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Brown's Chemistry
Theoretical–Physical
Advanced Level
Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 4.2
4.2
Solubility
Product Ksp and common ion effect
What do we mean by 'solubility product'?
What is the 'common ion effect'? How do you write Ksp equilibrium
expressions? How do you do solubility product calculations?
GCSE/IGCSE
Notes on reversible reactions and chemical equilibrium
Part 4 sub–index 4.1 Partition between two
phases * 4.2 Solubility product Ksp
& common ion effect *
4.3
Ion–exchange systems
Advanced Equilibrium Chemistry Notes Part 1. Equilibrium,
Le Chatelier's Principle–rules
* Part 2. Kc and Kp equilibrium expressions and
calculations * Part 3.
Equilibria and industrial processes * Part 4 sub–index
(this section): 4.1
Partition between two
phases * 4.2 Solubility product Ksp
and common ion effect *
4.3 Ion–exchange systems *
Part 5. pH, weak–strong acid–base theory and
calculations * Part 6. Salt hydrolysis,
Acid–base titrations–indicators, pH curves and buffers * Part 7.
Redox equilibria, half–cell electrode potentials,
electrolysis and electrochemical series
*
Part 8. Phase equilibria–vapour
pressure, boiling point and intermolecular forces
*
M = old fashioned shorthand for mol dm–3
4.2 Solubility
Product and the Common Ion Effect
-
When a
sparingly soluble salt is mixed with water a dynamic equilibrium
is established in which salt is constantly dissolving and
crystallising at the same rate when the solution is saturated, and the
maximum constant concentration is achieved.
-
e.g. for calcium
sulphate: CaSO4(s)
+ aq Ca2+(aq) + SO42–(aq)
-
However, since the
concentration of water and the solid is effectively constant the
equilibrium expression is simplified to:
-
Ksp
= [Ca2+(aq)] [SO42–(aq)]
= 2.4 x 10–5 mol2 dm–3
-
Ksp
is called the solubility product of the ions concerned and
is constant at constant temperature for a saturated solution
i.e. when no more will dissolve.
-
The solubility
product for a sparingly soluble strong electrolyte is defined as the
product of the concentration of the ions raised to their appropriate
powers in a saturated solution at a specific temperature.
-
At the saturation
solution point, controlled by the Ksp expression, it doesn't matter how much solid you add, no more can
dissolve.
-
The solubility of
the sparingly soluble salt is governed by the Ksp
expression, i.e. whatever ion concentrations are present of the
compound, then the expression must be obeyed. (see
common ion effect below).
-
If on mixing
solutions containing the two constituent ions the Ksp
expression is exceeded, precipitation will take place until the
product of the ion concentrations equals the Ksp value. If
the Ksp expression is not exceeded, no precipitation will take place.
-
–
-
Other
solubility product expressions
-
Note: In the Ksp
expression, the
ion concentrations are raised to powers equal to their molar ratio in the compound:
-
silver chloride :
AgCl(s) + aq Ag+(aq) + Cl–(aq)
-
calcium carbonate:
CaCO3(s) + aq Ca2+(aq) + CO32–(aq)
(ΔH slightly –ve)
-
Ksp
= [Ca2+(aq)] [CO32–(aq)]
= 5.0 x 10–10 mol2 dm–6
-
Sea shells are
mainly made of calcium carbonate which is a tough mineral and will not
dissolve appreciably in water.
-
However, when the
sea creatures die and the remains sink to great depth in the oceans,
the shells will then dissolve because of changes in the position of
two equilibria. At great depths the pressure is much higher and the
temperature lower, both factors favouring more CO2
dissolving, therefore the following forward reaction is promoted,
-
CaCO3(aq)
+ CO2(aq) + H2O(l) Ca2+(aq) + 2HCO3–(aq)
-
and secondly,
the lower temperature favours a higher Ksp value because
the dissolving process is slightly exothermic, so more of the
calcium carbonate will dissolve. There are no shells lying at the
bottom of the oceans! The limestone cliffs you see in the landscape
were formed in warm shallow tropical seas.
-
magnesium
hydroxide : Mg(OH)2(s) + aq Mg2+(aq) + 2OH–(aq)
-
silver
chromate(VI) : Ag2CrO4(aq) + aq
2Ag+(aq) + CrO42–(aq)
-
aluminium
hydroxide : Al(OH)3(s) + aq Al3+(aq) + 3OH–(aq)
-
antimony sulphide:
S2S3(s) + aq 2Sb3+(aq) + 3S2–(aq)
-
The
common ion effect
-
If a 2nd soluble
substance is added to the saturated solution of a salt, that has an
ion in common with the salt, this will cause precipitation of the
salt. This is because the Ksp expression is initially
exceeded, causing precipitation until the new concentrations satisfy
the Ksp expression Le Chatelier again!). This precipitation
is known as the 'common ion effect' and happens irrespective of
the solubility of the substances e.g.
-
If you add some
concentrated hydrochloric acid to a saturated sodium chloride
('salt' is very soluble) solution ('brine') sodium chloride is
precipitated.
-
NaCl(s)
Na+(aq) + Cl–(aq)
-
[Na+(aq)]
x [Cl–(aq)] is ~constant for a saturated
solution of sodium chloride at constant temperature. If you increase
one ion concentration, the other must decrease to maintain
equilibrium, and the only way this can happen is by precipitation of
the salt.
-
Applying Le
Chatelier's Principle, the conc. HCl(aq) increases the
chloride ion Cl–(aq) concentration so the
reaction 'net change' must initially move to the left to restore
equilibrium i.e. NaCl is precipitated.
-
Similarly if you
add dilute sulphuric acid OR saturated calcium chloride solution to
saturated calcium sulphate (sparingly soluble) solution then calcium
sulphate appears as a faint precipitate.
-
CaSO4(s)
Ca2+(aq) + SO42–(aq)
-
[Ca2+(aq)]
x [SO42–(aq)] is ~constant for
a saturated solution of calcium sulfate at constant temperature.
-
The sulphuric
acid increases the sulfate ion SO42–(aq)
concentration OR calcium chloride increases the calcium ion Ca2+(aq)
concentration. Either way, increasing one of the concentrations of
the salts constituent ions will cause precipitation of calcium
sulfate according to Le Chatelier's Principle.
-
Example
calculation 4.2.1
-
Example
calculation 4.2.2
-
If equal volumes
of aqueous 0.01 mol dm–3 sodium chloride and 0.005 mol dm–3
silver nitrate solution are mixed, show by calculation whether or not
a precipitate of silver chloride forms?
-
[NaCl(aq)]
= [Cl–(aq)] = 0.01 / 2 = 0.005 mol dm–3
-
[AgNO3(aq)]
= [Ag+(aq)] = 0.005 / 2 = 0.0025 mol dm–3
-
The ionic product
= [Ag+(aq)] x [Cl–(aq)] =
0.005 x 0.0025 = 1.25 x 10–5 mol2 dm–6
-
The ionic
product on mixing exceeds the Ksp value so precipitation
takes place.
-
Example calculation 4.2.3
-
For the
equilibrium: CaSO4(s) + aq Ca2+(aq) + SO42–(aq)
-
(a) What is
the solubility of calcium sulphate in a saturated solution of the salt
in g cm–3?
-
From the arguments
outlined in example 4.2.1(a)
-
[Ca2+(aq)]
= [CaSO4(aq)] =
√(2.0 x 10–5) =
4.47 x 10–3 mol dm–3
-
Ar's:
Ca = 40, S = 32, O = 16, Mr(CaSO4) = 136, 1 dm3
= 103 cm3
-
solubility = 4.47
x 10–3 x 136 = 0.608 g dm–3, so scaling down to
1 cm3
-
solubility
= 0.608 / 103 =
6.1 x 10-4 g/cm3
(answer to 2 sf)
-
(b)
Assuming equal volumes of solutions are mixed, what is the minimum
concentration of sodium sulphate (Na2SO4)
solution that when added to a 1.0 x 10–4 mol dm–3
solution of calcium sulphate, will just begin to cause precipitation
of calcium sulphate?
-
Let [NaaSO4(aq)]
= [SO42–(aq)] = the theoretical
concentration of sodium sulphate at the point of precipitation.
-
On mixing the
equal volumes of solutions, the calcium sulphate concentration is
halved.
-
Therefore on
substituting into the Ksp expression on the point of
precipitation ...
-
2.0 x 10–5
= ((1.0 x 10–4)/2) x [NaaSO4(aq)]
-
therefore [NaaSO4(aq)]
= 2.0 x 10–5/0.5 x 10–4 = 0.40 mol dm–3
-
However, on
mixing, the sodium sulphate solution concentration must also be
halved,
-
therefore the
original sodium sulphate solution must be at least
0.8 mol dm–3
-
Example
calculation 4.2.4
-
Equal volumes of
0.025 mol dm–3 potassium bromide (KBr) and 0.005 mol
dm–3 lead(II) nitrate (Pb(NO3)2)solutions
were mixed.
-
(a) Write out (i)
the Ksp expression for lead(II) bromide and (ii) the ionic
equation for its precipitation.
-
(b) Show by
calculation if lead(II) bromide precipitates after mixing the
solutions.
-
[Br–(aq)]
= [KBr(aq)] = 0.025 / 2 = 1.25 x 10–2 mol dm–3
-
[Pb2+(aq)]
= [Pb(NO3)2(aq)] = 0.005 / 2 = 2.5 x
10–3 mol dm–3
-
The ionic
product = [Pb2+(aq)] [Br–(aq)]2
= 2.5 x 10–3 x (1.25 x 10–2)2 =
3.91 x 10–7 mol3 dm–9
-
The Ksp
value of 7.9 x 10–5 is NOT exceeded, so no precipitation
takes place.
Part 4 sub–index
4.1 Partition between two
phases *
4.2 Solubility product Ksp
& common ion effect *
4.3
Ion–exchange systems
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