1.7.8

Solubility Equilibria Calculations

Test yourself

𝐾sp & Solubility

The 𝐾sp of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation.

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Example

  • The 𝐾sp of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation.
  • For example:
    • MpXq(s)⇌𝑝Mm+(aq)+𝑞Xn−(aq)
  • For cases such as these, one may derive 𝐾sp values from provided solubilities, or vice-versa.
    • These calculations are most conveniently performed using a compound’s molar solubility, measured as moles of dissolved solute per liter of saturated solution.
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Example calculation

  • Fluorite, CaF2, is a slightly soluble solid that dissolves according to the equation:

    • CaF2(s)⇌Ca2+(aq)+2F(aq)
  • The concentration of Ca2+ in a saturated solution of CaF2 is 2.15 × 10–4 M.

    • What is the solubility product of fluorite?
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Example continued

  • According to the stoichiometry of the dissolution equation, the fluoride ion molarity of a CaF2 solution is equal to twice its calcium ion molarity:

    • [F]=(2 mol F/1 mol Ca2+)=
    • (2) x (2.15×10−4𝑀) =
    • 4.30×10−4𝑀
  • Substituting the ion concentrations into the 𝐾sp expression gives:
    • 𝐾sp = [Ca2+][F]2 =
    • (2.15×10−4) x (4.30×10−4)2=
    • 3.98×10−11
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Example 2

  • Determination of molar solubility from 𝐾sp requires the construction of ICE tables.
  • Let's go through an example:
    • The 𝐾sp of copper(I) bromide, CuBr, is 6.3×10–9. Calculate the molar solubility of copper bromide.
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Example 2 - continued

  1. Construct the dissolution equation and solubility product expression:
    • CuBr(s) ⇌ Cu+(aq)+Br(aq)
    • 𝐾sp=[Cu+][Br]
  2. Following the ICE approach to this calculation yields the above table.
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Example 2 - continued

  • Substituting the equilibrium concentration terms into the solubility product expression and solving for x yields:

    • 𝐾sp=[Cu+][Br] = 6.3×10−9
    • 6.3×10−9=(𝑥)(𝑥) = 𝑥2
    • 𝑥 = √(6.3×10−9) = 7.9×10−5𝑀

Jump to other topics

1Physical Chemistry

1.1Atoms, Molecules & Stoichiometry

1.2Atomic Structure

1.3Chemical Bonding

1.4States of Matter

1.5Chemical Energetics

1.6Electrochemistry

1.7Equilibria

1.8Partition Coefficient

1.9Reaction Kinetics

2Inorganic Chemistry

3Organic Chemistry & Analysis

3.1Introduction to Organic Chemistry

3.2Hydrocarbons

3.3Halogen Derivatives

3.4Hydroxy Compounds

3.5Carbonyl Compounds

3.6Carboxylic Acids & Derivatives

3.7Nitrogen Compounds

3.8Polymerisation

3.9Analytical Techniques

3.10Organic Synthesis

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