Chapter 18 Part 2 Solubility and Complexati on Equilibria
Jan 13, 2016
Chapter 18 Part 2
Solubility and
Complexation Equilibria
TemperatureDependence of Molar
Solubility
Rationale’s for Insolubility
+ -( s) ( aq) ( aq)
-10SP
AgCl Ag +Cl
K =1.8×10
Solubility Product Constant DataSolubility Product Constants for Some Inorganic Compounds at 25C
Substance KSP Substance KSP
Aluminum compounds
Lead compounds
AlAsO4 1.6 10–16 Pb3(AsO4)2 4.1 x 10–36
Al(OH)3 1.9 x 10–33 PbBr2 6.3 x 10–6
AlPO4 1.3 x 10–20 PbCO3 1.5 x 10–13
Antimony compounds
PbCl2 1.7 x 10–5
Sb2S3 1.6 x 10–93 PbCrO4 1.8 x 10–14
Barium compounds
PbF2 3.7 x 10–8
Ba3(AsO4)2 1.1 x 10–13 Pb(OH)2 2.8 x 10–16
BaCO3 8.1 x 10–9 PbI2 8.7 x 10–9
BaC2O4·2H2O* 1.1 x 10–7 Pb3(PO4)2 3.0 x 10–44
BaCrO4 2.0 x 10–10 PbSeO4 .5 x 10–7
BaF2 1.7 x 10–6 PbSO4 1.8 x 10–8
Ba(OH)2·8H2O* 5.0 x 10–3 PbS 8.4 x 10–28
Examples of Precipitation Reactions2+ 2- -14
( s) ( aq) ( aq)4 4 SPPbCrO Pb +CrO K =1.8×10 2+ 2- -28
( s) ( aq) ( aq) SPPbS Pb +S K =8.4×10 3 38
( s ) ( aq) ( aq)3 SPFe( OH) Fe 3OH K 6.3 10 2 12
( s ) ( aq) ( aq)2 4 4 SPAg CrO 2Ag CrO K 9.0 10
The Solubility Curve
0.1 M [Cl-]
Strong complexes can solubilize salts
Complex Formation Equilibrium Constants
+ - - 52 F
+ 2- 2- 3- 132 3 2 3 2 F
+ - - 182 F
+ + 73 3 2 F
2+ 2+ 113 3 4 F
2+ - 2- 254 F
2+ -
Ag +2Cl [AgCl ] K =2.5×10
Ag +2S O [Ag( S O ) ] K =2.0×10
Ag +2CN [Ag( CN) ] K =5.6×10
Ag +2NH [Ag( NH ) ] K =1.6×10
Cu +4NH [Cu( NH ) ] K =9.0×10
Cu +4CN [Cu( CN) ] K =1.0×10
Cd +4CN
2- 164 F[Cd( CN) ] K =7.1×10
2+ - 2- 64 F
2+ - 2- 164 F
2+ - 2- 304 F
2+ - 2- 414 F
3+ 3+ 313 3 6 F
2+ 2+ 93 3 4 F
Cd +4I [CdI ] K =2.0×10
Hg +4Cl [HgCl ] K =1.7×10
Hg +4I [HgI ] K =2.0×10
Hg +4CN [Hg( CN) ] K =2.5×10
Co +6NH [Co( NH ) ] K =5.0×10
Zn +4NH [Zn( NH ) ] K =2.9×10
Solubility rules
Solubility product (Ksp)
Common ion effect
Effect on solubility of adding ions already in equilibrium
Effect on solubility of adding ligands
Effect on solubility of pH
Reaction quotients
Concepts from Chapter 18 Part 2