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Ksp and Solubility Calculations

Learn and apply Ksp and Solubility Calculations in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Ksp and Solubility Calculations: Orientation

In Kₛₚ questions, the marks come from doing the setup correctly: write the dissolution equilibrium, write the right Kₛₚ expression, then link ion concentrations to solubility s. This lesson gives a repeatable workflow and the Qₛₚ vs Kₛₚ precipitation test.

If the acid-base foundations feel rusty, revisit Acids and Bases (Theories) and keep the Aqueous Equilibria hub open for linked methods.

Definitions (Must Know)

A. Solubility product, Kₛₚ

For a sparingly soluble salt:

MaXb(s) ⇌ aMb⁺(aq) + bXa⁻(aq)
Kₛₚ = [Mb⁺]^a[Xa⁻]^b

B. Ionic product, Qₛₚ

The ionic product, Qₛₚ, is calculated using the current ion concentrations (not necessarily equilibrium). Comparing Qₛₚ with Kₛₚ tells you whether a precipitate forms.

Detailed Explanations

A. Workflow: writing the Kₛₚ expression (method marks)

  1. Write the dissolution equilibrium with correct charges and coefficients.
  2. Omit the pure solid from the expression.
  3. Raise each ion concentration to the power of its coefficient.

Mini example: CaF₂(s) ⇌ Ca²⁺(aq) + 2F-(aq) So Kₛₚ = [Ca²⁺][F⁻]².

B. Workflow: solubility from Kₛₚ (typical exam method)

If MX(s) ⇌ M + (aq) + X-(aq) and solubility is s:

Kₛₚ = s²

If MX₂(s) ⇌ M²⁺(aq) + 2X-(aq) and solubility is s:

Kₛₚ = s(2s)² = 4s³

Because Kₛₚ is an equilibrium constant at a fixed temperature, therefore once the solution is saturated you must have Qₛₚ = Kₛₚ using the equilibrium ion concentrations.

C. Workflow: precipitation prediction (Qₛₚ vs Kₛₚ)

  1. Find ion concentrations after mixing/dilution (use total volume).
  2. Calculate Qₛₚ from those concentrations.
  3. Compare:
    • because Qₛₚ > Kₛₚ means the ion product is too large, therefore ions must be removed → precipitate forms
    • if Qₛₚ < Kₛₚ, no precipitate forms

Worked Examples

Modelled example 1

Find molar solubility for a one-to-one salt

Core

Problem

Given Kₛₚ(AgCl) = 1.8 × 10⁻¹⁰, find the solubility of AgCl in pure water in mol dm⁻³.
Study the worked solution

  1. Write the dissolution equilibrium

    Method

    Dissociate one formula unit into its aqueous ions.

    Reason

    The coefficients determine how molar solubility maps to equilibrium concentrations.

    Working

    AgCl(s) ⇌ Ag + (aq) + Cl-(aq).

  2. Express concentrations using s

    Method

    Let the molar solubility be s.

    Reason

    Each mole of dissolved AgCl forms one mole of each ion.

    Working

    [Ag⁺] = s, [Cl⁻] = s, so Kₛₚ = s².

  3. Solve and check

    Method

    Take the positive square root.

    Reason

    Concentration cannot be negative.

    Working

    s = square root of (1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol dm⁻³.

Guided practice 2

Predict precipitation after mixing

About 8 min

Problem

50.0 cm³ of 1.0 × 10⁻³ mol dm⁻³ AgNO₃ is mixed with 50.0 cm³ of 1.0 × 10⁻³ mol dm⁻³ NaCl. Given Kₛₚ(AgCl) = 1.8 × 10⁻¹⁰, determine whether AgCl precipitates.

Try this before viewing the solution

Hints

Hint 1: account for total volume
Each 50.0 cm³ solution is diluted to a total of 100.0 cm³ on mixing.
Hint 2: compare the ion product
Calculate Qₛₚ = [Ag⁺][Cl⁻] using the diluted concentrations, then compare it with Kₛₚ.
View solution step by step

  1. Find concentrations immediately after mixing

    Method

    Apply the twofold dilution to both ions.

    Reason

    Each solute occupies twice its original volume before any precipitation is considered.

    Working

    [Ag⁺] = [Cl⁻] = 5.0 × 10⁻⁴ mol dm⁻³.

  2. Calculate the trial ion product

    Method

    Use the same concentration powers as the Kₛₚ expression.

    Reason

    Qₛₚ tests whether the mixed solution contains more free ions than saturation permits.

    Working

    Qₛₚ = (5.0 × 10⁻⁴)² = 2.5 × 10⁻⁷.

  3. Compare and conclude

    Method

    Compare Qₛₚ with Kₛₚ.

    Reason

    The ion product is much larger than the equilibrium value, so ions must be removed from solution.

    Working

    2.5 × 10⁻⁷ is greater than 1.8 × 10⁻¹⁰; therefore AgCl(s) precipitates.

Common misconception 3

Carry coefficients into the Ksp expression

Find and correct the mistake

Learner claim

Given Kₛₚ(CaF₂) = 3.9 × 10⁻¹¹, a learner sets both ion concentrations equal to solubility s, writes Kₛₚ = s², and takes a square root. Locate the first error and calculate the correct molar solubility.

Choose the concentration mapping

If CaF2 solubility is s

View solution step by step

  1. Correct the dissolution mapping

    Method

    Use the coefficient two for fluoride.

    Reason

    CaF₂(s) ⇌ Ca²⁺(aq) + 2F-(aq) forms twice as many fluoride ions.

    Working

    [Ca²⁺] = s and [F⁻] = 2s.

  2. Build the Ksp expression

    Method

    Raise fluoride concentration to its stoichiometric power.

    Reason

    The exponent comes from the balanced dissolution equation.

    Working

    Kₛₚ = [Ca²⁺][F⁻]² = s(2s)² = 4s³.

  3. Solve the cubic

    Method

    Divide by four and take the positive cube root.

    Reason

    The concentration relationship is cubic, not quadratic.

    Working

    s = ((3.9 × 10⁻¹¹)/4)^(1/3) = 2.14 × 10⁻⁴ mol dm⁻³.

Examiner practice 4

Calculate Ksp from measured solubility

4 marks

Problem

The solubility of Mg(OH)₂ in water is 1.20 × 10⁻⁴ mol dm⁻³. Calculate Kₛₚ. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Write the dissolution equation

    1 mark

    Method

    Balance the ions released by one formula unit.

    Reason

    The coefficients determine concentrations and powers.

    Working

    Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH-(aq).

  2. Map solubility to ions

    1 mark

    Reason

    One mole of solid produces one mole of magnesium ions and two moles of hydroxide ions.

    Working

    [Mg²⁺] = s and [OH⁻] = 2s.

  3. State the expression

    1 mark

    Reason

    The pure solid is omitted and hydroxide has power two.

    Working

    Kₛₚ = s(2s)² = 4s³.

  4. Calculate Ksp

    1 mark

    Method

    Substitute the measured molar solubility.

    Reason

    The equilibrium ion concentrations derive from that saturated-solution value.

    Working

    Kₛₚ = 4(1.20 × 10⁻⁴)³ = 6.91 × 10⁻¹².

Challenge 5

Find the concentration at precipitation onset

Minimal support

Problem

A solution contains 2.0 × 10⁻⁴ mol dm⁻³ Ag⁺. Given Kₛₚ(AgCl) = 1.8 × 10⁻¹⁰, calculate the free Cl⁻ concentration at which AgCl just begins to precipitate. State what happens above this concentration.

Try this before viewing the solution

Hints

Hint 1: use the boundary condition
At the instant precipitation begins, the ion product has just reached Kₛₚ.
Hint 2: make chloride the unknown
Set [Ag⁺][Cl⁻] = 1.8 × 10⁻¹⁰ and divide by the given silver-ion concentration.
View solution step by step

  1. Set the saturation boundary

    Method

    Use equality at precipitation onset.

    Reason

    Below saturation no solid forms; at the boundary Qₛₚ = Kₛₚ.

    Working

    [Ag⁺][Cl⁻] = 1.8 × 10⁻¹⁰.

  2. Solve for chloride

    Method

    Divide by the fixed silver-ion concentration.

    Reason

    The AgCl expression has one-to-one concentration powers.

    Working

    [Cl⁻] = (1.8 × 10⁻¹⁰)/(2.0 × 10⁻⁴) = 9.0 × 10⁻⁷ mol dm⁻³.

  3. Interpret concentrations above the boundary

    Method

    Compare a larger chloride concentration with the threshold.

    Reason

    It would make Qₛₚ exceed Kₛₚ.

    Working

    Above 9.0 × 10⁻⁷ mol dm⁻³, AgCl precipitates until the free-ion product returns to Kₛₚ.

Mind Stretchers

Mind stretcher 1Extension

Kₛₚ(AgCl) = 1.8 × 10⁻¹⁰ and Kₛₚ(Ag₂CrO₄) = 1.1 × 10⁻¹². Which salt is more soluble in pure water? (Show a solubility calculation for both.)

Show Answer

Mark scheme:

  • AgCl(s) ⇌ Ag + (aq) + Cl-(aq): Kₛₚ = s² ⇒ s = square root of (1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵
  • Ag₂CrO₄(s) ⇌ 2Ag + (aq) + CrO₄²⁻(aq): Kₛₚ = (2s)²(s) = 4s³
  • s = ((1.1 × 10⁻¹²)/4)^(1/3) = 6.50 × 10⁻⁵
  • Therefore Ag₂CrO₄ is more soluble (stoichiometry matters, not just Kₛₚ size).