SolubilityEquil 1
Solubility Product
If an “insoluble” or slightly soluble material placed in water then there is Equilibrium between solid and ions in solution for AgCl (s)  Ag + (aq) + Cl - (aq)
K = [Ag + ] [Cl
[AgCl]
] but there is no conc for pure solid [AgCl] so just write
K sp
= [Ag + ] [ Cl ]
K sp
is solubility product
K sp
is function of temperature table of values at 25°C
Concentration is in mol/L
K sp
(AgCl) = 1.7 x10
K sp
(PbCl
2
) = 1.6 x10
-10
-5
More soluble then more ions in solution and K sp
is larger value
Example - Bisthmium Sulfide
For Bi
2
S
3
 2 Bi 3+ + 3 S 2-
Write K sp
= [Bi 3+ ] 2 [S 2] 3
Must account for all ions present but not solid

SolubilityEquil 2
K sp
Problems
1. Know concentrations find K sp
2. Know K sp
find concentrations
Example (1) :
For Silver Chromate
Maximum 7.8 x10 -5 mol Ag
2
CrO
4
dissolves in 1 L of water
Find K sp
= ?
Ag
2
CrO
4
 2 Ag + + CrO
4
2-
] 2 [CrO
4
2] K sp
= [Ag +
K sp
= [ ( 2 ) ( 7.8 x10
K sp
= 1.9 x10 -12
-5 ) ] 2 [ 7.8 x10 -5 ]
Example (2):
For CaF
2
given that K sp
= 3.9 x10 -11 find concentrations in saturated solution [Ca 2+ ] = ? [F ] = ?
CaF
2
 Ca 2+ + 2 F -
x 2x
K sp
= [Ca
3.9 x10 -11
3.9 x10 -11
2+
= [x] [2x] x = 2.1 x10
] [F
= 4x
-4
]
3
2
2
So…
[Ca 2+
[F -
] = 2.1 x10 -4 mol/L
] = 4.2 x10 -4 mol/L

SolubilityEquil 3
Corrections to Model
Complications in this model occur and less correct if:
1.
at higher concentrations
2.
if other ions present
A more complex analysis uses activities in place of concentrations but we can ignore this correction.
Example when K sp
does not predict correctly due to salt effect – solubility of a salt is increased by presence of other ions in solution
AgCl 20% more soluble in 0.20 M KNO
3 more stable and tend to stay in solution
, other ions surround ions in solution so
Normally don’t have such complications for dilute solutions (can ignore)

SolubilityEquil 4
Precipitation
The formula for the concentrations used to calculate the K
[Ag
[Bi
+
3+
] [Cl
]
[Ag + ] 2
2 [S
]
2] 3
[CrO
4
2]
Ion product (IP) – actual concentrations present sp
is the ion product
Relation between K sp
and ion product
Ion product < K sp
Unsaturated solution
Ion product = K sp
Saturated solution
Ion product > K sp
Supersaturated solution expect precipitate will form
Supersaturated dissolve at one temperature where K sp
greater, then allow temp to come to desired value if ions stay in solution then IP > K sp
Normally,
Unless special information given
IP > K sp
Precipitate will form
IP ≤ K sp
No precipitate
IP Example-
Will precipitate form from
10 mL of 0.010M AgNO
3
and 10 mL 0.0001 M NaCl mixed if K sp
= 1.7 x10 -10 AgCl ?
(Conc.) (Vol.) = (mol) moles of Ag
Cl -
+ = (0.01 mol/L) (0.010 L) = 1.0 x10 -4
= (1.0 x10 -4
mol
mol/L) (0.010 L) = 1.0 x10 -6 mol
Conc. after dilution (1.0 x10 -4 mol)/ (0.020 L) = 5.0 x10 -3 M Ag +
(1.0 x10 -6 mol)/ (0.020 L) = 5.0 x10 -5 M Cl -
[Ag + ] [Cl ] = 2.5 x10 -7 > K sp
AgCl will precipitate!

SolubilityEquil 5
IP Example –
Will precipitate form from?
0.001M Mg(NO
3
)
2
in pH = 9.0 given K sp
= 8.98 x10 -12 Mg(OH)
2
So first find pOH = 5.0 then [OH-] = 1.0 x10 -5
IP = [Mg 2+] [OH
= [1.0 x10 -3
] 2
] [1.0 x10
= 1.0 x10 -13 < K sp
-5 ] 2
No precipitate