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Ksp : Solubility Product

Key Concepts

  • Solubility Product, Ksp, refers to an ionic compound dissolving to form ions

  • For the reaction:
    MA(s) M+(aq) + A-(aq)

    The equilibrium constant, K is:

    K = [M+][A-]

    [MA]

  • Since the concentration of the solid is a constant, it is effectively incorporated into the equilibrium constant:
    Ksp = [M+][A-]

  • [M+][A-] is referred to as the ion product.

  • At equilibrium, the ion product = solubility product (Ksp) and the solution is said to be saturated.

  • A precipitate will form when two solutions are mixed if the ion product is greater than the solubility product, ie, [M+][A-] > Ksp
Condition Nature of Solution Possibility of Precipitation
ion product < Ksp unsaturated solution no precipitation occurs
ion product = Ksp saturated solution saturated solution at equilibrium
ion product > Ksp supersaturated solution precipitation occurs

Example : Calculating the solubility of an ionic compound (MA).

Calculate how much silver bromide will dissolve in 1 L of water given Ksp = 5.0 x 10-13 at 25oC.
  1. Write the equation for dissolving AgBr in water:
    AgBr(s) Ag+(aq) + Br-(aq)

  2. Write the equilibrium expression:
    Ksp = [Ag+][Br-] = 5.0 x 10-13

  3. Determine the relative concentrations of each ion:
    at equilibrium [Ag+] = [Br-] (from the balanced chemical equation)
    So, [Ag+] = [Br-] = x

  4. Substitute these vales into the equilibrium expression:
    5.0 x 10-5 = [x]2

  5. Solve for x:
    x = √5.0 x 10-5 = 7 x 10-7 mol L-1
    So, [Ag+] = [Br-] = 7 x 10-7 mol L-1

    The solubility of AgBr at 25oC is 7 x 10-7 mol L-1
    7 x 10-7 moles AgBr will dissolve in 1L of water at 25oC.

Example : Calculating the solubility of an ionic compound (MA2).

Calculate how much strontium fluoride will dissolve in 1 L of water given Ksp = 2.5 x 10-9 at 25oC.
  1. Write the equation for dissolving SrF2 in water:
    SrF2(s) Sr2+(aq) + 2F-(aq)

  2. Write the equilibrium expression:
    Ksp = [Sr2+][F-]2 = 2.5 x 10-9

  3. Determine the relative concentrations of each ion:
    at equilibrium [Sr2+] = x and [F-] = 2x (from the balanced chemical equation)

  4. Substitute these vales into the equilibrium expression:
    2.5 x 10-9 = [x][2x]2 = 4x3

  5. Solve for x:
    2.5 x 10-9 ÷ 4 = x3 = 6.25 x 10-10
    x = 3√6.25 x 10-10 = 8.5 x 10-4 mol L-1

    The solubility of SrF2 at 25oC is 8.5 x 10-4 mol L-1
    8.5 x 10-4 moles of SrF2 will dissolve in 1L of water at 25oC.

Example : Deciding whether a precipitate will form

Will a precipitate form if 25.0 mL of 1.4 x 10-9 mol L-1 NaI and 35.0 mL of 7.9 x 10-7 mol L-1 AgNO3 are mixed? (Ksp for AgI at 25oC is 8.5 x 10-17)
  1. Write the net ionic equation for the dissolving of the precipitate:
    AgI Ag+ + I-

  2. Write the expression for the ion product:
    ion product = [Ag+][I-]

  3. Calculate the concentration each reactant after mixing:

    [I-] :
        c1 = [I-] before mixing = 1.4 x 10-9 mol L-1 (assuming full dissociaiton of NaI)
        V1 = initial volume = 25.0 mL = 25.0 x 10-3 L
        V2 = final volume after mixing = 25.0 + 35.0 = 60.0 mL = 60.0 x 10-3 L (assuming volumes are additive)
        c2 = [I-] after mixing = c1 x V1 ÷ V2 = (1.4 x 10-9 x 25.0 x 10-3) ÷ (60.0 x 10-3) = 5.8 x 10-10 mol L-1

    [Ag+]:
        c1 = [Ag+] before mixing = 7.9 x 10-7 mol L-1 (assuming full dissociaiton of AgNO3)
        V1 = initial volume = 35.0 mL = 35.0 x 10-3 L
        V2 = final volume after mixing = 25.0 + 35.0 = 60.0 mL = 60.0 x 10-3 L (assuming volumes are additive)
        c2 = [Ag+] after mixing = c1 x V1 ÷ V2 = (7.9 x 10-7 x 35.0 x 10-3) ÷ (60.0 x 10-3) = 4.6 x 10-7

  4. Calculate the ion product:
        ion product = [Ag+][I-] = 5.8 x 10-10 x 4.6 x 10-7 = 2.7 x 10-16

  5. Decide whether a precipitate forms:
        If ion product > Ksp a precipitate will form
        If ion product < Ksp a precipitate will not form
        In this case, 2.7 x 10-16 > Ksp (8.6 x 10-17) so a precipitate will form


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