Solubility Product

Introduced to this in AP Chemistry

Useful for determining solubility of ionic solids in an aqueous solution

BaF (s) Ba (aq) + 2 F (aq)

K = [F][Ba]

Solubility vs. Solubility Product §

• K = solubility product - has only one value for a specific temperature
• Solubility = equilibrium position concentration: [] = molarity
• K and molar solubility are not the same thing

Solving Solubility Problems §

The salt, BaF at 25 C has a measured solubility of 1.82 * 10 Calculate K

BaF (s) Ba + 2F

Solving Solubility with a Common Ion §

For the salt AgCrO at 25C, K = 9.0 * 10

What is the solubility in 0.050 M NaCrO

AgCrO (s)	2Ag (aq)	CrO (aq)
Initial	0	0.050
Change	+2s	+s
Equilibrium	2s	0.050+s

Predicting Formation of Precipitate §

Q is ion product expressed in the same way as K for a particular system

• K = Q saturated solution, but no percipitation
• K < Q saturated solution, with percipitation
• K > Q unsaturated solution

Example §

If the solutions are added together, predict if a precipitate of PbCl will form. (PbCl, K = 1.6 * 10)

30.0 mL of 0.10 M NaCl

20.0 mL of 0.025 M Pb(NO)

[Pb] = = 0.010 M

[Cl] = = 0.060 M

Q = [Pb][Cl] = (0.010 M)(0.060 M) = 3.6 * 10

K = 1.6 * 10

K < Q precipitate of PbCl will form