solubility solubility constant factors affecting solubility complex ion effect to solubility

1. 8–1
John A. Schreifels
Chemistry 212
Chapter 18-1
Chapter 18
Solubility and Complex-Ion Equilibria

2. 8–2
John A. Schreifels
Chemistry 212
Chapter 18-2
Overview
• Solubility Equilibria
– Solubility Product Constant
– Solubility and Common Ion Effect
– Precipitation Calculations
– Effect of pH on Solubility
• Complex-Ion Equilibria
– Complex Ion Formation
– Complex Ions and Solubility
• Application of Solubility Equilibria
– Qualitative analysis of metal ions

3. 8–3
John A. Schreifels
Chemistry 212
Chapter 18-3
Solubility Equilibria
• Solubility of a solid treated as with other equilibria. Solution is
saturated. No more solid will dissolve since dynamic
equilibrium.
AgCl(s)  Ag+(aq) + Cl(aq) Ksp = [Ag+][Cl]
• Solid not included in the equilibrium expression.
– MyXz(s)  yM+p(aq)+zXq(aq) Ksp=[M+p]y[Xq]z
where Ksp = solubility product.
E.g. determine the equilibrium expression for each: PbCl2,
Ag2SO4,Al(OH)3.
• Ksp can be determined if the solubility is known.
E.g. Determine Ksp for silver chromate (Ag2CrO4) if its solubility
in water is 0.0290 g/L at 25C.
– Determine molar solubility.
– Determine Ksp.
E.g. 2 Determine Ksp of CaF2 if its solubility is 2.20x104M.

4. 8–4
John A. Schreifels
Chemistry 212
Chapter 18-4
SOLUBILITY FROM Ksp
• Can be determined by using stoichiometry to express all quantities in
terms of one variable- solubility, x. i.e. for the reaction. Use equilibrium
table to write concentration of each in terms of the compound dissolving
E.g. determine the solubility of PbCl2 if its Ksp = 1.2x105
PbCl2(s)  Pb2+(aq) + 2Cl(aq)
E.g.1 Determine solubility of AgCl if its Ksp = 1.8x1010M2.
E.g.2 Determine solubility of Ag2CO3 if its Ksp = 8.1x1012M3.
E.g.3 Determine solubility of Fe(OH)3 if its Ksp = 4x1038M4
PbCl2(s)  Pb2+
+ 2Cl
Initial conc. 0 0
C +x +2x
Equilibrium conc. +x +2x
AgCl (s)  Ag+
+ Cl
Initial conc. 0 0
C +x +x
Equilibrium conc. +x +x

5. 8–5
John A. Schreifels
Chemistry 212
Chapter 18-5
Factors that Affect Solubility
• The common–Ion effect (Remember
LeChatelier’s Principle)
E.g. Determine solubility of PbCl2 (Ksp =
1.2x105)in 0.100M NaCl.
– Write equilibrium table in terms of x and [Cl]
a common–ion
reduces the
solubility of the
compound.
– Assume that [Cl]NaCl >>x
– Solve for x.
• E.g. determine the solubility of CaF2 in a solution of CaCl2. Ksp =
3.9x1011.
PbCl2(s)  Pb2+
+ 2Cl
Initial conc. 0 0.100 M
C +x +2x
Equilibrium conc. +x 0.100 M +2x

6. 8–6
John A. Schreifels
Chemistry 212
Chapter 18-6
Precipitation of Ionic Compounds
• Starting with two solutions, Qsp used to predict precipitation and even the extent of
it.
– Precipitation = reverse of dissolution
– Precipitation occurs when Qsp > Ksp until Qsp = Ksp
– If Qsp < Ksp, precipitation won’t occur.
E.g. determine if precipitation occurs after mixing 50.00 mL 3.00x103 M BaCl2 and
50.00 mL 3.00x103 M Na2CO3.
Solution:
– CBaCl2 = 1.50x103 M; CNa2CO3 = 1.50x103 M
– Qsp = 1.50x103 M1.50x103 M = 2.25x106
– Qsp >1.1x1010.= Ksp precipitation.
E.g. 2 determine equilibrium concentration of each after precipitation occurs.
Solution:
– assume complete precipitation occurs;
– set up equilibrium table; and solve for equilibrium concentration of barium and carbonate
ion concentrations.
BaSO4 (s)  Ba2+
+ 
2
4
SO
0 0
+x +x
x x

7. 8–7
John A. Schreifels
Chemistry 212
Chapter 18-7
Precipitation of Ionic Compounds
Eg. 3 determine the fraction of Ba2+ that has
precipitated.
Solution:
– Use the amount remaining in solution (results of E.g. 2)
divided by starting concentration to determine the fraction of
barium that is left in solution.
– Subtract from above.
E.g.4 determine the Br concentration when AgCl
starts to precipitate if the initial concentration of
bromide and chloride are 0.100 M. Ksp(AgBr) =
5.0x1013; Ksp(AgCl) = 1.8x1010.

8. 8–8
John A. Schreifels
Chemistry 212
Chapter 18-8
Factors that Affect Solubility-pH
• pH of the Solution: LeChatelier’s Principle again.
E.g. determine the solubility of CaF2 at a pH of 2.00. Ksp =
3.9x1011. Ka(HF) = 6.6x104.
Strategy:
– Determine the ratio of [F] and [HF] from the pH and Ka.
– Write an expression for solubility in terms of Ka and pH and
– Substitute into solubility equation to determine the solubility.
Solution:
– Ksp = 3.9x1011 = x[F]2 (pH changes the amount of free Fluoride.)
– Let x = solubility. Then 2x = [F] + [HF]
– From equilibrium equation:
– 2x = [F](1+1/0.066) = 16.15*[F] or
– [F] = 2*x/16.15 = 0.124*x
– 3.9x1011 = x(0.0124*x)2
– x = 1.36x103 M vs. 2.13x104 M (normal solubility)
066
0
10
00
1
10
6
6
2
4
3
.
x
.
x
.
]
O
H
[
K
]
HF
[
]
F
[ a








9. 8–9
John A. Schreifels
Chemistry 212
Chapter 18-9
Separation of Ions By Selective Precipitation
• Metal ions with very different Ksp can be separated.
• Divalent metal ions are often separated using solubility variations for the
metal sulfides.
• Solution is saturated with H2S at 0.100 M; pH adjusted to keep one
component soluble and the other insoluble.
• H2S is diprotic acid; the overall reaction to get to sulfide is:
• Combine with solubility equilibrium reaction to get the overall equilibrium
expression and constant.
E.g. determine the solubility of 0.00500 M Zn2+ in 0.100 M H2S at pH = 1. Ksp
= 1.10x1021.
H2S (aq)+ H2O HS
+ H3O+
Ka1 = 8.9x108
HS
+ H2O S2
+ H3O+
Ka2 = 1.2x1013
H2S + 2H2O S2
+ 2H3O+
K = 1.1x1020
MS + M2+
+ S2
Ksp
S2
+ 2H3O+
H2S + 2H2O K = 9.09x1019
MS + 2H3O+
M2+
+ H2S Kspa = KKsp
2
3
2
2
spa
]
O
H
[
]
S
H
][
M
[
K




10. 8–10
John A. Schreifels
Chemistry 212
Chapter 18-10
Complex Ions
• Formation of Complex Ions (Coordination Complexation ) =
an ion formed from a metal ion with a Lewis base attached to it
by a coordinate covalent bond.
Ag+(aq) + 2NH3(aq)  Ag(NH3)2(aq) Kf = 1.7x107
• Large equilibrium constant indicates that “free” metal is
completely converted to the complex.
Eg. What is the concentration of the silver amine complex
above in a solution that is originally 0.100 M Ag+ and 1.00 M
NH3?
E.g. determine the [Ag+] (free silver concentration) in 0.100 M
AgNO3 that is also 1.00 M NaCN.
Ag+
+ 2CN
 
2
Ag(CN) Kf = 5.6x1018
Initial 0 0.800 0.100M
Equil +x +2x x
Equil x 0.800+2x 0.100x

11. 8–11
John A. Schreifels
Chemistry 212
Chapter 18-11
Factors that Affect Solubility: Complexation
• Free metal ion concentration in solution is reduced
when complexing agent added to it;
• Free metal ion concentration needed in solubility
expression.
E.g. determine if precipitation will occur in a solution containing
0.010 M AgNO3 and 0.0100 M Nal in 1.00 M NaCN. Recall Kf =
5.6x1018
Agl(s)Ag+ + l Ksp = 8.5.x1017
Strategy:
– Determine the free metal concentration in the solution.
– Use free metal concentration with iodide concentration to get Qsp
• If Qsp < Ksp, no precipitation
• If Qsp > Ksp, precipitation
• If Qsp = Ksp, precipitation is starting.

12. 8–12
John A. Schreifels
Chemistry 212
Chapter 18-12
Solubility with Complexing Agent
E.g. Determine the solubility of AgI in 1.00 M
NaCN. Recall Kf = 5.6x1018
Agl(s)Ag+ + l Ksp = 8.5.x1017
Strategy:
– Combine to equilibria equations to find a single
equation describing the equilibrium.
Agl(s)  Ag+
+ l
Ksp = 8.5.x1017
Ag+
+ 2CN
 
2
Ag(CN) Kf = 5.6x1018
Agl(s) + 2CN
 
2
Ag(CN) + l
K = 476
the presence of a
complexing agent
increases the
solubility
– Setup equilibrium table and solve.
AgI(s) + 2CN
(aq)  
2
Ag(CN) I
(aq)
Starting 1.00 M 0 0
 2x x x
Equil 1.00 – 2x x x
2
2
2
00
1
476
)
x
.
(
x

