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Predict precipitation after mixing

Key idea: After mixing, divide each ion's amount by the total volume and calculate the ionic product Q. A precipitate forms if Q exceeds Ksp; equality marks saturation.

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  • GCE A-Level H2 Chemistry 2027
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Mix a solution of silver nitrate with a solution of sodium chloride and a white precipitate usually appears, but not always: if both solutions are dilute enough, nothing happens. The solubility product decides. This lesson calculates the ion concentrations in a mixture and compares them with Kₛₚ to predict whether a precipitate forms.

The ionic product

For any mixture, the ionic product, Q, has the same form as Kₛₚ but uses the ion concentrations actually present, before any solid forms. For silver chloride, Q = [Ag⁺][Cl⁻]; for calcium fluoride, Q = [Ca²⁺][F⁻]².

  • If Q > Kₛₚ, the solution holds more ions than a saturated solution can. Solid precipitates until the ionic product falls to Kₛₚ.
  • If Q < Kₛₚ, the solution is not saturated, and no precipitate forms.
  • If Q = Kₛₚ, the solution is exactly saturated.

Check your understanding 1

A solution contains no solid and has Q exactly equal to Ksp. Which conclusion follows from the ion product?

At Q = Ksp
Show answer

The solution is saturated. Precipitation is favoured when Q exceeds Ksp. Equality alone does not specify an observable amount of solid.

Concentrations after mixing

Use a fixed temperature, additive solution volumes and fully dissociated salts, with no other reaction consuming the relevant ions. Mixing two solutions dilutes both. For each ion, find the amount from its own solution and divide by the total volume:

[ion]ₘᵢₓₜᵤᵣₑ = (c × Vₛₒₗᵤₜᵢₒₙ)/Vₜₒₜₐₗ

If the volumes are equal, each concentration halves. If they are not, each ion is diluted by a different factor.

Worked example 1

Mixing unequal volumes

Problem

20.0 cm³ of 1.0 × 10⁻³ mol dm⁻³ AgNO₃ is mixed with 80.0 cm³ of 1.0 × 10⁻⁶ mol dm⁻³ NaCl. Kₛₚ(AgCl) = 1.8 × 10⁻¹⁰ mol² dm⁻⁶. Does a precipitate form?
Show full solution

  1. Find the amount of each ion

    Method

    Multiply concentration by volume for each solution.

    Reason

    Each salt gives one of the relevant ion per formula unit.

    Working

    Answer

    n(Ag⁺) = 0.0200 × 1.0 × 10⁻³ = 2.0 × 10⁻⁵ mol; n(Cl⁻) = 0.0800 × 1.0 × 10⁻⁶ = 8.0 × 10⁻⁸ mol

  2. Divide by the total volume

    Method

    The mixture has a volume of 0.1000 dm³.

    Reason

    Both ions are now spread through the whole mixture.

    Working

    Answer

    [Ag⁺] = 2.0 × 10⁻⁴; [Cl⁻] = 8.0 × 10⁻⁷ mol dm⁻³

  3. Compare Q with Ksp

    Method

    No precipitate forms.

    Reason

    The ionic product is below Kₛₚ.

    Working

    Answer

    Q = (2.0 × 10⁻⁴)(8.0 × 10⁻⁷) = 1.6 × 10⁻¹⁰ < 1.8 × 10⁻¹⁰

Guided practice 2

Mixing equal volumes

About 6 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. Kₛₚ(AgCl) = 1.8 × 10⁻¹⁰ mol² dm⁻⁶. Does a precipitate form?

Try this before viewing the solution

Show the steps Optional working
Unit: dm³
Unit: mol dm⁻³
Unit: mol dm⁻³
Unit: mol² dm⁻⁶
Does AgCl precipitate?

Hints

Hint 1

dilution

What is the total volume containing each ion after mixing?
Hint 2

compare

Calculate Q = [Ag⁺][Cl⁻] and compare it with Kₛₚ.
Show solution step by step

  1. Concentrations after mixing

    Method

    Halve each concentration.

    Reason

    Each solution doubles in volume.

    Working

    Answer

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

  2. Calculate Q

    Method

    Multiply the two concentrations.

    Reason

    Q has the same form as Kₛₚ.

    Working

    Answer

    Q = (5.0 × 10⁻⁴)² = 2.5 × 10⁻⁷

  3. Compare

    Method

    A precipitate of AgCl forms.

    Reason

    Q is far greater than Kₛₚ.

    Working

    Answer

    2.5 × 10⁻⁷ > 1.8 × 10⁻¹⁰

The concentration needed to start precipitation

The saturation boundary is Q = Kₛₚ. Above it, precipitation is favoured; exactly at it, the solution is saturated. Rearranging gives the threshold concentration of one ion for a known free concentration of the other. A visible precipitate also depends on nucleation and the amount of solid formed.

Try it yourself 3

When does AgCl start to precipitate?

Minimal support

Problem

Chloride ions are added slowly to a solution with [Ag⁺] = 2.0 × 10⁻⁴ mol dm⁻³. Ignore any change in volume. Kₛₚ(AgCl) = 1.8 × 10⁻¹⁰ mol² dm⁻⁶. Calculate the chloride concentration at the saturation threshold for AgCl.

Try this before viewing the solution

Unit: mol dm⁻³

Hints

Hint 1

the condition

Which ion-product condition marks saturation?
Hint 2

rearrange

Divide Kₛₚ by the silver-ion concentration.
Show solution step by step

  1. Set the condition

    Method

    Set the ionic product equal to Kₛₚ.

    Reason

    Below this point the solution is unsaturated; above it, solid forms.

    Working

    Answer

    [Ag⁺][Cl⁻] = 1.8 × 10⁻¹⁰

  2. Solve

    Method

    Divide by [Ag⁺].

    Reason

    The silver concentration is known.

    Working

    Answer

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

Check your understanding 2

Show the saturation condition, rearrangement and unit for your threshold answer. Explain the difference between a saturated solution and a visible precipitate.

Show answer

At the boundary [Ag⁺][Cl⁻] = Kₛₚ, so [Cl⁻] = (1.8 × 10⁻¹⁰)/(2.0 × 10⁻⁴) = 9.0 × 10⁻⁷ mol dm⁻³. The concentration-product unit divided by a concentration leaves a concentration. Below this boundary the solution is unsaturated; above it precipitation is favoured. Equality marks saturation, while visible solid requires enough precipitation and nucleation.

Check your understanding 3

50.0 cm³ of 2.0 × 10⁻³ mol dm⁻³ calcium nitrate is mixed with 50.0 cm³ of 2.0 × 10⁻⁴ mol dm⁻³ sodium fluoride. Kₛₚ(CaF₂) = 3.9 × 10⁻¹¹ mol³ dm⁻⁹. Does calcium fluoride precipitate?

Show the steps Optional working
Unit: mol dm⁻³
Unit: mol³ dm⁻⁹
Does CaF₂ precipitate?
Show answer

After mixing, [Ca²⁺] = 1.0 × 10⁻³ and [F⁻] = 1.0 × 10⁻⁴ mol dm⁻³. Then Q = [Ca²⁺][F⁻]² = (1.0 × 10⁻³)(1.0 × 10⁻⁴)² = 1.0 × 10⁻¹¹. This is less than 3.9 × 10⁻¹¹, so no precipitate forms. Forgetting to square [F⁻] would give 1.0 × 10⁻⁷ and the wrong conclusion.

Check your understanding 4

Show the full calcium fluoride mixture calculation: amounts or dilution factors, both concentrations, the ionic product with its power, and the conclusion.

Show answer

Each 50.0 cm³ solution is diluted to 100.0 cm³. Calcium becomes 1.0 × 10⁻³ and fluoride 1.0 × 10⁻⁴ mol dm⁻³. Sodium fluoride gives one fluoride ion per formula unit; the CaF₂ dissolution coefficient sets the power in Q, not a doubling of the initial fluoride concentration. Q = (1.0 × 10⁻³)(1.0 × 10⁻⁴)² = 1.0 × 10⁻¹¹ mol³ dm⁻⁹ < Kₛₚ, so no precipitate forms.

Common mistakes

  • Using the concentrations of the solutions before they were mixed.
  • Halving every concentration when the two volumes are not equal.
  • Leaving out the powers in Q, for example not squaring [F⁻] for CaF₂.
  • Comparing Q with the solubility, s, instead of with Kₛₚ.
  • Thinking a sparingly soluble salt precipitates whenever its ions meet. It precipitates only when Q > Kₛₚ.

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