One example of a dynamic equilibrium is where the rate of dissolution (dissociation) is
equal to the rate of crystallization. This is an example of a heterogeneous equilibrium
between the solid and dissolved ions. To determine the degree of dissociation, or the
conditions that result in the precipitation of slightly soluble salts, equilibrium constants
and equilibrium law equations are involved.
Question: What can be interpreted from the solubility product constant, Ksp, for silver
chloride?
AgCl (s) ↔ Ag+ (aq) + Cl- (aq)
Ksp = [Ag+] [Cl-] = 1.8 x 10-10 @ 25°C
Answer: The solubility product constant, Ksp, is the value obtained from the equilibrium
law for a saturated solution. This small Ksp value indicates that the concentration of ions
in a saturated solution of silver chloride is very small.
Ksp values are typically listed for ionic compounds with low solubility that form
precipitates. Highly soluble ionic compounds do not form precipitates and their
solutions are not saturated. Solubilities of highly soluble substances are listed in mol/L
or g/100 mL instead of Ksp values.
Example 1: Magnesium fluoride is a hard, slightly soluble salt. Calculate solubility
product of magnesium fluoride at 25°C given a solubility of 1.72 x 10-3 g/100 mL.
Example 2: Calculate the molar solubility of zinc hydroxide at 25°C where solubility
product constant is 7.7 x 10-17.
Using Trial Ion Product, Q, to Predict Solubility
Trial ion product Q = Ksp (saturated solution)
Precipitate will not form.
Trial ion product Q < Ksp (unsaturated solution)
Precipitate will not form.
Trial ion product Q > Ksp (supersaturated solution) Precipitate will form.
Example 1: If 100.0 mL of 0.100 mol/L CaCl2 (aq) and 100.0 mL of 0.0400 mol/L Na2SO4 (aq)
are mixed at 20°C, determine whether a precipitate will form. For CaSO4 (aq) at 20°C,
Ksp = 3.6 x 10-5.
Example 2: Would a precipitate of lead (II) sulfate, PbSO4 (aq) (Ksp = 1.8 x 10-8) form if
255 mL of 0.000 16 mol/L lead (II) nitrate, Pb(NO3)2 (aq) is poured into 456 mL of
0.000 23 mol/L sodium sulfate, Na2SO4 (aq)?
The Common Ion Effect
Equilibrium is usually established in pure water.
Question: How is molar solubility of lead (II) chloride affected when the solvent is sodium
chloride instead of water?
Answer:
solute:
PbCl2 (s) ↔ Pb2+ (aq) + 2 Cl- (aq)
solvent:
NaCl (s) ↔ Na+ (aq) + Cl- (aq)
The chlorine ions (Cl-) from sodium chloride solvent will chlorine ions to be in excess.
Equilibrium will shift to the left and more PbCl2 will precipitate out of solution. This lowering
of solubility of PbCl2, as seen by the greater amount of precipitate formed, by the addition of
a common ion (i.e. Cl-) is referred to as the common ion effect.
Example 1: What is the molar solubility of PbCl2 (s) in a 0.20 mol/L NaCl (aq) solution at
SATP? Lead (II) chloride has a Ksp = 1.7 x 10-5 at 25°C. (Note: If Ksp value is not given, it
can be found by looking it up in a reference table).