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Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 4.2

4.2 Solubility Product Ksp and common ion effect

What do we mean by 'solubility product'? What is the 'common ion effect'? How do you write Ksp equilibrium expressions? How do you do solubility product calculations?

(c) doc b GCSE/IGCSE Notes on reversible reactions and chemical equilibrium

Part 4 sub–index 4.1 Partition between two phases * 4.2 Solubility product Ksp & common ion effect * 4.3 Ion–exchange systems

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. Kc and Kp equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * Part 4 sub–index (this section): 4.1 Partition between two phases * 4.2 Solubility product Ksp and common ion effect * 4.3 Ion–exchange systems * Part 5. pH, weak–strong acid–base theory and calculations * Part 6. Salt hydrolysis, Acid–base titrations–indicators, pH curves and buffers * Part 7. Redox equilibria, half–cell electrode potentials, electrolysis and electrochemical series * Part 8. Phase equilibria–vapour pressure, boiling point and intermolecular forces * M = old fashioned shorthand for mol dm–3

4.2 Solubility Product and the Common Ion Effect

  • When a sparingly soluble salt is mixed with water a dynamic equilibrium is established in which salt is constantly dissolving and crystallising at the same rate when the solution is saturated, and the maximum constant concentration is achieved.

  • e.g. for calcium sulphate: CaSO4(s) + aq (c) doc b Ca2+(aq) + SO42–(aq)

    • the equilibrium expression is: K = [Ca2+(aq)] [SO42–(aq)] / [ CaSO4(s)]

  • However, since the concentration of water and the solid is effectively constant the equilibrium expression is simplified to:

    • Ksp = [Ca2+(aq)] [SO42–(aq)] = 2.4 x 10–5 mol2 dm–3

    • Ksp is called the solubility product of the ions concerned and is constant at constant temperature for a saturated solution i.e. when no more will dissolve.

    • The solubility product for a sparingly soluble strong electrolyte is defined as the product of the concentration of the ions raised to their appropriate powers in a saturated solution at a specific temperature.

    • At the saturation solution point, controlled by the Ksp expression, it doesn't matter how much solid you add, no more can dissolve.

    • The solubility of the sparingly soluble salt is governed by the Ksp expression, i.e. whatever ion concentrations are present of the compound, then the expression must be obeyed. (see common ion effect below).

    • If on mixing solutions containing the two constituent ions the Ksp expression is exceeded, precipitation will take place until the product of the ion concentrations equals the Ksp value. If the Ksp expression is not exceeded, no precipitation will take place.

    • –

  • Other solubility product expressions

    • Note: In the Ksp expression, the ion concentrations are raised to powers equal to their molar ratio in the compound:

      • This is analogous to the molar ratio from the equations for Kc or Kp expressions, and similarly, watch the units!

    • silver chloride : AgCl(s) + aq  (c) doc b Ag+(aq) + Cl–(aq)

      • Ksp = [Ag+(aq)] [Cl–(aq)] = 1.8 x 10–10 mol2 dm–6

    • calcium carbonate: CaCO3(s) + aq (c) doc b Ca2+(aq) + CO32–(aq) (ΔH slightly –ve)

      • Ksp = [Ca2+(aq)] [CO32–(aq)] = 5.0 x 10–10 mol2 dm–6

      • Sea shells are mainly made of calcium carbonate which is a tough mineral and will not dissolve appreciably in water.

      • However, when the sea creatures die and the remains sink to great depth in the oceans, the shells will then dissolve because of changes in the position of two equilibria. At great depths the pressure is much higher and the temperature lower, both factors favouring more CO2 dissolving, therefore the following forward reaction is promoted,

      • CaCO3(aq) + CO2(aq) + H2O(l) (c) doc b Ca2+(aq) + 2HCO3–(aq)

      • and secondly, the lower temperature favours a higher Ksp value because the dissolving process is slightly exothermic, so more of the calcium carbonate will dissolve. There are no shells lying at the bottom of the oceans! The limestone cliffs you see in the landscape were formed in warm shallow tropical seas.

    • magnesium hydroxide : Mg(OH)2(s) + aq (c) doc b Mg2+(aq) + 2OH–(aq)

      • Ksp = [Mg2+(aq)] [OH–(aq)]2 = 1.1 x 10–11 mol3 dm–9

    • silver chromate(VI) : Ag2CrO4(aq) + aq (c) doc b 2Ag+(aq) + CrO42–(aq)

      • Ksp = [Ag+(aq)]2 [ CrO42–(aq)] = 1.3 x 10–12 mol3 dm–9

    • aluminium hydroxide : Al(OH)3(s) + aq (c) doc b Al3+(aq) + 3OH–(aq)

      • Ksp = [Al3+(aq)] [OH–(aq)]3 = 1.0 x 10–33 mol4 dm–12

    • TOP and LINKSantimony sulphide: S2S3(s) + aq (c) doc b 2Sb3+(aq) + 3S2–(aq)

      • Ksp = [Sb3+(aq)]2 [S2–(aq)]3 = 1.7 x 10–93 mol5 dm–15

  • The common ion effect

    • If a 2nd soluble substance is added to the saturated solution of a salt, that has an ion in common with the salt, this will cause precipitation of the salt. This is because the Ksp expression is initially exceeded, causing precipitation until the new concentrations satisfy the Ksp expression Le Chatelier again!). This precipitation is known as the 'common ion effect' and happens irrespective of the solubility of the substances e.g.

      • If you add some concentrated hydrochloric acid to a saturated sodium chloride ('salt' is very soluble) solution ('brine') sodium chloride is precipitated.

        • NaCl(s) (c) doc b Na+(aq) + Cl–(aq)

        • [Na+(aq)] x [Cl–(aq)] is ~constant for a saturated solution of sodium chloride at constant temperature. If you increase one ion concentration, the other must decrease to maintain equilibrium, and the only way this can happen is by precipitation of the salt.

        • Applying Le Chatelier's Principle, the conc. HCl(aq) increases the chloride ion Cl–(aq) concentration so the reaction 'net change' must initially move to the left to restore equilibrium i.e. NaCl is precipitated.

      • Similarly if you add dilute sulphuric acid OR saturated calcium chloride solution to saturated calcium sulphate (sparingly soluble) solution then calcium sulphate appears as a faint precipitate.

        • CaSO4(s) (c) doc b Ca2+(aq) + SO42–(aq)

        • [Ca2+(aq)] x [SO42–(aq)] is ~constant for a saturated solution of calcium sulfate at constant temperature.

        • The sulphuric acid increases the sulfate ion SO42–(aq) concentration OR calcium chloride increases the calcium ion Ca2+(aq) concentration. Either way, increasing one of the concentrations of the salts constituent ions will cause precipitation of calcium sulfate according to Le Chatelier's Principle.

  • Example calculation 4.2.1

    • (a) What is the molarity of calcium carbonate in a saturated solution at 298K?

      • Since mole ratio of Ca2+:CaCO3:CO32– is a 1:1:1 ratio, so [Ca2+(aq)] = [CO32–(aq)] = [CaCO3(aq)]

      • Ksp = [Ca2+(aq)] [CO32–(aq)] = 5.0 x 10–10 mol2 dm–6

      • therefore: [CaCO3(aq)] = √(5.0 x 10–10) = 2.24 x 10–5 mol dm–3

    • (b) What is the solubility of calcium carbonate in 0.1 mol dm–3 sodium carbonate solution?

      • The tiny contribution of carbonate from the dissolved calcium carbonate can be ignored, so effectively the carbonate concentration is that of the sodium carbonate itself.

      • Ksp = [Ca2+(aq)] [CO32–(aq)] = 5.0 x 10–10 = [Ca2+(aq)] x 0.1

      • [Ca2+(aq)] = [CaCO3(aq)] = 5.0 x 10–10/0.1 = 5.0 x 10–9 mol dm–3

  • Example calculation 4.2.2

    • If equal volumes of aqueous 0.01 mol dm–3 sodium chloride and 0.005 mol dm–3 silver nitrate solution are mixed, show by calculation whether or not a precipitate of silver chloride forms?

      • Ksp(AgCl) = [Ag+(aq)] [Cl–(aq)] = 1.8 x 10–10 mol2 dm–6

    • [NaCl(aq)] = [Cl–(aq)] = 0.01 / 2 = 0.005 mol dm–3

      • (both original concentrations are halved because of mixing equal volumes i.e. diluted by a factor of 2)

    • [AgNO3(aq)] = [Ag+(aq)] = 0.005 / 2 = 0.0025 mol dm–3

    • The ionic product = [Ag+(aq)] x [Cl–(aq)] = 0.005 x 0.0025 = 1.25 x 10–5 mol2 dm–6

    • The ionic product on mixing exceeds the Ksp value so precipitation takes place.

  • TOP and LINKSExample calculation 4.2.3

    • For the equilibrium:  CaSO4(s) + aq (c) doc b Ca2+(aq) + SO42–(aq)

      • Ksp = [Ca2+(aq)] [SO42–(aq)] = 2.0 x 10–5 mol2 dm–6

    • (a) What is the solubility of calcium sulphate in a saturated solution of the salt in g cm–3?

      • From the arguments outlined in example 4.2.1(a)

      • [Ca2+(aq)] = [CaSO4(aq)] = √(2.0 x 10–5) = 4.47 x 10–3 mol dm–3

      • Ar's: Ca = 40, S = 32, O = 16, Mr(CaSO4) = 136, 1 dm3 = 106 cm3

      • solubility = 4.47 x 10–3 x 136 = 0.608 g dm–3, so scaling down to 100 cm3

      • solubility = 0.608 x 100/106 = 6.1 x 10–5 g/100 cm3

    • (b) Assuming equal volumes of solutions are mixed, what is the minimum concentration of sodium sulphate (Na2SO4) solution that when added to a 1.0 x 10–4 mol dm–3 solution of calcium sulphate, will just begin to cause precipitation of calcium sulphate?

      • Let [NaaSO4(aq)] = [SO42–(aq)] = the theoretical concentration of sodium sulphate at the point of precipitation.

        • This is ignoring the tiny contribution from the calcium sulphate solution.

      • On mixing the equal volumes of solutions, the calcium sulphate concentration is halved.

      • Therefore on substituting into the Ksp expression on the point of precipitation ...

      • 2.0 x 10–5 = ((1.0 x 10–4)/2)  x [NaaSO4(aq)]

      • therefore [NaaSO4(aq)] = 2.0 x 10–5/0.5 x 10–4 = 0.4 mol dm–3

      • However, on mixing, the sodium sulphate solution concentration must also be halved,

      • therefore the original sodium sulphate solution must be at least 0.8 mol dm–3

  • Example calculation 4.2.4

    • Equal volumes of 0.025 mol dm–3 potassium bromide (KBr) and 0.005 mol dm–3 lead(II) nitrate (Pb(NO3)2)solutions were mixed.

      • Ksp(PbBr2) = 7.9 x 10–5 mol3 dm–9

    • (a) Write out (i) the Ksp expression for lead(II) bromide and (ii) the ionic equation for its precipitation.

      • (i) Ksp(PbBr2) = [Pb2+(aq)] [Br–(aq)]2, (ii) Pb2+(aq) + 2Br–(aq) ==> PbBr2(s)

    • (b) Show by calculation if lead(II) bromide precipitates after mixing the solutions.

      • [Br–(aq)] = [KBr(aq)] = 0.025 / 2 = 1.25 x 10–2 mol dm–3

      • [Pb2+(aq)] = [Pb(NO3)2(aq)] = 0.005 / 2 = 2.5 x 10–3 mol dm–3

      • The ionic product = [Pb2+(aq)] [Br–(aq)]2 = 2.5 x 10–3 x (1.25 x 10–2)2 = 3.91 x 10–7 mol3 dm–9

      • The Ksp value of 7.9 x 10–5 is NOT exceeded, so no precipitation takes place.

TOP and LINKS

Part 4 sub–index 4.1 Partition between two phases * 4.2 Solubility product Ksp & common ion effect * 4.3 Ion–exchange systems

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