Chem. 1B – 10/13 Lecture

Announcements I

• Lab– Starting Wednesday: Experiment 5

(Acids/Bases and Buffers)– Report for Lab #3 due

• Mastering Assignments –16b due today• Bonus In-Class Problems:

– Can pick up after class (max of 0.25 to 0.5 pts awarded to each student in group)

Announcements II

• Today’s Lecture– Solubility and Precipitation:

• precipitation• selective precipitation• use in qualitative analysis (emphasized in lab)

– Complex Ion Formation– Thermodynamics (Start by reviewing Ch. 6)

Chem 1B – Aqueous ChemistrySolubility (Chapter 16)

• Precipitation– If we mix an ion pair from a sparingly soluble

salt together, how do we know if precipitation will occur?

– Example: 0.040 M Ca2+ + 0.0010 M F- – do we get CaF2 (s)?

– Using Ksp reaction (backwards of what is occurring), we can compare Q with Ksp

– If Q < Ksp, no precipitation occurs (solution stays clear)

– If Q > Ksp, either precipitation occurs or supersaturated solution [Example problem]

Chem 1B – Aqueous ChemistrySolubility (Chapter 16)

• Precipitation for selective ion removal– Example: An old battery plant had a leak of

lead and sulfuric acid that was neutralized by addition of CaCO3. The collected liquid has [Ca2+] = 0.20 M and [Pb2+] = 0.10 M. A chemist wants to save the lead (in form Pb2+) but not the Ca2+. Can she selectively precipitate out >98% of the Pb2+ without precipitating Ca2+ by addition of SO4

2-?Ksp(CaSO4) = 7.10 x 10-5 and Ksp (PbSO4) = 1.82 x 10-8

Chem 1B – Aqueous ChemistrySolubility (Chapter 16)

• Qualitative Analysis– Based on selective removing specific

(less soluble) ions– Anions already examined in lab– Text covers how to separate cations

(somewhat similar to experiment 8)– Schemes generally start with addition of

Cl- (most chlorides are soluble – except Ag+, Hg2

2+, and Pb2+ which precipitate)

Chem 1B – Aqueous ChemistrySolubility (Chapter 16)

• Qualitative Analysis – cont.– Sulfides are generally insoluble (except with

group I and some group II cations)– Sulfide is a strong base (Kb1 > 1), which

means that addition of acid increases solubility

– Ksp is defined somewhat differently:for CuS, rxn is CuS(s) + H2O(l) ↔ Cu2+(aq) + HS-(aq) + OH-(aq) and Ksp = [Cu2+][HS-][OH-]

– Because of basic nature, addition of acid increases solubility

Chem 1B – Aqueous ChemistrySolubility (Chapter 16)

• Qualitative Analysis – cont.– However, some Ksp values are so small,

that even addition of acid will not all significant dissolution

– Example Ni2+ (Ksp = 3 x 10-20) vs. Cu2+ (Ksp = 1.3 x 10-36)

– Solubility in 0.2 M H+ and 0.2 M H2S: show calculation

– This allows separation into acid insoluble (smaller group) and base insoluble sulfides (also precipitates out insoluble hydroxides)

Chem 1B – Aqueous ChemistrySolubility (Chapter 16)

• Qualitative Analysis – last step– Phosphate is usually added last (fewer

exceptions to the generally insoluble rule) to remove alkaline earth cations

– Remaining ions (alkali metals + NH4+ are

always soluble)

Chem 1B – Aqueous ChemistryComplex Ion Formation (Chapter 16)

• Complex Ions– Consists of a metal reacting with ligands– Metal is Lewis Acid (electron pair acceptor) while

Ligand is Lewis Base– Water is a common ligand (but often not included

in reaction)– Example (2 reactions are essentially the same):Cu2+(aq) + 4NH3(aq) ↔ Cu(NH3)4

2+(aq) orCu2+(H2O)4 + 4NH3(aq) ↔ Cu(NH3)4

2+(aq) + 4H2O(l)– Kf values tend to be large (Kf = 1.7 x 1013), but

stoichiometry limits importance to higher concentrations of ligand

Chem 1B – Aqueous ChemistryComplex Ion Formation (Chapter 16)

• Complex Ions – cont.– Example:Cu2+(aq) + 4NH3(aq) ↔ Cu(NH3)4

2+(aq)– Kf values tend to be large (Kf = 1.7 x 1013), but

stoichiometry limits importance to higher concentrations of ligand

Kf = [Cu(NH3)42+]/[Cu2+][NH3]4

Equil. [NH3] (M) [Cu(NH3)42+]/[Cu2+]

1.0 x 10-2 170,000

1.0 x 10-3 17

1.0 x 10-4 0.0017

Chem 1B – Aqueous ChemistryComplex Ion Formation (Chapter 16)

• Complex Ions – Example problem– The complex Zn(CN)4

2- forms (from Zn2+ + CN-) with a formation constant (Kf) of 2.1 x 1019. If 1.0 x 10-5 moles of ZnCl2 is added to 1.00 L of a 0.0010 M NaCN solution, what is the concentration of each species at equilibrium?

Chem 1B – Aqueous ChemistryComplex Ion Formation (Chapter 16)

• Complex Ions – Effects on Solubility– The strong binding possible with ligands allows

insoluble salts to become soluble in the presence of a ligand

– Examples: 1) AgCl(s) + NH3(aq)solubility rxn: AgCl(s) ↔ Ag+(aq) + Cl-(aq) Ksp = 1.77 x 10-

10

complex rxn: Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+(aq) Kf = 1.7

x 107

sum: AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+(aq) + Cl-(aq) K =

3.0 x 10-3

Chem 1B – Aqueous ChemistryComplex Ion Formation (Chapter 16)

• Complex Ions – Effects on Solubility– Examples: 2) Ca2+ + C2O4

2- (oxalate anion) – anion can lead to both precipitation and complex formation

solubility rxn: CaC2O4(s) ↔ Ca2+(aq) + C2O42-(aq) 1.3 x 10-8

complex rxn: Ca2+(aq) + 2C2O42-(aq) ↔ Ca(C2O4)2

2-(aq) Kf = 2.3 x 104

At low [C2O42-] (e.g. 1.0 x 10-7 M at equilibrium), Ca2+ is fairly

soluble (0.13 M) and almost no complex forms ([Ca(C2O4)22-] = 3

x 10-13 M and doesn’t contribute to solubility)At moderate [C2O4

2-] (e.g. 1.0 x 10-3 M), Ca2+ is less soluble (1.3 x 10-5 M), and complex still hasn’t formed much (3 x 10-7) to affect solubility

At high [C2O42-] (e.g. 0.5 M), very little Ca2+ is present (2.6 x 10-8

M), but complex starts to increase net solubility (1.5 x 10-4 M)

Complex Ions – “U” Shaped Solubility Curves

Calcium Oxalate

1.00E-10

1.00E-09

1.00E-08

1.00E-07

1.00E-06

1.00E-05

1.00E-04

1.00E-03

1.00E-02

1.00E-01

1.00E+00

1.00E-06 1.00E-05 1.00E-04 1.00E-03 1.00E-02 1.00E-01 1.00E+00

[C2O42-] (M)

Con

c. (M

) CaC2O4(aq)Ca2+Ca(C2O4)22-Total Ca

Solubility in water

Common ion effect

Complex ion effect

Note: looks “U” shaped if not on log scale (otherwise “V” shaped)

Chem 1B – Aqueous ChemistryComplex Ion Formation (Chapter 16)

• Complex Ions – Effects on Solubility– The second example also applies to metal

hydroxides (e.g. Zn(OH)2 = sparingly soluble salt, but solubility increases at high pH due to formation of Zn(OH)4

2-)– Calculate the solubility of Zn2+ in buffers at pH

= 7, 10 and 13. Ksp(Zn(OH)2) = 3 x 10-17 and Kf (Zn(OH)4

2-) = 2 x 1015

Chem 1B – ThermodynamicsChapter 17

• Chapter 6 – Review– Types of Energy:

• kinetic energy (associated with motion)• potential energy (stored energy – e.g. ball at the

top of a hill)• Chemical energy (a type of stored energy)• Heat (a molecular scale type of kinetic energy)

– Systems and Surroundings• used to define energy transfers• example: system with reaction that produces

heat (conversion from chemical energy) can heat surroundings

Chem 1B – ThermodynamicsChapter 17

• Chapter 6 – Review II– Enthalpy (H)

• Energy related to heat• H = qp (heat in a constant pressure

system)• Endothermic reaction means H > 0, means

heat from surrounding used for reaction• Exothermic reaction means H > 0, means

heat from reaction goes to surroundings