CHAPTER 7 - IONIC EQUILIBRIA

SOLUBILITY EQUILIBRIUM
• Some salts are soluble but most are

insoluble or slightly soluble in water.

• A saturated solution is a solution that
contains the maximum amount of solute
that can dissolve in a solvent.

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 The solubility of a salt is the amount of
solid that dissolved in a known value of
saturated solution.

 The unit of solubility used may be gL-1 or
molL-1

 Molar solubility is the maximum number of
moles of solute that dissolves in a certain
quantity of solvent at a specific temperature.

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THE SOLUBILITY PRODUCT CONSTANT, Ksp

Ksp is the product of the molar concentrations of
the ions involved in the equilibrium, each raised
to the power of its stoichiometric coefficient in
the equilibrium equation.

 Ksp is called the solubility product constant.

 The degree of solubility of a salt is shown by
the value of Ksp.

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Consider the equilibrium system below :

MX (s) M+ (aq) + X- (aq)

Kc = [M+] [X-]
[MX]

Kc [MX] = [M+] [X-]
*since [MX] is a constant ;

Ksp = [M+] [X-]

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 Soluble salt such NaCl and KNO3 has
an extremely high value of Ksp .

 The smaller the value of Ksp the less
soluble the compound in water.

 Temperature ↑ , solubility ↑, Ksp ↑

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Example

Write the solubility product expression for each
of the following ionic compounds.

a) Ca3(PO4)2 b) Ag2CO3

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Solution:

a) Ca3(PO4)2

Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq)

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

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Solution:

b) Ag2CO3 2Ag+(aq) + CO32-(aq)
Ag2CO3(s) [ Ag+]2 [ CO32-]
Ksp =

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Example

The solubility product of silver chromate(VI), Ag2CrO4 is
2.4 x10-12 mol3dm-9. Calculate the concentration of
Ag+(aq) and CrO42-(aq) in the saturated solution.
Ans: [Ag+] = 1.68 x 10-5 M

[CrO42-] = 8.43 x 10-5 M

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Solution:

Ag2CrO4(s) 2Ag+(aq) + CrO42-(aq)
2x x

Ksp = [ Ag+]2 [ CrO42-]
2.4 X 10-12 = (2x)2 (x)

X= 8.4 x 10-3 = [CrO42-]
[Ag+] = 2(8.4 x 10-3)
1.7 x 10-3 M
=

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Example

The solubility of silver sulphide, Ag2S is
5.0x10-17. Calculate the solubility product
of Ag2S.
Ans: 5.0 x 10-49

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Solution:

Ag2S(s) 2Ag+(aq) + S2-(aq)
Ksp 2x x

= [ Ag+]2 [ S2-]
= (2x)2 (x)
= 4X3
= 5.0 x 10-49

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Example

Calculate the solubility of copper (II) hydroxide,
Cu(OH)2, in g L-1.
Ksp Cu(OH)2 = 2 x 10-20 M3,
Mr Cu(OH)2 = 97.57g mol-1

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Solution:

Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)
x 2x

Ksp = [Cu2+][OH-]2

2 x 10-20 = (x) (2x2) = 4x3

x = 1.709 x 10-7mol/L

Solubility of = 1.709 x 10-7mol/LX 97.57g/mol
Cu(OH)2 = 1.667 x 10-5 g/L

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Predicting the formation of a precipitate

✔ Precipitate is an insoluble solid formed (in the
solution) and is separated from the solution.

✔ A mixture of two solutions may produce a
precipitate depends on the value of Q.

✔ Q has the same expression as Ksp but the
concentrations of ions involved are not in
equilibrium (at any given time).

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✔ If a solution containing M+ ions is mixed with a
solution contains A- ions, then the ion product
expression, Q is given as :
MA (s) ⇋ M+ (aq) + A- (aq)
Q = [M+] [A-]

✔ At equilibrium, the solubility product constant
expressioncan be written as
Ksp = [M+] [A-]

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Three possible situations:

Q < Ksp;
Solution is not saturated.
Solid will dissolve and no precipitate formed.

Q = Ksp;
Saturated solution formed.

System is in equilibrium.

Q > Ksp;
Solution is supersaturated;

Ions will form precipitate until the ionic concentration

product of the system equals the Ksp (until the

system reaches equilibrium). 167

Example

Will precipitate form when 50 mL 0.001 M
NaOH is added to 150 mL of 0.01 M MgCl2.
(Ksp Mg(OH)2 = 2 x 10-11)

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Solution:

No. of mole of NaOH = MV/1000 = (0.001x50)/1000
= 5 x 10-5 mol

[NaOH] = 5x10-5mol/(0.05 + 0.15)L = 2.5 x 10-4M

No. of mole of MgCl2 = MV/1000 = (0.01x150)/1000
= 1.5 x 10-3 mol

[MgCl2] = 1.5x10-3mol/(0.05 + 0.15)L = 7.5 x 10-3M

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Solution:

MgOH(s) ⇋ Mg2+(aq) + 2OH-(aq)
Q = [ Mg2+] [ OH-]2

= (7.5 x 10-3) (2.5 x 10-4)2

= 4.69x10-10

Q > Ksp. Solution is supersaturated. Ions
will form precipitate until the system reach

equilibrium. 170

Example

Exactly 200 mL of 0.0040 M AgNO3 are
added to 800 mL of 0.008 M K2SO4. Will a
precipitate form?
(Ksp Ag2SO4 = 1.4x10-5)

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Solution:

No. of mole of AgNO3 = MV/1000 = (0.004x2000)/1000
= 8.0 x 10-4 mol

[AgNO3] = 5x10-5mol/(0.2 + 0.8)L = 8.0 x 10-4 M

No. of mole of K2SO4 = MV/1000 = (0.08x800)/1000
= 6.4 x 10-3 mol

[K2SO4] = 6.4x10-3mol /(0.2 + 0.8)L = 6.4 x 10-3M

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Solution:

Ag2SO4(s) 2Ag+(aq) + SO42-(aq)
Q = [ Ag+]2 [ SO42-]
= (8.0x10-4)2(6.4x10-3)
= 4.096 x 10-9

Q < Ksp. Solution unsaturated solid will
dissolve and not precipitate formed.

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COMMON ION EFFECT

Common ion is an ion that is common to
two or more components in a mixture of
an ionic solution.

Common ion effect is the shift in
equilibrium caused by the addition of a
compound having an ion in common
with the dissolved substances.

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Consider a beaker contains silver chloride, AgCl with
its solid and ions in equilibrium.

(Ksp AgCl = 1.6 x 10-10)

The equilibrium system is:
AgCl (s) ⇋ Ag+ (aq) + Cl- (aq)

If AgNO3 is added to the saturated AgCl solution :
AgNO3 (aq) → Ag+ (aq) + NO3- (aq)
(common ion)

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 [Ag+] will increase.

 Equilibrium position will shift backward and more
precipitate of AgCl (s) formed.

 The solubility of AgCl in the solution decreases
(compare the solubility in water).

 So, the addition of a common ion will reduce the
solubility of a slightly soluble salt.

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Example

Calculate the solubility of AgCl (mol L-1) in :

a. liquid water
b. 0.05 M of silver nitrate solution.

(Ksp AgCl = 1.6 x 10-10)

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Solution:

a. liquid water

AgCl (s) Ag+ (aq) + Cl- (aq)

(x) (x)

Ksp = [Ag+][Cl-]

1.6 x 10-10 = x2

x = 1.26 x 10-5M
Solubility of AgCl in water is 1.26 x 10-5M

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b. 0.05M of silver nitrate solution.
(Ksp AgCl = 1.6 x 10-10)

AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp (0.05 + x) (x)

= [Ag+][Cl-]

1.6 x 10-10 = (0.05 + x) (x)

K ↓↓, x<<0.05 until 0.05 + x ≈ 0.05
sp

1.6 x 10-10 = 0.05x

x = 3.2 x 10-9M
Solubility of AgCl in AgNO3 is 3.2 x 10-9M.

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