18-4 Solubility Equilibrium
 An equilibrium expression can be written for the balanced chemical equation of a solid
compound dissolving in water
A. Solubility Product
I. Saturated Solution: solution that contains the maximum amount of solute possible at a given
temperature. Depends on properties of the solute.
II. Solutes are “soluble” if 1g or more solute can dissolve in 100g of water
III.Solutes of “insoluble” if 0.lg or less can dissolve in 100g of water
IV.
Solutes are “slightly soluble” if between 0.1g and 1g can dissolve in 100g of water
V. Silver Chloride disassociating into ions:
a. AgCl(s) ↔ Ag+(aq) + Cl-(aq)
b. K = [Ag+][Cl-]/[AgCl] = [Ag+][Cl-]
c. When [Ag+] and [Cl-] = their maximum solubilities in solution,
Ksp = [Ag+][Cl-] = “Solubility Product Constant”
d. Calculation of Ksp: Look up Solubility of AgCl in Appendix A-13.
i. AgCl Solubility: 8.9x10-5g solute / 100g of H2O
ii. Convert 8.9x10-5g AgCl into moles AgCl
iii. Convert 100g H2O into Liter H2O
iv. find Molarity of AgCl = [Ag+] = [Cl-] (1:1:1 ratio)
v. Multiply [Ag+] by [Cl-], this is the Ksp value.
VI.
Calcium Fluoride dissociation:
a. CaF2(s) ↔ Ca2+(aq) + 2F-(aq)
b.
Ksp = [Ca2+][F-]2
i. Note that [F-] is squared, but would also be DOUBLED if you are calculating
concentration from the original amount of CaF2(s)
c. Calculating Ksp for CaF2:
i. CaF2 solubility: 1.7x10-3g solute per 100g H2O
ii. Convert to molarity: (2.2x10-4M)
iii. [Ca2+] = 2.2x10-4M, [F-] = 4.4x10-4M (Why?)
iv.
VII.
B.
Ksp = [Ca2+][F-]2 = 4.3x10-11
Table of Ksp values is on Page 579
Calculating Solubilities from Ksp
I. Barium Carbonate Dissociation: BaCO3(s) ↔ Ba2+(aq) + CO32-(aq)
II.
Ksp = [Ba2+][CO32-] = 5.1x10-9
III.In words: BaCO3 will dissolve until the product of the two ion concentrations is equal to
5.1x10-9.
IV.
]
The Solubility of the system can be calculated backwards by solving for [Ba2+] or [CO32-
V. [Ba2+] = 7.14x10-5M
C. Precipitation Calculations
I. If Molar quantities of the two ions in a dissociation are known, their product can be found
and compared to Ksp.
II. If the product is greater than Ksp, the ions will precipitate out of solution as a solid
III.If the product is less than Ksp, the ions will stay dissolved.
Example Problems
1. Calculate Ksp for Copper(I) Chloride (CuCl), given that the solubility of this compound is
1.08x10-2g CuCl per 100g H2O. (Ksp = 1.19x10-6)
2. Calculate the Ksp of PbCl2, which has a solubility of 1.0g PbCl2 per 100g H2O. (Ksp = 1.9x10-4)
3. 5.0grams of Ag2SO4 will dissolve in 1L of water. Calculate Ksp of this salt.(Ksp = 2x10-5)
4. Calculate the solubility of Silver Acetate,Ag(CH3COO), given the Ksp listed in Table 18-3
(0.044M)
5. Determine the concentration of Strontium ions in a saturated solution of Strontium Sulfate if the
Ksp for SrSO4 = 3.2x10-7(5.7x10-4M)
6. Will a precipitate form if 20.0mL of 0.010M BaCl2 is mixed with 20.0mL of 0.0050M Na2SO4?
(BaSO4 is the species that may precipitate) (1.2x10-5 >> 1.1x10-10, so precipitation occurs)
7. Does a precipitate form when 20.mL of 0.038M Pb(NO3)2 and 30.mL of 0.018M KCl solutions
are mixed?(PbCl2 is the species that may precipitate)(No,4.1x10-7 is much smaller than 1.6x105
)