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Solubility product
The solubility product constant, ksp is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the ksp value it has. Consider the general dissolution reaction below (in aqueous solutions):
MmAa —► m M+ + a A~
To solve for the ksp it is necessary to take the molarities or concentrations of the products c(M+) c(A~) and multiply them. If there are coefficients in front of any of the products, it is necessary to raise the product to that coefficient power (and also multiply the concentration by that coefficient). This is shown below:
ksp = [m]m ■ [a]a
Note that the reactant, MmAa is not included in the ksp equation. Solids are not included when calculating equilibrium constant expressions, because their concentrations do not change the expression; any change in their concentrations are insignificant, and therefore omitted. Hence, ksp represents the maximum extent that a solid that can dissolved in solution.
Example 1. Calculate the solubility product of CaF2. Concentration of its saturated solution is 1.888 x 10~4 mol • dm-3.
The relevant equilibrium is:
CaF2 —► Ca2+ + 2 F~ so the associated equilibrium constant is
ksp = [Ca2+] • [Ft
If we consider the concentration of CaF2 is c, then concentration of Ca2+ is c and concentration of F~ is 2c. Based on this we can calculate
ksp  =  c ■ (2c)2 = c • 4c2 = 4c3 ksv  =  2.692 x 10-11
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Example 2. Calculate the solubility product of As2S3. Concentration of its saturated solution is 8.191 x 10~7 mol • dm-3.
The relevant equilibrium is:
As2S3 — 2 As3+ + 3 S2~ so the associated equilibrium constant is
ksp = [As3+]2 • [S2-]3
If we consider the concentration of As2S3 is c, then concentration of As is 2c and concentration of S2~ is 3c. Based on this we can calculate
ksp        (2c)2 • (3c)3 = 4c2 • 27c3 = 108c5
ksp       3.981 x 10
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Example 3. Calculate the solubility of BaS04. Solubility product is ksp = 1.096 x 10"
BaS04 — Ba2+ + S042~
ksp       [Ba2+] • [S042" i*£sp c c
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\Jk7p
c  =   1.047 x 10-5 mol • dm"3
Example 4. Calculate the solubility of Fe(OH)3. Solubility product is ksp = 3.715 x 10"40.
Fe(OH)3 — Fe3+ + 3 OH"
ksp       [Fe3+] • [OH-]3 ksp  =  c ■ (3c)3 = 27c4 4/ ksp
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6.090 x 10-11 mol • dm"3
Example 5. Calculate the solubility product of Bi(OH)3. Concentration of its saturated solution is 1.102 x 10"8 mol • dm-3. (3.98 x 10"31)
Example 6. Calculate the solubility of Cu2S if the solubility product of this salt is ksp = 2.512 x 10"48. (8.56 x 10~17 mol • dm"3)