an example. 10 marks — 2 marks each for rows 4.2 to 4.6

Salt	Balanced dissolution equation	K_sp expression
Example AgCl	AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)	K_sp = [Ag⁺][Cl⁻]
4.2 BaSO₄
4.3 CaCO₃
4.4 PbI₂
4.5 Ag₂CrO₄
4.6 Mg(OH)₂

Stuck? Revisit lesson Card 1 and the formula-type SVG.

5. Short recall questions

Answer each in 1–2 sentences using precise chemical terms. 10 marks — 2 marks each)

5.1 Why is the solid not included in the K_sp expression?

5.2 What is the formula relating molar solubility s and K_sp for a 1:1 salt such as AgCl?

5.3 State the K_sp comparison rule: when is it valid to rank two salts by K_sp alone, and when must you calculate s?

5.4 Fluoride ions protect tooth enamel by forming fluorapatite with a much lower K_sp than hydroxyapatite. What does a lower K_sp tell you about the solubility of fluorapatite?

5.5 Water in limestone cave systems dissolves CaCO₃. Write the formula for the K_sp expression of this dissolution. (CaCO₃ is a 1:1 salt.)

Stuck? Revisit lesson Cards 1, 2, 4 and 5.

Answers — Do not peek before attempting

Q1 — Cloze answers (in order)

(1) K_sp (2) heterogeneous (3) solid (4) concentration (5) ions (6) exponents (7) denominator (8) saturation (9) molar solubility (10) equilibrium

Q2 — Term–definition matches

2.1 F • 2.2 B • 2.3 G • 2.4 A • 2.5 D • 2.6 C • 2.7 E • 2.8 H

Q3 — True / false with corrections

3.1 False. The solid is excluded from the K_sp expression entirely — it does not appear in the numerator or denominator. Pure solids have constant activity (= 1) and are excluded from all equilibrium expressions.

3.2 True.

3.3 False. Direct K_sp comparison is only valid for salts with the same formula type (same ion ratio). For salts with different formula types (e.g. 1:1 vs 1:2), you must calculate molar solubility s for each and compare s values. A smaller K_sp does not guarantee lower solubility across formula types.

3.4 True. Mg(OH)₂ is a 1:2 type (one cation, two anions). [Mg²⁺] = s, [OH⁻] = 2s. K_sp = s(2s)² = 4s³.

Q4 — K_sp expressions

4.2 BaSO₄: BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq) K_sp = [Ba²⁺][SO₄²⁻]

4.3 CaCO₃: CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq) K_sp = [Ca²⁺][CO₃²⁻]

4.4 PbI₂: PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq) K_sp = [Pb²⁺][I⁻]²

4.5 Ag₂CrO₄: Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq) K_sp = [Ag⁺]²[CrO₄²⁻]

4.6 Mg(OH)₂: Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq) K_sp = [Mg²⁺][OH⁻]²

Q5 — Short recall

5.1 A pure solid has constant activity (= 1 by convention), so it does not change the value of the equilibrium expression. Including it would just multiply the expression by a constant. By convention it is excluded, leaving only the aqueous ion concentrations.

5.2 For a 1:1 salt AB ⇌ A⁺ + B⁻: [A⁺] = s, [B⁻] = s. K_sp = s², therefore s = √K_sp.

5.3 Direct K_sp comparison is valid only for salts of the same formula type (same stoichiometric ion ratio, e.g. both 1:1 or both 1:2). For different formula types, calculate s for each from its own K_sp expression, then compare s values.

5.4 A lower K_sp for fluorapatite means it has a lower molar solubility than hydroxyapatite — it dissolves to a smaller extent in saliva, making the enamel more resistant to acid dissolution.

5.5 CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq). K_sp = [Ca²⁺][CO₃²⁻]