Reactions in Aqueous Solution


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+, CO32-

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)  +  CO32-(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+, SO42-, Na+, PO43-

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)  +  PO43-(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

Brønsted-Lowry definition Acids and bases can be strong or weak.

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)
2.  A weak acid is only partially dissociated to H+(aq) ions in water
 
HF hydrofluoric acid
HC2H3O2 (or CH3COOH) acetic acid
HC7H5O2 benzoic acid

Example:

HC2H3O2(aq)   H+(aq)  +  C2H3O2-(aq)
3.  A strong base dissociates completely in water to release OH- ions
 
LiOH Ca(OH)2
NaOH Sr(OH)2
KOH Ba(OH)2
RbOH TlOH
CsOH

Example:

NaOH(aq)   Na+(aq)  +  OH-(aq)
4.  A weak base is partially dissociated to form OH- ions.
 
NH3 ammonia
CH3NH2 methylamine
(CH3)2NH dimethylamine
C6H5NH2 aniline (phenylamine)

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)
Reactions between acids and bases

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)

2.  Strong acid + weak base:  Produces an acidic solution and a salt.
Complete equation:

HCl(aq)   +   NH3(aq)   NH4Cl(aq)

Net ionic:

H+(aq)  +  NH3(aq)   NH4+(aq)

3.  Weak acid + strong base:  Produces a basic solution and a salt.
Complete equation:

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

Net ionic:

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


Oxidation - reduction (redox) reactions

Oxidation and reduction always occur together with no net change in the number of electrons.

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)  +  Cu2+(aq)   Cu(s)   +   Zn2+(aq)
 
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)  +  2 Ag+(aq)   2 Ag(s)   +   Cu2+(aq)
 
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)  +  2 H+(aq)   Mg2+(aq)   +   H2(g)
 
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)  +  4 H+(aq)  +  2 NO3-(aq)   2 NO2(g)   +   H2O(l)  +  Cu2+(aq)
 
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
Cr2O72-(aq)  +  3 C2H5OH(aq)  +  8 H+(aq)   2 Cr3+(aq)   +   3 C2H4O(aq)  +  7 H2O(l)
 
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:

Cr2O72-(aq)  +  6 I-(aq)  +  14 H+(aq)   3 I2(s)  +  2 Cr3+(aq)  +  7 H2O(l)

Calculate the percent I- in the sample.