Ksp
1. Solubility Equilbrium-The heterogeneous equilibrium involved in the dissolution or precipitation of ionic
compounds. This will tell us the amount of a substance that will dissolve (solubility).
2. Vocabularya. saturated solution- a solution that contains undissolved solute in contact with the solvent.
b. ionic solid- totally ionizes into its constituents.
3. Solubility Product Constant – aka solubility product, Ksp, indicates how soluble the solid is in water.
a. Given an ionic compound BA, there is an equilibrium condition that occurs even though the compound
completely ionizes. At the saturation point, ions are reforming the solid at the same rate that the solid is
forming ions:
B2A(s) ↔ 2B+(aq) + A- (aq)
Ksp = [B+]2[ A-] ; remember BA doesn’t appear because it is a solid.
b. Remember:
Solubility Constant –
has one value for a given solid at a given temperature.
is the molar concentration of ions raised to their stoichiometric powers.
Solubility – a equilibrium position.
Solubility changes as concentrations change.
Solubility is the amount (grams) of substance that dissolves to form a saturated solution
Molar Solubility- a new unit
Molar solubility is the number of moles of solute dissolving to form a liter of saturated solution.
4. Conversion between the units
a. To convert solubility to Ksp
solubility needs to be converted into molar solubility (via molar mass);
molar solubility is converted into the molar concentration of ions at equilibrium (equilibrium
calculation),
Ksp is the product of equilibrium concentration of ions.
Calculating Ksp from Solubility I
Copper(I) bromide has a measured solubility of 2.0 X 10-4 mol/L at 25°C. Calculate its Ksp value.
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Ksp = 4.0 X 10-8
Calculating Ksp from Solubility II
Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0 X 10-15 mol/L at 25°C.
Ksp = 1.1 X 10-73
5. Ranking Solubility using Ksp- make sure the number of ions are equivalent.
a. If the salts being ranked have the same number of ions, the larger the Ksp, the more soluble.
b. Example
AgI(s)
Ksp = 1.5 X 10-16
CuI(s)
Ksp = 5.0 X 10-12
CaSO4(s)
Ksp = 6.1 X 10-5
Solubility ranking :
CaSO4(s) > CuI(s) > AgI(s)
c. If the salts being ranked do not have the same number of ions, you must calculate solubility using RICE.
6. Factors that affect Solubility
a. Common Ion Effect- is defined as a solution that contains an ion that is common to the solute. This will
change the equilibrium position by Le Chatelier’s principle. You know, the solubility of a soluble salt is
decreased by the presence of a second solute that furnishes a common ion.
CaF2(s) ↔ Ca+2 + 2F- the reaction shifts to the left if NaF is added to the solution
Exercise 15
Solubility and Common Ions
Calculate the solubility of solid CaF2 (Ksp = 4.0 X 10-11) in a 0.025 M NaF solution.
= 6.4 X 10-8 mol/L
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b. pH – the solubility of a substance whose anion is basic will be affected by the pH of the solution.
Mg(OH)2 ↔ Mg+2 + 2OH- Ksp = 1.8 X 10-11
If you added acid [H+], the equilibrium will shift to the right to replace the OH- making the solution more
soluble.
7. Reaction Quotient and Ksp to determine precipitation
a. With some knowledge of the reaction quotient, we can decide
 whether a ppt will form AND
 what concentrations of ions are required to begin the ppt. of an insoluble salt.
1. Q < Ksp, the system is not at equil. (unsaturated)
2. Q = Ksp, the system is at equil. (saturated)
3. Q > Ksp, the system is not at equil. (supersaturated)
Precipitates form when the solution is supersaturated!!!
Exercise 16
Determining Precipitation Conditions
A solution is prepared by adding 750.0 mL of 4.00 X 10-3 M Ce(NO3)3 to 300.0 mL of 2.00 X 10-2 M KIO3.
Will Ce(IO3)3 (Ksp = 1.9 X 10-10) precipitate from this solution?
yes
Exercise 17
Precipitation
A solution is prepared by mixing 150.0 mL of 1.00 X 10-2 M Mg(NO3)2 and 250.0 mL of
1.00 X 10-1 M
NaF. Calculate the concentrations of Mg2+ and F- at equilibrium with solid MgF2 (Ksp = 6.4 X 10-9).
[Mg2+] = 2.1 X 10-6 M
[F-] = 5.50 X 10-2 M
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