• Precipitation reactions
• Acid-base reactions
• Oxidation-reduction (redox)

Precipitation reactions

Precipitation reactions are sometimes called "double displacement" reactions.  To determine whether a precipitate will form when aqueous solutions of two compounds are mixed:

1.  Write down all ions in solution.
2.  Combine them (cation and anion) to obtain all potential precipitates.
3.  Use the solubility rules to determine which (if any) combination(s) are insoluble and will precipitate.

Examples:

a.  What happens when Ba(NO₃)₂(aq) and Na₂CO₃(aq) are mixed?

Ions present in solution:  Ba²⁺, NO₃^-, Na⁺, CO₃^2-

Potential precipitates:  BaCO₃, NaNO₃

Solubility rules:  BaCO₃ is insoluble (rule 5), NaNO₃ is soluble (rule 1).

Complete chemical equation:  Ba(NO₃)₂(aq)   +   Na₂CO₃(aq)   BaCO₃(s)   +   2 NaNO₃(aq)

Net ionic equation:  Ba²⁺(aq)  +  CO₃^2-(aq)   BaCO₃(s)
 

b.  What happens when Pb(NO₃)₂(aq) and NH₄I(aq) are mixed?

Ions present in solution:  Pb²⁺, NO₃^-, NH₄⁺, I^-

Potential precipitates:  PbI₂, NH₄NO₃

Solubility rules:  PbI₂ is insoluble (rule 3), NH₄NO₃ is soluble (rule 1).

Complete chemical equation:  Pb(NO₃)₂(aq)   +   2 NH₄I(aq)   PbI₂(s)   +   2 NH₄NO₃(aq)

Net ionic equation:  Pb²⁺(aq)  +  2 I^-(aq)   PbI₂(s)
 

c.  What happens when CuSO₄(aq) and Na₃PO₄(aq) are mixed?

Ions present in solution:  Cu²⁺, SO₄^2-, Na⁺, PO₄^3-

Potential precipitates:  Cu₃(PO₄)₂, Na₂SO₄

Solubility rules:  Cu₃(PO₄)₂ is insoluble (rule 5), Na₂SO₄ is soluble (rule 1).

Complete chemical equation:  3 CuSO₄(aq)   +   2 Na₃PO₄(aq)   Cu₃(PO₄)₂(s)   +   3 Na₂SO₄(aq)

Net ionic equation:  3 Cu²⁺(aq)  +  PO₄^3-(aq)   Cu₃(PO₄)₂(s)
 

d.  What happens when CoCl₂(aq) and KOH(aq) are mixed?

Ions present in solution:  Co²⁺, Cl^-, K⁺, OH^-

Potential precipitates:  Co(OH)₂, KCl

Solubility rules:  Co(OH)₂ is insoluble (rule 5), KCl is soluble (rule 1).

Complete chemical equation:  CoCl₂(aq)   +   2 KOH(aq)   Co(OH)₂(s)   +   2 KCl(aq)

Net ionic equation:  Co²⁺(aq)  +  2 OH^-(aq)   Co(OH)₂(s)
 

Acid-Base Reactions

Arrhenius definition

• An acid produces H⁺ ions in water.
• A base produces OH^- ions in water.

• An acid donates a H⁺.
• A base accepts a H⁺.

Reactions in water

1.  A strong acid is completely dissociated in water
 

HCl	hydrochloric acid
HBr	hydrobromic acid
HI	hydroiodic acid
HNO3	nitric acid
HClO4	perchloric acid
HClO3	chloric acid
H2SO4	sulfuric acid

Example:

HCl(aq)   H⁺(aq)  +  Cl^-(aq)

Example:

HC₂H₃O₂(aq)   H⁺(aq)  +  C₂H₃O₂^-(aq)

Example:

NaOH(aq)   Na⁺(aq)  +  OH^-(aq)

Amines are a common class of weak bases which can be considered to be derivatives of NH₃.

Example:

Reactions of weak bases are considered from the point of view of the Brønsted-Lowry definition, with water acting as an acid.

NH₃(aq)  +  H₂O(l)   NH₄⁺(aq)  +  OH^-(aq)

1.  Strong acid + strong base:  Produces a neutral solution (neither acidic nor basic) and a salt (an ionic compound).

Complete equation:

HCl(aq)   +   NaOH(aq)   H₂O(l)   +   NaCl(aq)

Net ionic:

H⁺(aq)  +  OH^-(aq)   H₂O(l)

Complete equation:

HCl(aq)   +   NH₃(aq)   NH₄Cl(aq)

Net ionic:

H⁺(aq)  +  NH₃(aq)   NH₄⁺(aq)

Complete equation:

HC₂H₃O₂(aq)   +   NaOH(aq)   H₂O(l)   +   NaC₂H₃O₂(aq)

Net ionic:

HC₂H₃O₂(aq)  +  OH^-(aq)   H₂O(l)   +  C₂H₃O₂^-(aq)

Oxidation - reduction (redox) reactions

• One species loses electrons - and is oxidized
• Another species gains electrons - and is reduced.

1.  Formation of an ionic compound from its elements (not usually aqueous).

Example:

2 Na(s)  +  Cl₂(g)   NaCl(s)

Na(s)   Na+(aq)  +  e-	oxidation (Each Na atom loses one electron)
2 e-  +  Cl2(g)  2 Cl-(aq)	reduction (Each Cl atom gains one electron)

Each of the reactions above is called a "half-reaction".  They are useful in sorting out the oxidation and reduction parts of a reaction and in counting electrons being transfered..

Example:

6 Li(s)  +  N₂(g)   2 Li₃N(s)

Li(s)   Li+(aq)  +  e-	oxidation (Each Li atom loses one electron)
6 e-  +  N2(g)   2 N3-(aq)	reduction (Each N atom gains three electrons)

The word oxidation come from the fact that oxygen causes many things to be oxidized.  Oxygen is reduced in the process.

Example:

2 Mg(s)  +  O₂(g)   2 MgO(s)

Mg(s)   Mg2+(aq)  +  2 e-	oxidation (Each Mg atom loses two electrons)
4 e-  +  O2(g)   2 O2-(aq)	reduction (Each O atom gains two electrons)

Example:

4 Fe(s)  +  3 O₂(g)   2 Fe₂O₃(s)

Fe(s)   Fe3+(aq)  +  3 e-	oxidation (Each Fe atom loses three electrons)
4 e-  +  O2(g)   2 O2-(aq)	reduction (Each O atom gains two electrons)

This is the rusting of iron.

2.  Oxidation-reduction in aqueous solution.

Redox reactions in aqueous solution are often complex.

One type involves a metal reacting with a cation to produce a new metal and cation.  These are sometimes called "single displacement" reactions.  They are usually written in net ionic form.

Example:

Zn(s)  +  Cu(NO₃)₂(aq)   Cu(s)  +  Zn(NO₃)₂(aq)

or

Zn(s)   Zn2+(aq)  +  2 e-	oxidation (Each Zn atom loses two electrons)
2 e-  +  Cu2+(aq)   Cu(s)	reduction (Each Cu atom gains two electrons)

Example:

Cu(s)  +  2 AgNO₃(aq)   Ag(s)  +  Cu(NO₃)₂(aq)

or

Cu(s)   Cu2+(aq)  +  2 e-	oxidation (Each Cu atom loses two electrons)
e-  +  Ag+(aq)   Ag(s)	reduction (Each Ag atom gains one electron)

Example:

Zn(s)  +  2 HCl(aq)   ZnCl₂(aq)  +  H₂(g)

or

Zn(s)   Zn2+(aq)  +  2 e-	oxidation (Each Zn atom loses two electrons)
2 e-  +  2 H+(aq)   H2(g)	reduction (Each H atom gains one electron)

Many other redox reactions are more difficult to understand in terms of electron transfer.

Example:

Cu(s)  +  4 HNO₃(aq)   2 NO₂(g)  +  2 H₂O(l)  +  Cu(NO₃)₂(aq)

or

Cu(s)   Cu2+(aq)  +  2 e-	oxidation (Each Cu atom loses two electrons)
e-  +  2 H+(aq)  +  NO3-(aq)   NO2(g)  +  H2O(l)	reduction (Each "N" gains one electron)

Example:

K₂Cr₂O₇(aq)  +  3 C₂H₅OH(aq)  +  4 H₂SO₄(aq)   Cr₂(SO₄)₃(aq)  +  K₂SO₄(aq)  +  3 C₂H₄O(aq)  +  7 H₂O(aq)

or

C2H5OH(aq)   C2H4O(aq)  +  2 H+(aq)  +  2 e-	oxidation (Each "C"  loses one electron)
6 e-  +  Cr2O72-(aq)  +  14 H+(aq)   2 Cr3+(aq)  +  7 H2O(l)	reduction (Each "Cr"  gains three electrons)

Solution Stoichiometry
 

1.  What volume of 0.200 M CuSO₄ solution is required to react exactly with 50.0 mL of 0.100 M NaOH?

CuSO₄(aq)   +   2 NaOH(aq)   Cu(OH)₂(s)   +   Na₂SO₄(aq)

2.  Calculate the mass of precipitate formed when 45.00 mL of 0.200 M NaOH and 22.50 mL of 0.150 M Cr(NO₃)₃ are mixed.

Cr(NO₃)₃(aq)   +   3 NaOH(aq)   Cr(OH)₃(s)   +   3 NaNO₃(aq)

3.  22.0 mL of 0.150 M K₂Cr₂O₇ is required to react with an iodide sample weighing 5.00 g according to:

Cr₂O₇^2-(aq)  +  6 I^-(aq)  +  14 H⁺(aq)   3 I₂(s)  +  2 Cr³⁺(aq)  +  7 H₂O(l)

Calculate the percent I^- in the sample.