AN INTRODUCTION TO
SOLUBILITY
PRODUCTS
KNOCKHARDY PUBLISHING
2015
SPECIFICATIONS
KNOCKHARDY PUBLISHING
SOLUBILITY PRODUCTS
INTRODUCTION
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SOLUBILITY PRODUCTS
CONTENTS
• Solubility of ionic compounds
• Saturated solutions
• Solubility products
• Calculations
• Common Ion Effect
SOLUBILITY OF IONIC COMPOUNDS
• ionic compounds tend to be insoluble in non-polar solvents
• ionic compounds tend to be soluble in water
• water is a polar solvent and stabilises the separated ions
SOLUBILITY OF IONIC COMPOUNDS
• ionic compounds tend to be insoluble in non-polar solvents
• ionic compounds tend to be soluble in water
• water is a polar solvent and stabilises the separated ions
• some ionic compounds are very insoluble (AgCl, PbSO4, PbS)
• even soluble ionic compounds have a limit as to how much dissolves
SATURATED SOLUTIONS
• solutions become saturated when solute no longer dissolves in a solvent
• solubility varies with temperature
• most solutes are more soluble at higher temperatures
SATURATED SOLUTIONS
• solutions become saturated when solute no longer dissolves in a solvent
• solubility varies with temperature
• most solutes are more soluble at higher temperatures
Ionic crystal lattices can dissociate
(break up) when placed in water. The
ions separate as they are stabilised
by polar water molecules.
SATURATED SOLUTIONS
• solutions become saturated when solute no longer dissolves in a solvent
• solubility varies with temperature
• most solutes are more soluble at higher temperatures
Ionic crystal lattices can dissociate
(break up) when placed in water. The
ions separate as they are stabilised
by polar water molecules.
Eventually, no more solute dissolves
and the solution becomes saturated.
There is a limit to the concentration
of ions in solution.
SATURATED SOLUTIONS
• solutions become saturated when solute no longer dissolves in a solvent
• solubility varies with temperature
• most solutes are more soluble at higher temperatures
Ionic crystal lattices can dissociate
(break up) when placed in water. The
ions separate as they are stabilised
by polar water molecules.
Eventually, no more solute dissolves
and the solution becomes saturated.
There is a limit to the concentration
of ions in solution.
SOLUBILITY PRODUCT
Even the most insoluble ionic compounds dissolve to a small extent.
An equilibrium exists between the undissolved solid and its aqueous ions;
M+(aq)
(i)
MX(s)
+
X¯(aq)
(ii)
BaSO4(s)
Ba2+(aq) +
SO42-(aq)
(iii)
PbCl2(s)
Pb2+(aq) +
2Cl¯(aq)
SOLUBILITY PRODUCT
Even the most insoluble ionic compounds dissolve to a small extent.
An equilibrium exists between the undissolved solid and its aqueous ions;
M+(aq)
(i)
MX(s)
+
X¯(aq)
(ii)
BaSO4(s)
Ba2+(aq) +
SO42-(aq)
(iii)
PbCl2(s)
Pb2+(aq) +
2Cl¯(aq)
Applying the equilibrium law to (i) and assuming the concentration of
MX(s) is constant in a saturated solution.
[M+(aq)] [X¯(aq)] = a constant, Ksp
[ ] is the concentration in mol dm-3
Ksp is known as the SOLUBILITY PRODUCT
SOLUBILITY PRODUCT
Ag+(aq) + Cl¯(aq)
Ksp = [Ag+(aq)] [Cl¯(aq)]
BaSO4(s)
Ba2+(aq) + SO42-(aq)
Ksp = [Ba2+(aq)] [SO42-(aq)]
PbCl2(s)
Pb2+(aq) + 2Cl¯(aq)
Ksp = [Pb2+(aq)] [Cl¯(aq)]2
AgCl(s)
Notice that the concentration of Cl¯(aq) is raised to the
power of 2 because there are two Cl¯(aq) ions in the equation
SOLUBILITY PRODUCT
Ag+(aq) + Cl¯(aq)
Ksp = [Ag+(aq)] [Cl¯(aq)]
BaSO4(s)
Ba2+(aq) + SO42-(aq)
Ksp = [Ba2+(aq)] [SO42-(aq)]
PbCl2(s)
Pb2+(aq) + 2Cl¯(aq)
Ksp = [Pb2+(aq)] [Cl¯(aq)]2
AgCl(s)
Notice that the concentration of Cl¯(aq) is raised to the power of 2
Complete the equilibrium equation and write an expression for Ksp for…
Pb2+(aq) + S2-(aq)
Ksp = [Pb2+(aq)] [S2-(aq)]
Fe(OH)2(s)
Fe2+(aq) + 2OH¯(aq)
Ksp = [Fe2+(aq)] [OH¯(aq)]2
Fe(OH)3(s)
Fe3+(aq) + 3OH¯(aq)
Ksp = [Fe3+(aq)] [OH¯(aq)]3
PbS(s)
SOLUBILITY PRODUCT
Ag+(aq) + Cl¯(aq)
Ksp = [Ag+(aq)] [Cl¯(aq)]
BaSO4(s)
Ba2+(aq) + SO42-(aq)
Ksp = [Ba2+(aq)] [SO42-(aq)]
PbCl2(s)
Pb2+(aq) + 2Cl¯(aq)
Ksp = [Pb2+(aq)] [Cl¯(aq)]2
AgCl(s)
Notice that the concentration of Cl¯(aq) is raised to the power of 2
Complete the equilibrium equation and write an expression for Ksp for…
Pb2+(aq) + S2-(aq)
Ksp = [Pb2+(aq)] [S2-(aq)]
Fe(OH)2(s)
Fe2+(aq) + 2OH¯(aq)
Ksp = [Fe2+(aq)] [OH¯(aq)]2
Fe(OH)3(s)
Fe3+(aq) + 3OH¯(aq)
Ksp = [Fe3+(aq)] [OH¯(aq)]3
PbS(s)
SOLUBILITY PRODUCT
Ag+(aq) + Cl¯(aq)
Ksp = [Ag+(aq)] [Cl¯(aq)]
BaSO4(s)
Ba2+(aq) + SO42-(aq)
Ksp = [Ba2+(aq)] [SO42-(aq)]
PbCl2(s)
Pb2+(aq) + 2Cl¯(aq)
Ksp = [Pb2+(aq)] [Cl¯(aq)]2
AgCl(s)
Notice that the concentration of Cl¯(aq) is raised to the power of 2
Complete the equilibrium equation and write an expression for Ksp for…
Pb2+(aq) + S2-(aq)
Ksp = [Pb2+(aq)] [S2-(aq)]
Fe(OH)2(s)
Fe2+(aq) + 2OH¯(aq)
Ksp = [Fe2+(aq)] [OH¯(aq)]2
Fe(OH)3(s)
Fe3+(aq) + 3OH¯(aq)
Ksp = [Fe3+(aq)] [OH¯(aq)]3
PbS(s)
SOLUBILITY PRODUCT
Ag+(aq) + Cl¯(aq)
Ksp = [Ag+(aq)] [Cl¯(aq)]
BaSO4(s)
Ba2+(aq) + SO42-(aq)
Ksp = [Ba2+(aq)] [SO42-(aq)]
PbCl2(s)
Pb2+(aq) + 2Cl¯(aq)
Ksp = [Pb2+(aq)] [Cl¯(aq)]2
AgCl(s)
Notice that the concentration of Cl¯(aq) is raised to the power of 2
Complete the equilibrium equation and write an expression for Ksp for…
Pb2+(aq) + S2-(aq)
Ksp = [Pb2+(aq)] [S2-(aq)]
Fe(OH)2(s)
Fe2+(aq) + 2OH¯(aq)
Ksp = [Fe2+(aq)] [OH¯(aq)]2
Fe(OH)3(s)
Fe3+(aq) + 3OH¯(aq)
Ksp = [Fe3+(aq)] [OH¯(aq)]3
PbS(s)
SOLUBILITY PRODUCT
Units The value of Ksp has units and it varies with temperature
units of…
AgCl
Ksp = [Ag+(aq)] [Cl¯(aq)]
mol2 dm-6
BaSO4
Ksp = [Ba2+(aq)] [SO42-(aq)]
mol2 dm-6
PbCl2
Ksp = [Pb2+(aq)] [Cl¯(aq)]2
mol3 dm-9
SOLUBILITY PRODUCT
Units The value of Ksp has units and it varies with temperature
units of…
AgCl
Ksp = [Ag+(aq)] [Cl¯(aq)]
BaSO4
Ksp = [Ba2+(aq)] [SO42-(aq)]
mol2 dm-6
PbCl2
Ksp = [Pb2+(aq)] [Cl¯(aq)]2
mol3 dm-9
Work out the units of Ksp for the following…
PbS
Ksp = [Pb2+(aq)] [S2-(aq)]
Fe(OH)2
Ksp = [Fe2+(aq)] [OH¯(aq)]2
Fe(OH)3
Ksp = [Fe3+(aq)] [OH¯(aq)]3
mol2 dm-6
SOLUBILITY PRODUCT
Units The value of Ksp has units and it varies with temperature
units of…
AgCl
Ksp = [Ag+(aq)] [Cl¯(aq)]
mol2 dm-6
BaSO4
Ksp = [Ba2+(aq)] [SO42-(aq)]
mol2 dm-6
PbCl2
Ksp = [Pb2+(aq)] [Cl¯(aq)]2
mol3 dm-9
Work out the units of Ksp for the following…
PbS
Ksp = [Pb2+(aq)] [S2-(aq)]
mol2 dm-6
Fe(OH)2
Ksp = [Fe2+(aq)] [OH¯(aq)]2
mol3 dm-9
Fe(OH)3
Ksp = [Fe3+(aq)] [OH¯(aq)]3
mol4 dm-12
CALCULATIONS
Solubility products can be used to calculate the solubility of compounds.
At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6.
Calculate the solubility of lead(II) sulphide.
CALCULATIONS
Solubility products can be used to calculate the solubility of compounds.
At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6.
Calculate the solubility of lead(II) sulphide.
The equation for its solubility is
PbS(s)
Pb2+(aq) + S2-(aq)
CALCULATIONS
Solubility products can be used to calculate the solubility of compounds.
At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6.
Calculate the solubility of lead(II) sulphide.
The equation for its solubility is
The expression for the solubility product is
Pb2+(aq) + S2-(aq)
PbS(s)
Ksp =
[Pb2+(aq)] [S2-(aq)]
CALCULATIONS
Solubility products can be used to calculate the solubility of compounds.
At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6.
Calculate the solubility of lead(II) sulphide.
The equation for its solubility is
Pb2+(aq) + S2-(aq)
PbS(s)
The expression for the solubility product is
Ksp =
[Pb2+(aq)] [S2-(aq)]
According to the equation, you get one Pb2+(aq) for
every one S2-(aq); the concentrations will be equal
[Pb2+(aq)] = [S2-(aq)]
Substituting and rewriting the expression for Ksp
Ksp =
[Pb2+(aq)]2
CALCULATIONS
Solubility products can be used to calculate the solubility of compounds.
At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6.
Calculate the solubility of lead(II) sulphide.
The equation for its solubility is
Pb2+(aq) + S2-(aq)
PbS(s)
The expression for the solubility product is
Ksp =
[Pb2+(aq)] [S2-(aq)]
According to the equation, you get one Pb2+(aq) for
every one S2-(aq); the concentrations will be equal
[Pb2+(aq)] = [S2-(aq)]
Substituting and rewriting the expression for Ksp
Ksp =
Re-arranging;
[Pb2+(aq)] =
Ksp
=
4 x 10-28 =
[Pb2+(aq)]2
2 x 10-14 mol dm-3
CALCULATIONS
Solubility products can be used to calculate the solubility of compounds.
At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6.
Calculate the solubility of lead(II) sulphide.
The equation for its solubility is
Pb2+(aq) + S2-(aq)
PbS(s)
The expression for the solubility product is
[Pb2+(aq)] [S2-(aq)]
Ksp =
According to the equation, you get one Pb2+(aq) for
every one S2-(aq); the concentrations will be equal
[Pb2+(aq)] = [S2-(aq)]
Substituting and rewriting the expression for Ksp
Ksp =
Re-arranging;
[Pb2+(aq)] =
Ksp
=
4 x 10-28 =
As you get one Pb2+ from one PbS, the solubility of PbS
[Pb2+(aq)]2
2 x 10-14 mol dm-3
=
2 x 10-14 mol dm-3
CALCULATIONS
Solubility products can be used to calculate the solubility of compounds.
At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6.
Calculate the solubility of lead(II) sulphide.
The equation for its solubility is
Pb2+(aq) + S2-(aq)
PbS(s)
The expression for the solubility product is
[Pb2+(aq)] [S2-(aq)]
Ksp =
According to the equation, you get one Pb2+(aq) for
every one S2-(aq); the concentrations will be equal
[Pb2+(aq)] = [S2-(aq)]
Substituting and rewriting the expression for Ksp
Ksp =
Re-arranging;
[Pb2+(aq)] =
Ksp
=
4 x 10-28 =
As you get one Pb2+ from one PbS, the solubility of PbS
Mr for PbS is 239; the solubility is 239 x 2 x 10-14 g dm-3
[mass = moles x molar mass]
[Pb2+(aq)]2
2 x 10-14 mol dm-3
=
=
2 x 10-14 mol dm-3
5.78 x 10-12 g dm-3
CALCULATIONS
Solubility products can be used to calculate the solubility of compounds.
At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6.
Calculate the solubility of lead(II) sulphide.
The equation for its solubility is
Pb2+(aq) + S2-(aq)
PbS(s)
The expression for the solubility product is
[Pb2+(aq)] [S2-(aq)]
Ksp =
According to the equation, you get one Pb2+(aq) for
every one S2-(aq); the concentrations will be equal
[Pb2+(aq)] = [S2-(aq)]
Substituting and rewriting the expression for Ksp
Ksp =
Re-arranging;
[Pb2+(aq)] =
Ksp
=
4 x 10-28 =
As you get one Pb2+ from one PbS, the solubility of PbS
Mr for PbS is 239; the solubility is 239 x 2 x 10-14 g dm-3
[mass = moles x molar mass]
[Pb2+(aq)]2
2 x 10-14 mol dm-3
=
=
2 x 10-14 mol dm-3
5.78 x 10-12 g dm-3
CALCULATIONS
The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative
mass of MY is 200. Calculate the solubility product of the salt MY at 25°C.
CALCULATIONS
The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative
mass of MY is 200. Calculate the solubility product of the salt MY at 25°C.
Solubility of MY in mol dm-3
= solubility in g
molar mass
=
5 x 10-10 g dm-3
200
CALCULATIONS
The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative
mass of MY is 200. Calculate the solubility product of the salt MY at 25°C.
Solubility of MY in mol dm-3
= solubility in g
molar mass
=
=
5 x 10-10 g dm-3
200
2.5 x 10-12 mol dm-3
CALCULATIONS
The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative
mass of MY is 200. Calculate the solubility product of the salt MY at 25°C.
Solubility of MY in mol dm-3
= solubility in g
molar mass
=
=
The equation for its solubility is
MY(s)
5 x 10-10 g dm-3
200
2.5 x 10-12 mol dm-3
M+(aq) + Y-(aq)
CALCULATIONS
The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative
mass of MY is 200. Calculate the solubility product of the salt MY at 25°C.
Solubility of MY in mol dm-3
= solubility in g
molar mass
=
=
The equation for its solubility is
The expression for the solubility product is
MY(s)
Ksp =
5 x 10-10 g dm-3
200
2.5 x 10-12 mol dm-3
M+(aq) + Y-(aq)
[M+(aq)] [Y-(aq)]
CALCULATIONS
The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative
mass of MY is 200. Calculate the solubility product of the salt MY at 25°C.
Solubility of MY in mol dm-3
= solubility in g
molar mass
=
=
The equation for its solubility is
The expression for the solubility product is
According to
the equation;
MY(s)
Ksp =
5 x 10-10 g dm-3
200
2.5 x 10-12 mol dm-3
M+(aq) + Y-(aq)
[M+(aq)] [Y-(aq)]
moles of M+(aq) = moles of Y-(aq) = moles of dissolved MY(s)
CALCULATIONS
The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative
mass of MY is 200. Calculate the solubility product of the salt MY at 25°C.
Solubility of MY in mol dm-3
= solubility in g
molar mass
=
=
The equation for its solubility is
The expression for the solubility product is
According to
the equation;
MY(s)
Ksp =
5 x 10-10 g dm-3
200
2.5 x 10-12 mol dm-3
M+(aq) + Y-(aq)
[M+(aq)] [Y-(aq)]
moles of M+(aq) = moles of Y-(aq) = moles of dissolved MY(s)
Substituting values
Ksp = [2.5 x 10-12 ] [2.5 x 10-12 ]
The value of the solubility product
Ksp = 6.25 x 10-24 mol2 dm-6
CALCULATIONS
The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative
mass of MY is 200. Calculate the solubility product of the salt MY at 25°C.
Solubility of MY in mol dm-3
= solubility in g
molar mass
=
=
The equation for its solubility is
The expression for the solubility product is
According to
the equation;
MY(s)
Ksp =
5 x 10-10 g dm-3
200
2.5 x 10-12 mol dm-3
M+(aq) + Y-(aq)
[M+(aq)] [Y-(aq)]
moles of M+(aq) = moles of Y-(aq) = moles of dissolved MY(s)
Substituting values
Ksp = [2.5 x 10-12 ] [2.5 x 10-12 ]
The value of the solubility product
Ksp = 6.25 x 10-24 mol2 dm-6
THE COMMON ION EFFECT
Adding a common ion, (one which is present in the solution), will result in the
precipitation of a sparingly soluble ionic compound.
eg
Adding a solution of sodium chloride to a saturated solution of
silver chloride will result in the precipitation of silver chloride.
THE COMMON ION EFFECT
Adding a common ion, (one which is present in the solution), will result in the
precipitation of a sparingly soluble ionic compound.
eg
Adding a solution of sodium chloride to a saturated solution of
silver chloride will result in the precipitation of silver chloride.
Adding the ionic compound MA to a
solution of MX increases the
concentration of M+(aq). M+(aq) is a
common ion as it is already in solution.
THE COMMON ION EFFECT
Adding a common ion, (one which is present in the solution), will result in the
precipitation of a sparingly soluble ionic compound.
eg
Adding a solution of sodium chloride to a saturated solution of
silver chloride will result in the precipitation of silver chloride.
Adding the ionic compound MA to a
solution of MX increases the
concentration of M+(aq). M+(aq) is a
common ion as it is already in solution.
The extra M+ ions means that the
solubility product is exceeded. To
reduce the value of [M+(aq)][X-(aq)]
below the Ksp, some ions are removed
from solution by precipitating.
THE COMMON ION EFFECT
Adding a common ion, (one which is present in the solution), will result in the
precipitation of a sparingly soluble ionic compound.
eg
Adding a solution of sodium chloride to a saturated solution of
silver chloride will result in the precipitation of silver chloride.
Silver chloride dissociates in water as follows
The solubility product at 25°C is
Ag+(aq) + Cl¯(aq)
AgCl(s)
Ksp =
[Ag+(aq)] [Cl¯(aq)]
= 1.2 x 10-10 mol2 dm-6
If the value of the solubility product is exceeded, precipitation will occur.
THE COMMON ION EFFECT
Adding a common ion, (one which is present in the solution), will result in the
precipitation of a sparingly soluble ionic compound.
eg
Adding a solution of sodium chloride to a saturated solution of
silver chloride will result in the precipitation of silver chloride.
Silver chloride dissociates in water as follows
The solubility product at 25°C is
Ag+(aq) + Cl¯(aq)
AgCl(s)
Ksp =
[Ag+(aq)] [Cl¯(aq)]
= 1.2 x 10-10 mol2 dm-6
If the value of the solubility product is exceeded, precipitation will occur.
The value can be exceeded by adding EITHER of the two soluble ions.
If sodium chloride solution is added, the concentration of Cl¯(aq) will increase and
precipitation will occur.
Likewise, addition of silver nitrate solution AgNO3(aq) would produce the same
effect as it would increase the concentration of Ag+(aq).
THE COMMON ION EFFECT
If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3)
solutions are mixed, will AgCl be precipitated?
Ksp = 1.2 x10-10 mol2 dm-6
THE COMMON ION EFFECT
If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3)
solutions are mixed, will AgCl be precipitated?
Ksp = 1.2 x10-10 mol2 dm-6
• in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3
in NaCl the concentration of Cl¯ is 2 x 10-5 mol dm-3
THE COMMON ION EFFECT
If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3)
solutions are mixed, will AgCl be precipitated?
Ksp = 1.2 x10-10 mol2 dm-6
• in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3
in NaCl the concentration of Cl¯ is 2 x 10-5 mol dm-3
• when equal volumes are mixed, the concentrations are halved
[Ag+] = 1 x 10-5 mol dm-3
[Cl¯] = 1 x 10-5 mol dm-3
THE COMMON ION EFFECT
If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3)
solutions are mixed, will AgCl be precipitated?
Ksp = 1.2 x10-10 mol2 dm-6
• in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3
in NaCl the concentration of Cl¯ is 2 x 10-5 mol dm-3
• when equal volumes are mixed, the concentrations are halved
[Ag+] = 1 x 10-5 mol dm-3
[Cl¯] = 1 x 10-5 mol dm-3
• [Ag+] [Cl¯] = [1 x 10-5 mol dm-3] x [1 x 10-5 mol dm-3]
= 1 x 10-10 mol2 dm-6
THE COMMON ION EFFECT
If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3)
solutions are mixed, will AgCl be precipitated?
Ksp = 1.2 x10-10 mol2 dm-6
• in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3
in NaCl the concentration of Cl¯ is 2 x 10-5 mol dm-3
• when equal volumes are mixed, the concentrations are halved
[Ag+] = 1 x 10-5 mol dm-3
[Cl¯] = 1 x 10-5 mol dm-3
• [Ag+] [Cl¯] = [1 x 10-5 mol dm-3] x [1 x 10-5 mol dm-3]
= 1 x 10-10 mol2 dm-6
• because this is lower than the Ksp for AgCl... NO PRECIPITATION OCCURS
THE COMMON ION EFFECT
If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3)
solutions are mixed, will AgCl be precipitated?
Ksp = 1.2 x10-10 mol2 dm-6
• in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3
in NaCl the concentration of Cl¯ is 2 x 10-5 mol dm-3
• when equal volumes are mixed, the concentrations are halved
[Ag+] = 1 x 10-5 mol dm-3
[Cl¯] = 1 x 10-5 mol dm-3
• [Ag+] [Cl¯] = [1 x 10-5 mol dm-3] x [1 x 10-5 mol dm-3]
= 1 x 10-10 mol2 dm-6
• because this is lower than the Ksp for AgCl... NO PRECIPITATION OCCURS
AN INTRODUCTION TO
SOLUBILITY
PRODUCTS
THE END
© 2015 JONATHAN HOPTON & KNOCKHARDY PUBLISHING