chap16

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SOLUBILITY AND PRECIPITATION
EQUILIBRIA
CHAPTER
16.1 The Nature of Solubility Equilibria
16.2 Ionic Equilibria between Solids and Solutions
16.3 Precipitation and the Solubility Product
16.4 The Effects of pH on Solubility
16.5 Complex Ions and Solubility
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16.1 THE NATURE OF SOLUBILITY EQUILIBRIA
 General Features of Solubility Equilibria
 Saturation
~ Dissolution-precipitation equilibrium
 Recrystallization ~ Purification of solids
Solvent of crystallization
2 Li+(aq) + SO42-(aq) + H2O(l) → Li2SO4H2O(s)
~ different chemical formula & mass
 Supersaturation ~ Slow equilibrium
Fig. 16.1 Deposit of K2PtCl4 from the saturated
aqueous solution as the water evaporates.
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 The solubility of Ionic Solids
Solubility at 25°C,
AgClO4 ; 5570 g/L, AgCl; 0.0018 g/L
 Temperature dependence
- Mostly endothermic
→ Solubility increases with T
- CaSO4 exothermic
→ Solubility decreases with T
 Classification (at 25 °C)
Soluble > 10 g/L,
Slightly soluble 0.1~10 g/L,
Insoluble < 0.1 g/L
Fig. 16.3
Temperature
dependence of solubility.
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16.2 IONIC EQUILIBRIA BETWEEN SOLIDS AND
SOLUTIONS
Highly soluble salt: Nonideal solution, CsCl(s)  Cs+(aq) + Cl-(aq)
Fig. 16.5 The dissolution of the ionic solid CsCl in water
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 Solubility and Ksp
AgCl(s)  Ag+(aq) + Cl-(aq)
 Solubility product:
Ksp = [Ag+][Cl-] = 1.610-10 at 25°C
Solubility (S) of AgCl at 25°C calculated from Ksp
Ksp = [Ag+][Cl-] = S2 = 1.610-10
S = 1.2610-5 M
Gram solubility = (1.2610-5 mol/L)  (143.3 g/mol)
= 1.810-3 g/L
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EXAMPLE 16.1
Calculation of [Ca2+] and [F-] in a saturated solution
of CaF2 at 25°C: Ksp → Solubility
CaF2(s)  Ca2+(aq) + 2 F-(aq)
Ksp = [Ca2+][F-]2 = 3.910-11 at 25°C
[Ca2+] = S, [F-] = 2S
Ksp = [Ca2+][F-]2 = S (2S)2 = 4S3 → S = 2.1 10-4 M
Gram solubility = (2.110-4 mol/L)  (78.1 g/mol) = 0.017 g/L
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EXAMPLE 16.2
Solubility (0.029 g/L) → Ksp
Ag2CrO4(s)  2 Ag+(aq) + CrO42-(aq)
Ksp = [Ag+]2[CrO42-] = 2.7 10-12
Gram solubility: 0.029 g/L
Molar solubility: 0.029 g/L = 8.74 10-5 mol/L = S
[Ag+]= 2S, [CrO42-] = S
Ksp = [Ag+]2[CrO42-] = 4S3 = 2.7 10-12
→ 42 % greater than the tabulated value, 1.9 10-12
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16.3 PRECIPITATION AND THE SOLUBILITY
PRODUCT
 Precipitation from Solution
Ksp = [Ag+][Cl-]
Q0 = [Ag+]0[Cl-]0
~ initial reaction quotient
Q0 > Ksp
precipitation
Q0 < Ksp
dissolution
Fig. 16.6 A plot of precipitation and dissolution equilibrium for AgCl in water.
The slope of the path toward equilibrium represented by red or blue arrow is 1.
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EXAMPLE 16.4
[Ag+]0 = 0.0015 M, [Cl-]0 = 5.010-6 M
Equilibrium concentrations?
Cl- is the limiting reactant → complete precipitation first
Remaining [Ag+] = 0.0015  5.0 10-6  0.0015 M
AgCl(s)  Ag+(aq) + Cl-(aq)
---------------------------------------------------------------------Initial
0.0015
0
Change
+y
+y
--------------- -----Equilibrium
0.0015 + y
y
---------------------------------------------------------------------Ksp = 1.60 10-10 = (0.0015 + y) y  0.0015 y
y = [Cl-] = 1.1 10-7 M,
[Ag+] = 0.0015 M
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 The Common-Ion Effect
~ Solubility decreases in the presence of a common ion
AgCl
NaCl or AgNO3
Solubility of AgCl(s) in 1.00 L of 0.100 M NaCl solution
[Ag+]NaCl = S, [Cl-]NaCl = 0.100 + S
Ksp = 1.60  10-10
= [Ag+] NaCl [Cl-] NaCl
= S (0.100 + S)  0.100 S
(S < Swater =1.3 10-5 << 0.100)
[Ag+] NaCl = S = 1.60 10-9 M
[Cl-] NaCl = 0.100 M
[Ag  ]H 2O
[Ag

]H2O  1.3 105 M 
1.3 105
 8.1103

9

[Ag
]
1.6
10
0.1M
NaCl
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Fig. 16.7 Common-ion effect for the solubility
of AgCl in AgNO3 solution and in NaCl solution.
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16.4 THE EFFECTS OF pH ON SOLUBILITY
CaCO3(s) + H3O+(aq)  Ca2+(aq) + HCO3-(aq) + H2O(l)
Fig. 16.8 Damage due to increased acidity from air pollution.
On the east pier of Stanford White's Washington Square Arch
is Herma A. MacNeil's Washington in War (1916)
(Washington Square Park in the Greenwich Village neighborhood
of Lower Manhattan in New York City)
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 Solubility of Hydroxides
Zn(OH)2(s)  Zn2+(aq) + 2 OH-(aq)
Ksp = [Zn2+][OH-]2 = 4.5 10-17
In acidic solution, [OH-] decreases. → reaction goes to the right
EXAMPLE 16.6
Comparison of solubilities of Zn(OH)2(s) in pure water
and in a buffer with pH 6.00.
In pure water, [Zn2+] = S, [OH-] = 2S
Ksp = S(2S)2
S = [Zn2+] = 2.2 10-6 M, [OH-] = 2S = 4.5 10-6 M, pH = 8.65
In a pH = 6.00 buffer, [OH-] = 1.0 10-8 M (fixed).
[Zn2+] = Ksp / [OH-]2 = 0.45 M
Metal hydroxides are basic → more soluble in acidic solution
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 Solubility of Salts of Bases
CaF2(s)  Ca2+(aq) + 2 F-(aq),
Ksp = 3.9 10-11
- Solubility of CaF2(s) at low pH :
F-(aq) + H3O+(aq)  HF(aq) + H2O(l),
K = 2.9 103
 more soluble in acidic solution (large K)
[H3O+]  → [F-]  → more CaF2(s) dissolves (Le Chatelier)
- Solubility of AgCl(s) at low pH :
AgCl(s)  Ag+(aq) + Cl-(aq)
- Even in acidic solution,
Cl-(aq) + H3O+(aq)  HCl(aq) + H2O(l)
→ negligible effect of pH on the solubility of AgCl
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16.5 COMPLEX IONS AND SOLUBILITY
 Complex-Ion Equilibria
Ag+(aq) + NH3(aq)  Ag(NH3)+(aq)
K1 = [Ag(NH3) +] / ([Ag+][NH3]) = 2.1 103
Ag(NH3)+(aq) + NH3(aq)  Ag(NH3)2+(aq)
K2 = [Ag(NH3)2+] / ([Ag(NH3)+][NH3]) = 8.2 103
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
Kf = K1 K2 = [Ag(NH3)2+] / ([Ag+][NH3]2) = 1.7 107
Kf : Formation constant
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EXAMPLE 16.7
0.100 mol of AgNO3 dissolved in 1.00 L of 1.00 M NH3
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[Ag+] and [Ag(NH3)+] at equilibrium ?
Assume that Ag+ is present as Ag(NH3)2+.
[Ag(NH3)2+]0 = 0.100 M; [NH3] 0 = 1.00 M – (2  0.100) M = 0.80 M
Ag(NH3)2+(aq)  Ag(NH3)+(aq) + NH3(aq),
K–2 = K2–1
Ag(NH3)+(aq)  Ag+(aq) + NH3(aq),
K–1 = K1–1
Ag(NH3)2+(aq)  Ag(NH3)+(aq) + NH3(aq)
-----------------------------------------------------------------------------------------Initial
0.100
0
0.80
Change
–y
+y
+y
----------------------------Equilibrium
0.100 – y
y
0.80 + y
-----------------------------------------------------------------------------------------K 2 =
1 [Ag NH3  ][NH3 ] y  0.80 + y 
1
=
=
=
+
0.10  y
8.2  103
K2
[Ag NH3 2 ]
K 1 =
+
1 [Ag+ ][NH3 ] [Ag ]  0.80 
1
=
=
=
+
5
K 1 [Ag NH3  ]
1.5  10
2.1 103
+
y = [Ag(NH3)+]
= 1.5 10–5 M
[Ag+] = 9 10–9 M
<< [Ag(NH3)2+]  0.100 M
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 Formation of coordination of complexes
→ increases solubilities
AgBr(s)  Ag+(aq) + Br–(aq)
Ksp = 7.7 10–13
AgBr(s) + 2 S2O32– (aq)
 Ag(S2O3)23– (aq) + Br–(aq)
thiosulfate ion, S2O32–
sulfate ion, SO42–
Fig. 16.12 Effect of complex ion formation on solubility.
AgBr in thiosulfate solution and in pure water.
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EXAMPLE 16.8
Solubility of AgBr in 1.00 M aqueous solution of NH3?
AgBr(s) + 2 NH3(aq)  Ag(NH3)2+ (aq) + Br–(aq), K
AgBr(s)  Ag+(aq) + Br–(aq),
Ksp = 7.7 10–13
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq), Kf = 1.7 107
K = KspKf = 1.3 10–5
Solubility of AgBr: S
S = [Br–]  [Ag(NH3)2+] → [NH3] = 1.00 – 2S
K =
S2
[Ag(NH3 )2+ ][Br  ]

 1.3  105
2
2
[NH3 ]
1.00 - 2S 
S = 3.6 10–3 = [Br–]  [Ag(NH3)2+]
[Ag+] = Ksp/ [Br–] = Ksp/ S = 2.1 10–10 << [Ag(NH3)2+]
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 Re-dissolving by forming complex ions
The opposite of
common ion effects
Hg2+(aq) + 2I–(aq)  HgI2 (s)
HgI2(s) + I–(aq)  HgI3–(aq)
HgI3–(aq) + I–(aq)  HgI4–(aq)
Fig. 16.13 "Orange Tornado"
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 Separation of cations
In a strongly basic solution of Mg2+ and Zn2+
Mg(OH)2(s)
Zn(OH)42–(aq)
In a strongly basic solution of Al3+ and Fe3+
Fe(OH)3(s)
Al(OH)42–(aq)
Fig.16.14 AlCl3(s) + H2O(l)
 Chemistry
Al(OH)4–(aq)
General
II + HCl(aq)
Fig. 16.15 Solubility of Zn(OH)2
in acid, water, base.
10 Problem Sets
For Chapter 16,
14, 22, 30, 34, 44, 48, 66, 74, 76, 82
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