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Oceans

Predicting precipitation and the common ion effect

  • 2 things to know
  • Components 01, 02 and 03

Overview

Recall questions on comparing the ionic product with Ksp, halving concentrations on mixing, and why a common ion reduces solubility.

What you need to know

Answer each one before you reveal it.

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Numbers

Why must the concentrations be halved when equal volumes of two solutions are mixed?

Process

How does a fall in ocean pH threaten shells made of calcium carbonate?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Ksp for calcium carbonate is 5.0 × 10⁻⁹ mol² dm⁻⁶. Calculate its solubility in mol dm⁻³ and in g dm⁻³, given Mᵣ = 100.1.

    Model answer

    s = √(5.0 × 10⁻⁹) = 7.1 × 10⁻⁵ mol dm⁻³, which is 7.1 × 10⁻⁵ × 100.1 = 7.1 × 10⁻³ g dm⁻³.

  2. 2

    Equal volumes of 2.0 × 10⁻³ mol dm⁻³ silver nitrate and 2.0 × 10⁻³ mol dm⁻³ sodium chloride are mixed. Will silver chloride precipitate, given Ksp = 1.7 × 10⁻¹⁰ mol² dm⁻⁶?

    Model answer

    After mixing each ion is at 1.0 × 10⁻³ mol dm⁻³, giving an ionic product of 1.0 × 10⁻⁶ mol² dm⁻⁶, which exceeds Ksp, so a precipitate forms.

  3. 3

    Why must the concentrations be halved when equal volumes of two solutions are mixed?

    Model answer

    Each solution is diluted into twice its original volume, so the amount of each ion is spread through double the volume.

  4. 4

    What does it mean if the ionic product of a mixture is smaller than Ksp?

    Model answer

    The solution is unsaturated, so no precipitate forms and any solid present would dissolve.

  5. 5

    Calculate the solubility of silver chloride in 0.10 mol dm⁻³ sodium chloride, given Ksp = 1.7 × 10⁻¹⁰ mol² dm⁻⁶.

    Model answer

    [Ag⁺] = Ksp/[Cl⁻] = 1.7 × 10⁻¹⁰/0.10 = 1.7 × 10⁻⁹ mol dm⁻³, far below the value of 1.3 × 10⁻⁵ mol dm⁻³ in pure water.

  6. 6

    Why does adding a common ion reduce the solubility of a sparingly soluble salt?

    Model answer

    Increasing the concentration of one ion shifts the dissolving equilibrium back towards the solid so that the product of the ion concentrations still equals Ksp.

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