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Solubility Equilibria Calculations

Chemistry: Cambridge International A Level, topic 1.7.8

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.

Example

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)β‡Œπ‘Mm+(aq)+π‘žXnβˆ’(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.
Example calculation

Example calculation

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

  • CaF2(s)β‡ŒCa2+(aq)+2Fβˆ’(aq)

  • The concentration of Ca2+ in a saturated solution of CaF2 is 2.15 Γ— 10–4 M.

  • What is the solubility product of fluorite?

Example continued

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:

  • [Fβˆ’]=(2 mol Fβˆ’/1 mol Ca2+)=

  • (2) x (2.15Γ—10βˆ’4𝑀) =

  • 4.30Γ—10βˆ’4𝑀

  • Substituting the ion concentrations into the 𝐾sp expression gives:

  • 𝐾sp = [Ca2+][Fβˆ’]2 =

  • (2.15Γ—10βˆ’4) x (4.30Γ—10βˆ’4)2=

  • 3.98Γ—10βˆ’11

Example 2

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 6.3Γ—10–9. Calculate the molar solubility of copper bromide.
Example 2 - continued

Example 2 - continued

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

Example 2 - continued

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

  • 𝐾sp=[Cu+][Brβˆ’] = 6.3Γ—10βˆ’9

  • 6.3Γ—10βˆ’9=(π‘₯)(π‘₯) = π‘₯2

  • π‘₯ = √(6.3Γ—10βˆ’9) = 7.9Γ—10βˆ’5𝑀

Test yourself on Solubility Equilibria Calculations

  • Mg(OH)2 is dissolved in solution and the [Mg2+] is determined to be 0.05𝑀. What is [OH-]?

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