Since the concentration of the solid is a constant, it is effectively incorporated into the equilibrium constant:

K_{sp} = [M^{+}][A^{-}]

[M^{+}][A^{-}] is referred to as the ion product.

At equilibrium, the ion product = solubility product (K_{sp}) 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^{-}] > K_{sp}

Condition

Nature of Solution

Possibility of Precipitation

ion product < K_{sp}

unsaturated solution

no precipitation occurs

ion product = K_{sp}

saturated solution

saturated solution at equilibrium

ion product > K_{sp}

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 K_{sp} = 5.0 x 10^{-13} at 25^{o}C.

Write the equation for dissolving AgBr in water:
AgBr(s) Ag^{+}(aq) + Br^{-}(aq)

Write the equilibrium expression:
K_{sp} = [Ag^{+}][Br^{-}] = 5.0 x 10^{-13}

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

Substitute these vales into the equilibrium expression:
5.0 x 10^{-5} = [x]^{2}

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 25^{o}C is 7 x 10^{-7} mol L^{-1} 7 x 10^{-7} moles AgBr will dissolve in 1L of water at 25^{o}C.

Example : Calculating the solubility of an ionic compound (MA_{2}).

Calculate how much strontium fluoride will dissolve in 1 L of water given K_{sp} = 2.5 x 10^{-9} at 25^{o}C.

Write the equation for dissolving SrF_{2} in water:
SrF_{2}(s) Sr^{2+}(aq) + 2F^{-}(aq)

Write the equilibrium expression:
K_{sp} = [Sr^{2+}][F^{-}]^{2} = 2.5 x 10^{-9}

Determine the relative concentrations of each ion:
at equilibrium [Sr^{2+}] = x and [F^{-}] = 2x (from the balanced chemical equation)

Substitute these vales into the equilibrium expression:
2.5 x 10^{-9} = [x][2x]^{2} = 4x^{3}

Solve for x:
2.5 x 10^{-9} ÷ 4 = x^{3} = 6.25 x 10^{-10} x = ^{3}√6.25 x 10^{-10} = 8.5 x 10^{-4} mol L^{-1}

The solubility of SrF_{2} at 25^{o}C is 8.5 x 10^{-4} mol L^{-1} 8.5 x 10^{-4} moles of SrF_{2} will dissolve in 1L of water at 25^{o}C.

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} AgNO_{3} are mixed? (K_{sp} for AgI at 25^{o}C is 8.5 x 10^{-17})

Write the net ionic equation for the dissolving of the precipitate:
AgI Ag^{+} + I^{-}

Write the expression for the ion product:
ion product = [Ag^{+}][I^{-}]

Calculate the concentration each reactant after mixing:

[I^{-}] :
c_{1} = [I^{-}] before mixing = 1.4 x 10^{-9} mol L^{-1} (assuming full dissociaiton of NaI)
V_{1} = initial volume = 25.0 mL = 25.0 x 10^{-3} L
V_{2} = final volume after mixing = 25.0 + 35.0 = 60.0 mL = 60.0 x 10^{-3} L (assuming volumes are additive)
c_{2} = [I^{-}] after mixing = c_{1} x V_{1} ÷ V_{2} = (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^{+}]:
c_{1} = [Ag^{+}] before mixing = 7.9 x 10^{-7} mol L^{-1} (assuming full dissociaiton of AgNO_{3})
V_{1} = initial volume = 35.0 mL = 35.0 x 10^{-3} L
V_{2} = final volume after mixing = 25.0 + 35.0 = 60.0 mL = 60.0 x 10^{-3} L (assuming volumes are additive)
c_{2} = [Ag^{+}] after mixing = c_{1} x V_{1} ÷ V_{2} = (7.9 x 10^{-7} x 35.0 x 10^{-3}) ÷ (60.0 x 10^{-3}) = 4.6 x 10^{-7}

Calculate the ion product:
ion product = [Ag^{+}][I^{-}] = 5.8 x 10^{-10} x 4.6 x 10^{-7} = 2.7 x 10^{-16}

Decide whether a precipitate forms:
If ion product > K_{sp} a precipitate will form
If ion product < K_{sp} a precipitate will not form
In this case, 2.7 x 10^{-16} > K_{sp} (8.6 x 10^{-17}) so a precipitate will form