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Chapter 18. Solubility and Complex-Ion Equilibria. Overview. Solubility Equilibria Solubility Product Constant Solubility and Common Ion Effect Precipitation Calculations Effect of pH on Solubility Complex-Ion Equilibria Complex Ion Formation Complex Ions and Solubility

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Chapter 18

Chapter 18

Solubility and Complex-Ion Equilibria


  • Solubility Equilibria

    • Solubility Product Constant

    • Solubility and Common Ion Effect

    • Precipitation Calculations

    • Effect of pH on Solubility

  • Complex-Ion Equilibria

    • Complex Ion Formation

    • Complex Ions and Solubility

  • Application of Solubility Equilibria

    • Qualitative analysis of metal ions

Solubility equilibria
Solubility Equilibria

  • Solubility of a solid treated as with other equilibria. Solution is saturated. No more solid will dissolve since dynamic equilibrium.

    AgCl(s)  Ag+(aq) + Cl(aq) Ksp = [Ag+][Cl]

  • Solid not included in the equilibrium expression.

    • MyXz(s)  yM+p(aq)+zXq(aq) Ksp=[M+p]y[Xq]z

      where Ksp = solubility product.

      E.g. determine the equilibrium expression for each: PbCl2, Ag2SO4,Al(OH)3.

  • Ksp can be determined if the solubility is known.

    E.g. Determine Ksp for silver chromate (Ag2CrO4) if its solubility in water is 0.0290 g/L at 25C.

    • Determine molar solubility.

    • Determine Ksp.

      E.g. 2 Determine Ksp of CaF2 if its solubility is 2.20x104M.

Solubility from k sp

  • Can be determined by using stoichiometry to express all quantities in terms of one variable- solubility, x. i.e. for the reaction. Use equilibrium table to write concentration of each in terms of the compound dissolving

    E.g. determine the solubility of PbCl2 if its Ksp = 1.2x105

    PbCl2(s)  Pb2+(aq) + 2Cl(aq)

    E.g.1 Determine solubility of AgCl if its Ksp = 1.8x1010M2.

    E.g.2 Determine solubility of Ag2CO3 if its Ksp = 8.1x1012M3.

    E.g.3 Determine solubility of Fe(OH)3 if its Ksp = 4x1038M4

Factors that affect solubility
Factors that Affect Solubility

  • The common–Ion effect(Remember LeChatelier’s Principle)

    E.g. Determine solubility of PbCl2 (Ksp = 1.2x105)in 0.100M NaCl.

    • Write equilibrium table in terms of x and [Cl]

a common–ion reduces the solubility of the compound.

  • Assume that [Cl]NaCl >>x

  • Solve for x.

  • E.g. determine the solubility of CaF2 in a solution of CaCl2. Ksp = 3.9x1011.

  • Precipitation of ionic compounds
    Precipitation of Ionic Compounds

    • Starting with two solutions, Qsp used to predict precipitation and even the extent of it.

      • Precipitation = reverse of dissolution

      • Precipitation occurs when Qsp > Ksp until Qsp = Ksp

      • If Qsp < Ksp, precipitation won’t occur.

        E.g. determine if precipitation occurs after mixing 50.00 mL 3.00x103 M BaCl2 and 50.00 mL 3.00x103 M Na2CO3.


      • CBaCl2 = 1.50x103 M; CNa2CO3 = 1.50x103 M

      • Qsp = 1.50x103 M1.50x103 M = 2.25x106

      • Qsp >1.1x1010.= Ksp precipitation.

        E.g. 2 determine equilibrium concentration of each after precipitation occurs.


      • assume complete precipitation occurs;

      • set up equilibrium table; and solve for equilibrium concentration of barium and carbonate ion concentrations.

    Precipitation of ionic compounds1
    Precipitation of Ionic Compounds

    Eg. 3 determine the fraction of Ba2+ that has precipitated.


    • Use the amount remaining in solution (results of E.g. 2) divided by starting concentration to determine the fraction of barium that is left in solution.

    • Subtract from above.

      E.g.4 determine the Br concentration when AgCl starts to precipitate if the initial concentration of bromide and chloride are 0.100 M. Ksp(AgBr) = 5.0x1013; Ksp(AgCl) = 1.8x1010.

    Factors that affect solubility ph
    Factors that Affect Solubility-pH

    • pH of the Solution: LeChatelier’s Principle again.

      E.g. determine the solubility of CaF2 at a pH of 2.00. Ksp = 3.9x1011. Ka(HF) = 6.6x104.


      • Determine the ratio of [F] and [HF] from the pH and Ka.

      • Write an expression for solubility in terms of Ka and pH and

      • Substitute into solubility equation to determine the solubility.


      • Ksp = 3.9x1011 = x[F]2 (pH changes the amount of free Fluoride.)

      • Let x = solubility. Then 2x = [F] + [HF]

      • From equilibrium equation:

      • 2x = [F](1+1/0.066) = 16.15*[F] or

      • [F] = 2*x/16.15 = 0.124*x

      • 3.9x1011 = x(0.0124*x)2

      • x = 1.36x103 M vs. 2.13x104 M (normal solubility)

    Separation of ions by selective precipitation
    Separation of Ions By Selective Precipitation

    • Metal ions with very different Ksp can be separated.

    • Divalent metal ions are often separated using solubility variations for the metal sulfides.

    • Solution is saturated with H2S at 0.100 M; pH adjusted to keep one component soluble and the other insoluble.

    • H2S is diprotic acid; the overall reaction to get to sulfide is:

    • Combine with solubility equilibrium reaction to get the overall equilibrium expression and constant.

      E.g. determine the solubility of 0.00500 M Zn2+ in 0.100 M H2S at pH = 1. Ksp = 1.10x1021.

    Complex ions
    Complex Ions

    • Formation of Complex Ions (Coordination Complexation ) = an ion formed from a metal ion with a Lewis base attached to it by a coordinate covalent bond.

      Ag+(aq) + 2NH3(aq)  Ag(NH3)2(aq) Kf = 1.7x107

    • Large equilibrium constant indicates that “free” metal is completely converted to the complex.

      Eg. What is the concentration of the silver amine complex above in a solution that is originally 0.100 M Ag+ and 1.00 M NH3?

      E.g. determine the [Ag+] (free silver concentration) in 0.100 M AgNO3 that is also 1.00 M NaCN.

    Factors that affect solubility complexation
    Factors that Affect Solubility: Complexation

    • Free metal ion concentration in solution is reduced when complexing agent added to it;

    • Free metal ion concentration needed in solubility expression.

    E.g. determine if precipitation will occur in a solution containing 0.010 M AgNO3 and 0.0100 M Nal in 1.00 M NaCN. Recall Kf = 5.6x1018

    Agl(s)Ag+ + l Ksp = 8.5.x1017


    • Determine the free metal concentration in the solution.

    • Use free metal concentration with iodide concentration to get Qsp

      • If Qsp < Ksp, no precipitation

      • If Qsp > Ksp, precipitation

      • If Qsp = Ksp, precipitation is starting.

    Solubility with complexing agent
    Solubility with Complexing Agent

    E.g. Determine the solubility of AgI in 1.00 M NaCN. Recall Kf = 5.6x1018

    Agl(s)Ag+ + l Ksp = 8.5.x1017


    • Combine to equilibria equations to find a single equation describing the equilibrium.

    the presence of a complexing agent increases the solubility

    • Setup equilibrium table and solve.