📚 Solubility Product Notes
CAPE Chemistry Unit 1 · Module 2: Kinetics and Equilibria · Section 5
5.1 Define Ksp 5.2 Common Ion Effect 5.3 Calculations 5.4 Selective Precipitation
1
💧 Solubility — Concepts and Terminology Objective 5.1 ▼
Solubility is the maximum mass (or moles) of a solute that will dissolve in a given volume of solvent at a specific temperature to produce a saturated solution. It is typically measured in mol dm⁻³ or g/100 g water.
Many ionic compounds dissolve in water, but some are only slightly soluble or appear to be insoluble. Even so-called 'insoluble' ionic compounds form some ions in solution — just to a very small extent. A solution is saturated when no more solute can dissolve.
Classification of Solubility
  • Soluble: Potassium chloride (KCl) — saturated solution contains 35.9 g/100 g water.
  • Sparingly soluble: Calcium hydroxide Ca(OH)₂ — saturated solution contains 0.113 g/100 g water.
  • 'Insoluble' (sparingly soluble): Silver chloride (AgCl) — saturated solution contains only 1.93 × 10⁻⁴ g/100 g water.
ℹ️
Note: No ionic compound is truly "zero-solubility." Even the least soluble salts release some ions into solution. Ksp quantifies this tiny but measurable dissolution.
2
⚖️ Solubility Product Ksp — Definition and Equilibrium Objective 5.1 ▼
When a sparingly soluble or 'insoluble' salt is added to water, an equilibrium is established between the ions in solution and the ions in the solid. Ions move from the solid to solution at the same rate they move from solution to solid.
Equilibrium for Silver Iodide
\[ \text{AgI}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{I}^-(aq) \]
The general equilibrium expression would be:
\[ K = \frac{[\text{Ag}^+(aq)][\text{I}^-(aq)]}{[\text{AgI}(s)]} \]
Since the concentration of a pure solid is constant (and therefore incorporated into K), we write the simplified expression:
\[ K_{sp} = [\text{Ag}^+][\text{I}^-] \]
Solubility Product (Ksp) is the product of the concentrations of each ion in a saturated solution of a sparingly soluble salt at 298 K (25°C), with each concentration raised to the power of its stoichiometric coefficient in the dissolution equation.
General Formula
For a salt that dissolves as: \(\text{M}_m\text{A}_n(s) \rightleftharpoons m\text{M}^{n+}(aq) + n\text{A}^{m-}(aq)\)
\[ K_{sp} = [\text{M}^{n+}]^m[\text{A}^{m-}]^n \]
Further Examples
Co(OH)₂
Ion ratio: Co²⁺ and 2OH⁻
\( K_{sp} = [\text{Co}^{2+}][\text{OH}^-]^2 \)
Al₂O₃
Ion ratio: 2Al³⁺ and 3O²⁻
\( K_{sp} = [\text{Al}^{3+}]^2[\text{O}^{2-}]^3 \)
Ca₃(PO₄)₂
Ion ratio: 3Ca²⁺ and 2PO₄³⁻
\( K_{sp} = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2 \)
PbI₂
Ion ratio: Pb²⁺ and 2I⁻
\( K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 \)
⚠️
Important: Ksp only applies when there is excess undissolved solid in equilibrium with the saturated solution. If all solid dissolves, it is not a Ksp situation.
3
📐 Units of Ksp Objective 5.1 ▼
Units for Ksp are calculated in the same way as for general equilibrium expressions involving Kc — by multiplying the units of each concentration term raised to its appropriate power.
Unit Examples
AgI: Ksp = [Ag⁺][I⁻]
Units = (mol dm⁻³) × (mol dm⁻³) = mol² dm⁻⁶
Co(OH)₂: Ksp = [Co²⁺][OH⁻]²
Units = (mol dm⁻³) × (mol dm⁻³)² = mol³ dm⁻⁹
Ca₃(PO₄)₂: Ksp = [Ca²⁺]³[PO₄³⁻]²
Units = (mol dm⁻³)³ × (mol dm⁻³)² = mol⁵ dm⁻¹⁵
General Rule
For Ksp = [cation]m[anion]n, the units are mol(m+n) dm−3(m+n)
📌
Examiner's note: Some textbooks and exam mark schemes treat Ksp as dimensionless (standard equilibrium constants have no units when activities are used). In CAPE, however, you are expected to include units in your Ksp expressions. Always check mark schemes.
4
🔗 The Common Ion Effect Objective 5.2 ▼
The common ion effect is the reduction in the solubility of a dissolved salt caused by adding a solution of a compound which has an ion in common with the dissolved salt.
The Principle
Ksp is a constant at a given temperature. If the concentration of one of the ions is increased by adding a common-ion source, the equilibrium shifts to the left (Le Chatelier's Principle), causing more precipitate to form and reducing the solubility of the salt.
Classic Example: BaSO₄ in H₂SO₄
\[ K_{sp}(\text{BaSO}_4) = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = 1.08 \times 10^{-10} \text{ mol}^2\text{dm}^{-6} \]
  • Solubility of BaSO₄ in pure water: \( \sqrt{1.08 \times 10^{-10}} \approx 1.04 \times 10^{-5} \) mol dm⁻³
  • In 0.10 mol dm⁻³ H₂SO₄, [SO₄²⁻] ≈ 0.10 mol dm⁻³
  • So [Ba²⁺] = 1.08 × 10⁻¹⁰ ÷ 0.10 = 1.08 × 10⁻⁹ mol dm⁻³ — about 10,000× less soluble!
Equilibrium Explanation
\[ \text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \]
Adding SO₄²⁻ ions shifts the equilibrium to the left. More BaSO₄ precipitates. The [Ba²⁺] must decrease to maintain Ksp.
Practical Importance in Chemistry
  • Gravimetric Analysis: In the quantitative determination of sulphate ions, BaSO₄ is precipitated by adding barium chloride. The precipitate is then washed with dilute sulphuric acid (rather than water) to exploit the common ion effect — preventing re-dissolution of BaSO₄ during washing.
  • Analytical Chemistry: The common ion effect is used to ensure completeness of precipitation reactions in quantitative work.
  • Industrial applications: Controlling crystallisation and separation processes in manufacturing.
🔬
The common ion effect is a direct consequence of Le Chatelier's Principle: adding a product to an equilibrium system shifts the equilibrium away from that product, increasing the reverse reaction rate.
5
🌧️ Predicting Precipitation — The Ionic Product Q Objective 5.4 ▼
Solubility product can be used to predict whether a precipitate will form when two solutions are mixed. We compare the ionic product (Q) with Ksp.
Ionic Product (Q) — also called the reaction quotient — is the product of the ion concentrations at any point, not just at equilibrium. It has the same mathematical form as Ksp.
Decision Rules
🌧️
Q > Ksp — Solution is supersaturated → precipitation occurs until Q = Ksp
⚖️
Q = Ksp — Solution is exactly saturated → system is at equilibrium
✅
Q < Ksp — Solution is unsaturated → no precipitation, more solid can dissolve
Selective Precipitation
When two or more ions in solution have different Ksp values for a common precipitating reagent, it is possible to selectively precipitate one ion while keeping the other in solution. This is the basis of many qualitative analysis techniques.
Example: Testing for Halide Ions with AgNO₃
  • Silver nitrate (AgNO₃) is added to an unknown solution containing halide ions.
  • Ag⁺ ions react with Cl⁻, Br⁻, and I⁻ to form precipitates of AgCl, AgBr, AgI.
  • These precipitates can be distinguished by their characteristic colours: AgCl (white), AgBr (cream/pale yellow), AgI (yellow).
  • Because silver halides have very low Ksp values, even a few drops of AgNO₃ causes precipitation.
\[ \text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s) \quad K_{sp} = 1.77 \times 10^{-10} \]
Kidney Stone Formation — A Biological Application
🫀
Kidney stones form when the solubility product of certain salts naturally present in urine is exceeded. If urine is too concentrated, calcium phosphate (Ca₃(PO₄)₂) and calcium oxalate (CaC₂O₄) may precipitate. Elevated calcium ion concentration shifts the equilibrium, favouring precipitation of insoluble calcium salts — particularly if conditions become too alkaline.
6
🧪 Determining Ksp by Experiment Objective 5.2 ▼
The experimental method is based on making a saturated solution of a particular substance and analysing it to find the concentration of one of the constituent ions.
Procedure — Determining Ksp for Ca(OH)₂
  • Add enough calcium hydroxide to a known volume of water so that solid is present as well as a solution. Shake and leave for 24 hours to reach equilibrium.
  • Filter off the solid calcium hydroxide carefully using filter paper.
  • Titrate measured samples of the calcium hydroxide solution with hydrochloric acid of known concentration.
  • Calculate the concentration of the calcium hydroxide solution from the titration results.
  • Substitute into the equilibrium expression: \( K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 \)
Calculation from Titration Data
If titration shows [Ca(OH)₂] = s mol dm⁻³, then:
\[ [\text{Ca}^{2+}] = s \qquad [\text{OH}^-] = 2s \]
\[ K_{sp} = s \times (2s)^2 = 4s^3 \]
📌
CAPE Practical Note: The syllabus suggests conducting a simple experiment to determine the solubility product of a substance. You should be prepared to plan, execute, and evaluate such an experiment in both school-based and external assessments.
7
✅ Key Points Summary All Objectives ▼
  • Ksp is an expression showing the equilibrium concentration of ions in a saturated solution of a sparingly soluble salt. It takes into account the relative number of each ion.
  • The value of Ksp can be calculated using the relevant equilibrium expression and the concentration of ions present at equilibrium.
  • The solubility of a sparingly soluble salt can be calculated using the value of Ksp and the formula of the salt.
  • The common ion effect refers to the reduction in solubility of a dissolved salt by adding a solution containing an ion in common with the dissolved salt.
  • When the ionic product Q exceeds Ksp, a salt is precipitated.
  • The solubility product of a salt can be found by determining the concentration of the ions in solution by titration or other methods.
  • Ksp is temperature-dependent — it only applies at the temperature specified (usually 298 K).
Ksp → Solubility
Given Ksp, find s:
Express ions in terms of s
Solve algebraically
Solubility → Ksp
Given s, find Ksp:
Find [ion] from s
Substitute into expression
🧮 Worked Calculations
Step-by-step solutions covering all CAPE Ksp calculation types
A
📊 Type A: Calculating Ksp from Solubility Obj 5.3 ▼
Q1 A saturated solution of lead iodide, PbI₂, contains 0.076 g of PbI₂ in 100 g solution. Calculate Ksp for lead iodide. (Mr of PbI₂ = 461.0 g mol⁻¹)
  • 1
    Calculate the concentration of PbI₂ in mol dm⁻³:
    \[ c = \frac{0.076}{461.0} \times \frac{1000}{100} = 1.65 \times 10^{-3} \text{ mol dm}^{-3} \]
  • 2
    Write the dissolution equation and identify ion concentrations:
    \[ \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \]
    \([\text{Pb}^{2+}] = 1.65 \times 10^{-3} \text{ mol dm}^{-3}\)
    \([\text{I}^-] = 2 \times 1.65 \times 10^{-3} = 3.30 \times 10^{-3} \text{ mol dm}^{-3}\)
  • 3
    Write the Ksp expression:
    \[ K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 \]
  • 4
    Substitute and calculate:
    \[ K_{sp} = (1.65 \times 10^{-3}) \times (3.30 \times 10^{-3})^2 = (1.65 \times 10^{-3}) \times (1.089 \times 10^{-5}) \]
    Ksp = 1.80 × 10⁻⁸ mol³ dm⁻⁹
Q2 The solubility of calcium fluoride, CaF₂, is 2.15 × 10⁻⁴ mol dm⁻³ at 25°C. Calculate Ksp.
  • 1
    Write the dissolution equation:
    \[ \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq) \]
    Let solubility = s = 2.15 × 10⁻⁴ mol dm⁻³
    \([\text{Ca}^{2+}] = s\) and \([\text{F}^-] = 2s\)
  • 2
    Ksp expression: \(K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = s \times (2s)^2 = 4s^3\)
  • 3
    Calculate:
    \[ K_{sp} = 4 \times (2.15 \times 10^{-4})^3 = 4 \times 9.94 \times 10^{-12} \]
    Ksp = 3.98 × 10⁻¹¹ mol³ dm⁻⁹ ≈ 3.45 × 10⁻¹¹ (literature value)
B
📈 Type B: Calculating Solubility from Ksp Obj 5.3 ▼
Q1 Calculate the solubility of calcium sulphate, CaSO₄, in mol dm⁻³. Ksp(CaSO₄) = 4.93 × 10⁻⁵ mol² dm⁻⁶.
  • 1
    Write the equilibrium: \(\text{CaSO}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{SO}_4^{2-}(aq)\)
  • 2
    Let solubility = s. Then [Ca²⁺] = [SO₄²⁻] = s, so Ksp = s × s = s²
  • 3
    \[ s = \sqrt{K_{sp}} = \sqrt{4.93 \times 10^{-5}} \]
    s = 7.02 × 10⁻³ mol dm⁻³
Q2 Calculate the solubility of magnesium hydroxide, Mg(OH)₂, in mol dm⁻³. Ksp(Mg(OH)₂) = 5.61 × 10⁻¹² mol³ dm⁻⁹.
  • 1
    Dissolution: \(\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq)\)
  • 2
    Let solubility = s: [Mg²⁺] = s, [OH⁻] = 2s
    Ksp = (s)(2s)² = 4s³
  • 3
    \[ s = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{5.61 \times 10^{-12}}{4}} = \sqrt[3]{1.40 \times 10^{-12}} \]
    s = 1.12 × 10⁻⁴ mol dm⁻³
C
🌧️ Type C: Predicting Precipitation using Ionic Product Q Obj 5.4 ▼
Q1 Will a precipitate form when 6.00 × 10⁻⁵ mol dm⁻³ BaCl₂ is added to an equal volume of 1.20 × 10⁻⁵ mol dm⁻³ Na₂SO₄? Ksp(BaSO₄) = 1.08 × 10⁻¹⁰ mol² dm⁻⁶.
  • 1
    After mixing equal volumes, concentrations halve:
    [Ba²⁺] = 6.00 × 10⁻⁵ ÷ 2 = 3.00 × 10⁻⁵ mol dm⁻³
    [SO₄²⁻] = 1.20 × 10⁻⁵ ÷ 2 = 6.00 × 10⁻⁶ mol dm⁻³
  • 2
    Calculate ionic product Q:
    \[ Q = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = (3.00 \times 10^{-5})(6.00 \times 10^{-6}) = 1.80 \times 10^{-10} \]
  • 3
    Compare Q with Ksp:
    Q = 1.80 × 10⁻¹⁰ > Ksp = 1.08 × 10⁻¹⁰
    ✓ Q > Ksp → A precipitate of BaSO₄ WILL form.
Q2 What minimum concentration of Ag⁺ ions is needed to precipitate AgCl from a solution containing [Cl⁻] = 1.0 × 10⁻⁴ mol dm⁻³? Ksp(AgCl) = 1.77 × 10⁻¹⁰ mol² dm⁻⁶.
  • 1
    Ksp expression: \(K_{sp} = [\text{Ag}^+][\text{Cl}^-]\)
  • 2
    Rearrange:
    \[ [\text{Ag}^+] = \frac{K_{sp}}{[\text{Cl}^-]} = \frac{1.77 \times 10^{-10}}{1.0 \times 10^{-4}} \]
    [Ag⁺]min = 1.77 × 10⁻⁶ mol dm⁻³
D
🔗 Type D: Solubility with the Common Ion Effect Obj 5.2 & 5.3 ▼
Q1 Calculate the solubility of BaSO₄ in 0.10 mol dm⁻³ H₂SO₄ solution. Ksp(BaSO₄) = 1.08 × 10⁻¹⁰ mol² dm⁻⁶.
  • 1
    Write the equilibrium: \(\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)\)
    Ksp = [Ba²⁺][SO₄²⁻] = 1.08 × 10⁻¹⁰
  • 2
    The H₂SO₄ provides [SO₄²⁻] = 0.10 mol dm⁻³ (the common ion). Let solubility of BaSO₄ = s mol dm⁻³.
    Since s ≪ 0.10, total [SO₄²⁻] ≈ 0.10 mol dm⁻³
  • 3
    \[ 1.08 \times 10^{-10} = [\text{Ba}^{2+}] \times 0.10 \]
    \[ [\text{Ba}^{2+}] = s = \frac{1.08 \times 10^{-10}}{0.10} \]
    s = 1.08 × 10⁻⁹ mol dm⁻³ (compare to 1.04 × 10⁻⁵ mol dm⁻³ in pure water — ~10,000× reduction!)
Q2 Calculate the solubility of AgCl in 0.050 mol dm⁻³ NaCl solution. Ksp(AgCl) = 1.77 × 10⁻¹⁰ mol² dm⁻⁶.
  • 1
    NaCl provides [Cl⁻] = 0.050 mol dm⁻³. Solubility of AgCl = s.
    Total [Cl⁻] ≈ 0.050 mol dm⁻³ (since s ≪ 0.050)
  • 2
    \[ s = [\text{Ag}^+] = \frac{K_{sp}}{[\text{Cl}^-]} = \frac{1.77 \times 10^{-10}}{0.050} \]
    s = 3.54 × 10⁻⁹ mol dm⁻³ (vs 1.33 × 10⁻⁵ mol dm⁻³ in pure water)
E
⚗️ Type E: Selective Precipitation Obj 5.4 ▼
Q1 A solution contains [Cl⁻] = 0.10 mol dm⁻³ and [Br⁻] = 0.10 mol dm⁻³. AgNO₃ solution is added dropwise. Which halide precipitates first? Ksp(AgCl) = 1.77 × 10⁻¹⁰, Ksp(AgBr) = 5.35 × 10⁻¹³ mol² dm⁻⁶.
  • 1
    Find [Ag⁺] needed to just begin precipitating each halide:
    \[ [\text{Ag}^+]_{\text{for AgCl}} = \frac{K_{sp}(\text{AgCl})}{[\text{Cl}^-]} = \frac{1.77 \times 10^{-10}}{0.10} = 1.77 \times 10^{-9} \text{ mol dm}^{-3} \]
    \[ [\text{Ag}^+]_{\text{for AgBr}} = \frac{K_{sp}(\text{AgBr})}{[\text{Br}^-]} = \frac{5.35 \times 10^{-13}}{0.10} = 5.35 \times 10^{-12} \text{ mol dm}^{-3} \]
  • 2
    AgBr requires a lower [Ag⁺] to begin precipitation (5.35 × 10⁻¹² < 1.77 × 10⁻⁹).
    AgBr precipitates first. AgCl remains in solution until [Ag⁺] reaches 1.77 × 10⁻⁹ mol dm⁻³.
  • 3
    When AgCl just begins to precipitate, find remaining [Br⁻]:
    \[ [\text{Br}^-] = \frac{K_{sp}(\text{AgBr})}{[\text{Ag}^+]} = \frac{5.35 \times 10^{-13}}{1.77 \times 10^{-9}} = 3.02 \times 10^{-4} \text{ mol dm}^{-3} \]
    97% of Br⁻ has been removed before Cl⁻ begins to precipitate — good separation is possible!
Q2 A solution contains [Ca²⁺] = 0.050 mol dm⁻³ and [Mg²⁺] = 0.050 mol dm⁻³. NaOH is added. Which hydroxide precipitates first? Ksp(Ca(OH)₂) = 5.02 × 10⁻⁶, Ksp(Mg(OH)₂) = 5.61 × 10⁻¹² mol³ dm⁻⁹.
  • 1
    Find [OH⁻] needed to begin precipitating each:
    \[ [\text{OH}^-]_{\text{for Ca(OH)}_2} = \sqrt{\frac{K_{sp}}{[\text{Ca}^{2+}]}} = \sqrt{\frac{5.02 \times 10^{-6}}{0.050}} = \sqrt{1.004 \times 10^{-4}} = 1.00 \times 10^{-2} \text{ mol dm}^{-3} \]
    \[ [\text{OH}^-]_{\text{for Mg(OH)}_2} = \sqrt{\frac{5.61 \times 10^{-12}}{0.050}} = \sqrt{1.12 \times 10^{-10}} = 1.06 \times 10^{-5} \text{ mol dm}^{-3} \]
  • 2
    Mg(OH)₂ requires less OH⁻ to precipitate.
    Mg(OH)₂ precipitates first. This allows separation of Mg²⁺ from Ca²⁺ by careful pH control.
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CAPE Chemistry Unit 1 — Module 2: Kinetics and Equilibria
Specific Objectives 5.1 – 5.4 · Solubility Product
Objectives 5.1–5.4 100+ Ksp Values Interactive Simulator Step-by-Step Solver

Ksp values sourced from IUPAC Solubility Data Series, NIST Thermochemical Tables, and peer-reviewed literature. Temperature: 25°C (298 K) unless stated.