Chapter 18 Part 2 Solubility and Complexation Equilibria

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Chapter 18 Part 2 Solubility and Complexation Equilibria. Temperature Dependence of Molar Solubility. Rationale’s for Insolubility. Solubility Product Constant Data. Examples of Precipitation Reactions. 0.1 M [Cl - ]. The Solubility Curve. Strong complexes can solubilize salts. - PowerPoint PPT Presentation

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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

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