Solubility
Product and
Common Ion
Effect
CAPE CHEMISTRY UNIT 1
Solubility
• Solubility is defined as the number of
grams or moles of compound needed
to saturate 100 g of water, or it can also be defined in
terms of 1 kg of water, at a given temperature
• For example, sodium chloride (NaCl) is
considered to be a soluble salt as a saturated
solution contains 36 g of NaCl per 100 g of water
• Lead chloride (PbCl2) on the other hand is
an insoluble salt as a saturated solution only
contains 0.99 g of PbCl2 per 100 g of water
Solubility Product
• The solubility product (Ksp) is:
• The product of the concentrations of each ion in a saturated solution of
a relatively soluble salt
• At 298 K
• Raised to the power of their relative concentrations
• C (s) ⇌ aAx+ (aq) + bBy- (aq)
• Ksp = [Ax+ (aq)]a [By- (aq)]b
Solubility Product
• When an undissolved ionic compound is in contact with
a saturated solution of its ions, an equilibrium is established
• The ions move from the solid to the saturated solution at the
same rate as they move from the solution to the solid
• For example, the undissolved magnesium chloride (MgCl2) is in
equilibrium with a saturated solution of its ions
• MgCl2 (s) ⇌ Mg2+ (aq) + 2Cl- (aq)
When the undissolved MgCl2 salt gets in contact with its ions in a saturated
solution, an equilibrium between the salt and ions is established
Solubility Product
The Solubility Product
• The solubility product for this
equilibrium is:
• Ksp = [Mg2+ (aq)] [Cl- (aq)]2
• The Ksp is only useful
for sparingly soluble salts
• The smaller the value of Ksp, the
lower the solubility of the salt
Expressing K sp
• The general equilibrium expression for
the solubility product (Ksp) is:
• cC (s) ⇌ aAx+ (aq) + bBy- (aq)
• Ksp = [Ax+ (aq)]a [By- (aq)]b
Worked Example: Expressing Ksp of ionic
compounds
• Calculations involving the solubility product (Ksp) may
include::
• Calculating the solubility product of a compound from its solubility
• Calculating the solubility of a compound from the solubility product
• Worked example: Calculating the solubility product of a
compound from its solubility
• Answer
• Step 1: Write down the equilibrium equation
• PbBr2 (s) ⇌ Pb2+ (aq) + 2Br- (aq)
• Step 2: Write down the equilibrium expression
• Ksp = [Pb2+(aq)] [Br- (aq)]2
• Step 3: Calculate the ion concentrations in the solution
• [PbBr2(s)] = 1.39 x 10-3 mol dm-3
• The ratio of PbBr2 to Pb2+ is 1:1
• [Pb2+(aq)] = [PbBr2(s)] = 1.39 x 10-3 mol dm-3
• The ratio of PbBr2 to Br- is 1:2
• [Br-(aq)] = 2 x [PbBr2(s)] = 2 x 1.39 x 10-3 mol dm-3
• = 2.78 x 10-3 mol dm-3
• Step 4: Substitute the values into the expression to find the solubility product
• Ksp = (1.39 x 10-3) x (2.78 x 10-3)2
• = 1.07 x 10-8
• Step 6: Determine the correct units of Ksp
• Ksp = (mol dm-3) x (mol dm-3)2
• = mol3 dm-9
• The solubility product is therefore 1.07 x 10-8 mol3 dm-9
Worked example: Calculating the
solubility of a compound from its
solubility product
• Step 1: Write down the equilibrium equation
• CuO (s) ⇌ Cu2+ (aq) + O2- (aq)
• Step 2: Write down the equilibrium expression
• Ksp = [Cu2+ (aq)] [O2- (aq)]
• Step 3: Simplify the equilibrium expression
• The ratio of Cu2+ to O2- is 1:1
• [Cu2+(aq)] = [O2-(aq)] so the expression can be simplified to:
• Ksp = [Cu2+ (aq)]2
• Step 4: Substitute the value of Ksp into the expression to find the
concentration
• 5.9 x 10-36 = [Cu2+ (aq)]2
Since [CuO(s)] = [Cu2+ (aq)] the solubility of copper oxide is 2.4 x 10-18 mol dm-3
Exam Tip
• Remember that the solubility product is only applicable to very
slightly soluble salts and cannot be used for soluble salts such as:
• Group 1 element salts
• All nitrates salts
• All ammonium salts
• Many sulfate salts
• Many halide salts (except for lead(II) halides and silver halides)
The Common Ion Effect
Solubility Product and The Common Ion Effect
• A saturated solution is a solution that contains
the maximum amount of dissolved salt
• If a second compound, which has an ion in common with the
dissolved salt, is added to the saturated solution, the solubility of the
salt reduces and a solid precipitate will be formed
• This is also known as the common ion effect
• For example, if a solution of potassium chloride (KCl) is added to
a saturated solution of silver chloride (AgCl) a precipitate of silver
chloride will be formed
• The chloride ion is the common ion
• The solubility product can be used to predict whether a precipitate
will actually form or not
• A precipitate will form if the product of the ion concentrations is greater
than the solubility product (Ksp)
Common Ion Effect in Silver Chloride
• When a KCl solution is added to a saturated solution of AgCl, an AgCl
precipitate forms
• In a saturated AgCl solution, the silver chloride is in equilibrium with
its ions
• AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
• When a solution of potassium chloride is added:
• Both KCl and AgCl have the common Cl- ion
• There is an increased Cl- concentration so the equilibrium position shifts to
the left
• The increase in Cl- concentration also means that [Ag+ (aq)] [Cl-(aq)]
is greater than the Ksp for AgCl
• As a result, the AgCl is precipitated
Common Ion Effect in Silver Chloride
The addition of potassium chloride to a saturated solution of silver chloride results in the
precipitate of silver chloride
Worked Example: Calculations using
the Ksp values and the concentration of
the common ion
• Step 1: Determine the equilibrium reaction of CaSO4
• CaSO4 (s) ⇌ Ca2+ (aq) + SO42- (aq)
• Step 2: Write down the equilibrium expression for Ksp
• Ksp = [Ca2+ (aq)] [SO42- (aq)]
• Step 3: Determine the concentrations of the ions
• There are equal volumes of each solution
• This means that the total solution was diluted by a factor of 2
• The new concentrations of the Ca2+ ion is halved
Ca
2+ = 1 𝑥 10 −3
2
= 5.0 x 10-4 mol dm-3
• The sulfate ion concentration remains the same as it is a
common ion and its concentration is the same in both solutions
• Step 4: Substitute the values into the expression
• Product of the ion concentrations = [Ca2+ (aq)] x [SO42- (aq)]
• = (5.0 x 10-4) x (1.0 x 10-3)
• = 5.0 x 10-7 mol2 dm-6
• Step 5: Determine if a precipitate will form
• As the product of the ion concentration (5.0 x 10-7 mol dm-3 ) is smaller
than the Ksp value (2.0 x 10-5 mol2 dm-6), the CaSO4 precipitate
will not be formed