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Solubility Equilibria. Will it all dissolve, and if not, how much will?. SOLUBILITY EQUILIBRIA. Solubility : Relative term used to describe how much of a particular substance dissolves in a certain amount of solvent. Substances that dissolve very well are said to be soluble

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solubility equilibria

Solubility Equilibria

Will it all dissolve, and if not, how much will?

  • Solubility: Relative term used to describe how

much of a particular substance dissolves in a

certain amount of solvent.

  • Substances that dissolve very well are said to

be soluble

  • Insoluble species don’t dissolve well.
  • All substances are “soluble” to some extent
  • We will look at slightly soluble substances
  • All dissolving is an equilibrium.
  • If there is not much solid it will all dissolve.
  • As more solid is added the solution will become saturated.
  • Solid ↔ dissolved
  • The solid will precipitate as fast as it dissolves, forming an equilibrium.
watch out
Watch out
  • Solubility is not the same as solubility product.
  • Solubility product is an equilibrium constant.
  • It doesn’t change except with temperature.
  • Solubility is an equilibrium position for how much can dissolve.
  • A common ion can change this.
  • Consider the following reaction
  • The equilibrium constant expression is

Ksp = [Pb2+][Cl-]2

  • Ksp is called the solubility product constant or

simply solubility product

  • For a compound of general formula, MyXz (next page)
Ksp = [Mz+]y[Xy-]z

Ksp = [Mg2+][NH4+][PO43-]

Ksp = [Zn2+][OH-]2

Ksp = [Ca2+]3[PO43-]2

Molar solubility: the number of moles that

dissolve to give 1 liter of saturated solution

  • As with any equilibrium constant the numerical

value must be determined from experiment

  • The Ksp expression is useful because it applies

to all saturated solutions

- the origins of the ions are not relevant

  • Consider that @ 25C Ksp AgI = 1.5 x 10-16
Solving Solubility Problems

For the salt AgI at 25C, Ksp = 1.5 x 10-16

AgI(s)  Ag+(aq) + I-(aq)







1.5 x 10-16 = x2

x = solubility of AgI in mol/L = 1.2 x 10-8 M

Solving Solubility Problems

For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5

PbCl2(s)  Pb2+(aq) + 2Cl-(aq)







1.6 x 10-5 = (x)(2x)2 = 4x3

x = solubility of PbCl2 in mol/L = 1.6 x 10-2 M

relative solubilities
Relative Solubilities
  • Ksp will only allow us to compare the solubility of solids the that fall apart into the same number of ions.
  • The bigger the Ksp of those the more soluble.
  • If they fall apart into different number of pieces you have to do the math.
the common ion effect
The Common Ion Effect
  • When the salt with the anion of a weak acid is added to that acid:
    • it reverses the dissociation of the acid.
    • lowers the percent dissociation of the acid.
  • The same principle applies to salts with the cation of a weak base..
  • The calculations are the same as with acid base equilibrium.
Solving Solubility with a Common Ion

For the salt AgI at 25C, Ksp = 1.5 x 10-16

What is its solubility in 0.05 M NaI?

AgI(s)  Ag+(aq) + I-(aq)







1.5 x 10-16 = (x)(0.05+x)  (x)(0.05)

x = solubility of AgI in mol/L = 3.0 x 10-15 M

ph and solubility
pH and solubility
  • OH- can be a common ion.
  • More soluble in acid.
  • For other anions if they come from a weak acid they are more soluble in a acidic solution than in water.
  • CaC2O4↔Ca+2 + C2O4-2
  • H+ + C2O4-2 ↔HC2O4-
  • Reduces C2O4-2 in acidic solution.
  • The reaction quotient (called ion product) may be applied to solubility equilibria - determines if a substance will precipitate from solution
  • Ion Product, Q =[M+]a[Nm-]b
  • If Ksp
  • If Ksp=Q equilibrium solution is just saturated
  • If Ksp>Q No precipitate, forward process occurs
precipitation example
Precipitation Example
  • A solution of 75.0 mL of 0.020 M BaCl2 is added to 125.0 mL of 0.040 M Na2SO4. Will a precipitate form? (Ksp= 1.5 x 10-9M BaSO4)

BaSO4 could form if Ksp

For Q you need initial concentrations:

[Ba2+] = mmol Ba2+ / total mL

= (0.0750L)(0.020 M)/(0.0750L + 0.125L) = 0.0075 M

[SO42-] = mmol SO42- / total mL

= (0.1250L)(0.040 M)/(0.0750L + 0.125L) = 0.025 M

Q = [Ba2+] [SO42-] = (0.0075 M)(0.025 M) = 1.9 x 10-4


To figure out concentrations set up an ice table.

Complex Ions

A Complex ion is a charged species composed of:

1. A metallic cation

2. Ligands – Lewis bases that have a lone electron pair that can form a covalent bond with an empty orbital belonging to the metallic cation

the addition of each ligand has its own equilibrium
The Addition Of Each Ligand Has Its Own Equilibrium
  • Usually the ligand is in large excess.
  • And the individual K’s will be large so we can treat them as if they go to equilibrium.
  • The complex ion will be the biggest ion in solution.
Coordination Number
  • Coordination number refers to the number of ligands attached to the cation
  • 2, 4, and 6 are the most common coordination numbers
Complex Ions and Solubility

AgCl(s)  Ag+ + Cl- Ksp = 1.6 x 10-10

Ag+ + NH3 Ag(NH3)+ K1 = 1.2 x 10-3

Ag(NH3)+ NH3 Ag(NH3)2+ K2 = 9.6 x 10-4

K = KspK1K2

AgCl + 2NH3 Ag(NH3)2+ + Cl-

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