Exchange or Metathesis Reaction or Double-Replacement Reaction
An exchange reaction occurs between compounds that, when written as a molecular equation, appear to involve the exchange of parts between the two reactants. An exchange reaction will occur when ions in solution form insoluble products, weak electrolytes, or nonelectrolytes. These are also called metathesis reactions. The word metathesis comes from Greek and means, "to change positions".

Exchange reactions have the general form AB + CD -----> AD + CB.

Exchange reactions can be categorized as:

1. Precipitations: one of the products is insoluble (e.g., AgCl(s)).
NaCl(aq) + AgNO₃(aq) ---> AgCl(s) + NaNO₃(aq)

2. Neutralizations: most often one of the products is water, H₂O(l).
HCl(aq) + KOH(aq) --> KCl(aq) + H₂O(l)

3. Gas-forming: one of the products is a gas (e.g., SO₂(g)).
Na₂SO₃(aq) + 2 HCl(aq) -----> 2 NaCl(aq) + H₂O(l) + SO₂(g)

Precipitation Reactions
A precipitate is an insoluble solid compound formed during a chemical reaction in solution (e.g., CaCO₃(s), AgCl(s), BaSO₄(s)). Precipitation reactions are a type of chemical reaction between ions that produce a precipitate.

Based upon the solubility rules, a reaction can be predicted as to whether it will take place or not. When solutions of electrolytes are mixed, a reaction will take place if one of the products is insoluble.

Reactions can be viewed as rearrangement of atoms. Therefore, swap cation and anions to determine what the possible products might be. If one of the products is insoluble a can be predicted to occur. If both predicted products are soluble then no reaction occurs.

Sample Problems
1) BaCl₂(aq) + ZnSO₄(aq) ---> ?
BaCl₂(aq) + ZnSO₄(aq) ---> BaSO₄(s) + ZnCl₂(aq)
BaSO₄ is an insoluble solid (precipitate) therefore a reaction takes place when aqueous solutions of BaCl₂ and ZnSO₄ are mixed together.

2) Na₂S(aq) + ZnCl₂(aq) ---> ?
Na₂S(aq) + ZnCl₂(aq) ---> 2 NaCl(aq) + ZnS(s)
ZnS is an insoluble solid (precipitate) therefore a reaction takes place when aqueous solutions of Na₂S and ZnCl are mixed together.

3) K₃PO₄(aq) + NaNO₃(aq) ---> ?
K₃PO₄(aq) + NaNO₃(aq) ---> KNO₃(aq) + Na₃PO₄(aq)
All of the proposed products are soluble. Therefore a reaction does not take place when aqueous solutions of K₃PO₄ and NaNO₃ are mixed.

To help get a better grasp of what is actually occurring in a precipitation reaction, the reaction equation can be written in three different forms.

First, the molecular equation shows compounds as if they exist in solution in the form of molecules, even though they are soluble ionic compounds and dissociate in solution.
(NH₄)₂CO₃(aq) + CaCl₂(aq) ---> 2 NH₄Cl(aq) + CaCO₃(s)

Second, the complete ionic equation shows the soluble ionic compounds as ions in solution and molecules are shown as nondissociated molecules in solution.
2NH₄¹⁺_(aq) + CO₃^2-_(aq) + Ca²⁺_(aq) + 2Cl^1-_(aq) ---> 2NH₄¹⁺_(aq) + 2Cl^1-_(aq) + CaCO_3(s)

Finally, the net ionic equation shows only the species that take part in a reaction.
Ca²⁺(aq) + CO₃^2-(aq) ---> CaCO₃(s)

Spectator ions are the ions that do not take part in a reaction and therefore remain in the same form and state on either side of the equation arrow. In the reaction just cited the sodium cation, Na¹⁺(aq), and the nitrate anion, NO₃^1-(aq), are spectator ions.

The molecular equation is useful because it tells us the identity of the reagents. It is necessary in practical work in the lab and industry and useful when you are deciding what chemicals to obtain from the stockroom to carry out a reaction and are usually used for stoichiometric calculations. It identifies the actual reactants and lets us work out the amounts to use.

The complete ionic equation better describes the reaction between electrolytes than the molecular equations. It aids in visualization of all that is happening in the solution when the reaction is taking place.

The net ionic equation is a generalization of the reaction. It focuses the attention most sharply on the chemical change that is actually taking place in the solution. The chemical behavior of a strong electrolyte solution can be attributed to the various kinds of ions it contains. Therefore the net ionic equation focuses on the essential feature of the precipitation reaction. It can be useful in illustrating the similarities between large number of reactions involving electrolytes.

Soluble chloride + soluble silver rxns with same net ionic eqn:
NaCl(aq) + AgNO₃(aq) ---> AgCl(s) + NaNO₃(aq)
MgCl₂(aq) + AgC₂H₃O₂(aq) ---> AgCl(s) + Mg(C₂H₃O₃)₂(aq)
Ag¹⁺(aq) + Cl^1-(aq) ---> AgCl(s)

Soluble carbonate + soluble calcium rxns with same net ionic eqn:
Na₂CO₃(aq) + Ca(NO₃)₂(aq) ---> 2 NaNO₃(aq) + CaCO₃(s)
K₂CO₃(aq) + Ca(C₂H₃O₂)₂(aq) ---> 2 NaC₂H₃O₂(aq) + CaCO₃(s)
Ca²⁺(aq) + CO₃^2-(aq) ---> CaCO₃(s)

Acid formulas are most often written with the donatable proton at the beginning of the formula.
    HCl; H₂SO₃; HC₇H₅O₂

A base is a compound that:
    1.  neutralizes the effects of an acid.
    2.  increases the concentration of hydroxide ions (OH^1-) when dissolved in water.
    3.  can accept a proton (H¹⁺).

Base formulas are most often written with the hydroxide ion at the end of the formula.
    NaOH; Ca(OH)₂; Al(OH)₃

A salt is a compound:
1. composed of metal cations or ammonium ion (NH₄¹⁺), and nonmetal anions or polyatomic anions other than hydroxide.
2. formed when an acid and a base react. ACID + BASE ---> SALT
    NaCl; MgBr₂; K₂SO₄; NH₄NO₃

The proton, H¹⁺, cannot exist in water. When an acid dissolves in water the hydronium ion is formed, H₃O¹⁺.
    H¹⁺ + H₂O -----> H₃O¹⁺

    H¹⁺ = H₃O¹⁺

For convenience, the hydronium ion, H₃O¹⁺, will be indicated by H¹⁺.

Acids can be divided into two broad groups, based upon the degree which they dissociate into ions, strong and weak.
Strong acids (HZ) are completely dissociated in solution and are therefore strong electrolytes.
    HZ(aq) ---> H^l+(aq) + Z^1-(aq)

Weak acids (HA) dissociate very little (i.e., exist primarily in the form of a molecular species) and are therefore weak electrolytes.
    HA(aq) <=====> H^l+(aq) + A^1-(aq)

There are six (6) strong acids. All other acids are weak acids.
    HCl        Hydrochloric acid
    HNO₃    Nitric acid
    H₂SO₄   Sulfuric acid
    HClO₄   Perchloric acid
    HBr        Hydrobromic acid
    HI           Hydroiodic acid

The vast majority of acids are weak acids.

Acids can also be classified according to composition and structure.
Monoprotic acids have only one acidic hydrogen (donatable proton).
    HCl, HNO₃, HClO₄, HBr, HI, HCN, HClO, HC₂H₃O₂

Polyprotic acids have more than one acidic hydrogen.
    H₂SO₄, H₃PO₄, H₂C₂O₄

Sulfuric acid is a diprotic acid and is a strong acid in its first ionization and a weak acid in its second ionization.
    First ionization: H₂SO₄(aq) ---> H^l+(aq) + HSO₄^1-(aq)
    Second ionization: HSO₄^1-(aq) <=====> H^l+(aq) + SO₄^2-(aq)
 
Phosphoric acid is weak in all three of its ionization steps.
    First ionization: H₃PO₄(aq) <=====> H^l+(aq) + H₂PO₄^1-(aq)
    Second ionization: H₂PO₄^1-(aq) <=====> H^l+(aq) + HPO₄^2-(aq)
    Third ionization: HPO₄^2-(aq) <=====> H^l+(aq) + PO₄^3-(aq)

Strong and Weak Bases
Strong bases are completely dissociated in solution and are therefore strong electrolytes. Most strong bases are inorganic.
    NaOH(aq) ---> Na^l+(aq) + OH^1-(aq)

Some bases such as Ca(OH)₂ are not very soluble in water, but the part that does dissolve completely dissociates. Ca(OH)₂ can be considered to be a strong electrolyte although it is not very soluble.

There are six (6) strong bases. All other bases are weak bases.
    LiOH       Lithium hydroxide
    NaOH     Sodium hydroxide
    KOH       Potassium hydroxide
    Ca(OH)₂ Calcium hydroxide
    Ba(OH)₂ Barium hydroxide
    Sr(OH)₂ Strontium hydroxide

Ammonia is a weak base. Gaseous ammonia dissolves in water and then the dissolved ammonia reacts with water to produce hydroxide ions, but most remains as molecules. Ammonia is less than 5% ionized.
    NH₃(g) + H₂O(l) ---> NH₃(aq)
    NH₃(aq) + H₂O(l) <=====> NH₄¹⁺(aq) + OH^1-(aq)

Amines are organic derivatives of ammonia and like ammonia are weak bases.
    CH₃NH₂(aq) + H₂O(l) <=====> CH₃NH₃¹⁺(aq) + OH^1-(aq)

ACID + BASE ----> SALT + WATER
    HClO₄(aq) + LiOH(aq) --> LiClO₄(aq) + H₂O(l)
    H₂SO₄(aq) + 2 NaOH(aq) --> Na₂SO₄(aq) + 2 H₂O(l)
    HI(aq) + KOH(aq) --> KI(aq) + H₂O(l)
    HBr(aq) + Ca(OH)₂(aq) --> CaBr₂(aq) + 2 H₂O(l)
    HCl(aq) + Sr(OH)₂(aq) --> SrCl₂(aq) + 2 H₂O(l)
    HNO₃(aq) + Ba(OH)₂(aq) --> Ba(NO₃)₂(aq) + 2 H₂O(l)

Net Ionic Equations of Acid-Base Reactions
    HClO₄(aq) + LiOH(aq) --> LiClO₄(aq) + H₂O(l)
    H^l+(aq) + ClO₄^1-(aq) + Li^l+(aq) + OH^1-(aq) ----> Li¹⁺(aq) + ClO₄^1-(aq) + H₂O(l)
    H^l+(aq) + OH^1-(aq) ----> H₂O(l)

The net ionic equation for the reaction between most strong acids and strong bases are identical
    H₂SO₄(aq) + 2 NaOH(aq) ---> Na₂SO₄(aq) + 2 H₂O(l)
    2 HNO₃(aq) + Ca(OH)₂(aq) ---> Ca(NO₃)₂(aq) + 2 H₂O(l)
    H¹⁺(aq) + OH^1-(aq) ---> H₂O(l)

In the net ionic eqn of a weak acid and a strong base, the acid is written in its undissociated form.
    HC₂H₃O₂(aq) + NaOH(aq) ----> Na C₂H₃O₂(aq) + H₂O(l)
    HC₂H₃O₂(aq) + Na^l+(aq) + OH^1-(aq) ----> Na¹⁺ + C₂H₃O₂^1-(aq) + H₂O(l)
    HC₂H₃O₂(aq) + OH^1-(aq) ----> C₂H₃O₂^1-(aq) + H₂O(l)

In the net ionic eqn of a strong acid and a weak base, the base is written in its undissociated form.
    Mg(OH)₂(s) + 2 H¹⁺(aq) ---> Mg²⁺(aq) + 2 H₂O(l)

The net ionic equation for the reaction of ammonia with hydrochloric acid is as follows.
    H¹⁺(aq) + NH₃(aq) ---> NH₄¹⁺(aq)

Some Common Gas-Forming Reactions
Anion          Reaction with H¹⁺
HCO₃^1-        HCO₃^1- + H¹⁺ ---> CO₂(g) + H₂O(l)
CO₃^2-           CO₃^2- + 2 H¹⁺ ---> CO₂(g) + H₂O(l)
HSO₃^1-        HSO₃^1- + H¹⁺ ---> SO₂(g) + H₂O(l)
SO₃^2-           SO₃^2- + 2 H¹⁺ ---> SO₂(g) + H₂O(l)
HS^1-             HS^1- + H¹⁺ ---> H₂S(g)
S^2-                S^2- + 2 H¹⁺ ---> H₂S(g)

Carbonates
Na₂CO₃(aq) + 2 HCl(aq) ----> 2 NaCl(aq) + H₂O(l) + CO₂(g)

Sulfites
Na₂SO₃(aq) + 2 HCl(aq) ----> 2 NaCl(aq) + H₂O(l) + SO₂(g)

Sulfides
Na₂S(aq) + 2 HCl(aq) ----> 2 NaCl(aq) + H₂S(g)

Ammonium salts react with strong bases to yield ammonia gas.
NaOH(aq) + NH₄Cl(aq) ---> NaCl(aq) + H₂O(l) + NH₃(g)

Redox reactions involve the transfer of electrons from one reactant to another.

The reducing agent is the species that is oxidized. It gains electrons.
The oxidizing agent is the species being reduced. It loses electrons.
The oxidizing agent brings about the reduction of another element and the reducing agent brings about the oxidation of another element.

 SnO₂(s) + C(s) ----> Sn(s) + 2 CO(g)

In this reaction carbon brings about the reduction of the tin ore to tin metal so it is referred to as the reducing agent.

The combustion of fuels and the reactions of metals with oxygen to give oxides were described by the word oxidation. The burning of magnesium in air is a redox reaction.
2 Mg(s) + O₂(g) ----> 2 MgO(s)

In this instance, oxygen is the oxidizing agent because it brings about the oxidation of the metal.

Rules for Assigning Oxidation Numbers
1.  The oxidation number of an atom of a pure element is 0.
2.  The oxidation number of a monatomic ion equals its charge.
3.  Some elements have the same oxidation number in the majority of compounds they form and can be used as a reference.
            Hydrogen is +1 unless combined with a metal, in this case it is –1.
            Fluorine is -1.
            Oxygen is –2, except in peroxides it is –1.
            In binary compounds with metals, Group 7A (halogens) elements have an oxidation number of –1.
            In binary compounds, Group 6A elements have an oxidation number of –2.
            In binary compounds, Group 5A elements have an oxidation number of -3.
            Group 1A metal cations all have an oxidation number of +1
            Group 2A metal cations all have an oxidation number of +2.
4.  For the atoms in a neutral species (an isolated atom, a molecule, or a formula unit) the total of all the oxidation numbers is 0; the sum of the oxidation numbers in a polyatomic ion equals the charge on the ion.

Example
What is the oxidation number of nitrogen in sodium nitrate, NaNO₃?
    Sodium is a group 1A metal so it has an oxidation number of +1.
    The oxidation number of nitrogen is calculated as follows:
    (1 N)(X) + (3 O)(-2) = 1-
    X = +5, therefore the oxidation number of nitrogen in +5.

Example
What is the oxidation number of phosphorus in magnesium phosphate, Mg₃(PO₄)₂?
    Magnesium is a group 2A metal so it has an oxidation number of +2.
    The oxidation number of phosphorus is calculated as follows:
    (1 P)(X) + (4 O)(-2) = 3-
    X = +5, therefore the oxidation number of phosphorus in +5.
 
 

OXIDATION	REDUCTION	"OIL RIG" acronym
Involves the loss of electrons.	Involves the gain of electrons.	Oxidation Involves Loss of electrons. Reduction Involves Gain of electrons.
Is characterized by an increase in oxidation number.	Is characterized by a decrease in oxidation number.

Identification of a redox reaction is carried out by assigning oxidation numbers of the reactants and products. When an oxidation number of an element changes then a redox reaction has taken place.

A redox reaction take place when oxidation number change for the reactants.

Example: ZnSO₄ + Mg --> MgSO₄ + Zn

Oxidation Number: +2                         0         +2                          0
                                Zn²⁺ + SO₄^2- + Mg --> Mg²⁺ + SO₄^2- + Zn

As a reactant the oxidation number of zinc was +2 but as a product the oxidation number of zinc is zero. As a reactant the oxidation number of magnesium was 0 but as a product the oxidation number of magnesium is +2.

Example: 2 HCl + Fe ---> FeCl₂ + H₂

Oxidation Number:      +1           -1         0          +2             -1        0
                                     2 H¹⁺ + 2 Cl^1- + Fe ---> Fe²⁺ + 2 Cl^1- + H₂

As a reactant the oxidation number of hydrogen was +1 but as a product the oxidation number of hydrogen is zero. As a reactant the oxidation number of iron was 0 but as a product the oxidation number of iron is +2.

Since elemental metals have an oxidation number of zero, metals are reducing agents. Strength of reducing agent varies.

The strength of a metal as a reducing agent gives rise to the Activity Series. A metal will displace from solution the ions of any metal that lies below in it the activity series.

Activity Series for Metals (Table 5.5; p. 187)

A metal with a higher activity will displace another metal in a salt.
    2 Na(s) + 2 H₂O ---> 2 NaOH(aq) + H₂(g)
    Ca(s) + 2 H₂O ---> Ca(OH)₂(aq) + H₂(g)
    Mg(s) + H₂O ---> NR
    Fe(s) + 2 HCl(aq) ---> FeCl₂(aq) + H₂(g)
    FeCl₂(aq) + H₂(g) ----> NR
    Cu(s) + 2 AgNO₃(aq) ---> Cu(NO₃)₂(aq) + 2 Ag(s)
    Cu(NO₃)₂(aq) + 2 Ag(s) --> NR

Nonmetal Reactivity
    2 Cl₂(aq) + Na₂S(aq) --> S(s) + 2 NaCl(aq)
    Cl₂(aq) + 2 NaBr(aq) --> 2 NaCl(aq) + Br₂(aq)
    Br₂(aq) + 2 NaCl(aq) ---> NR.

Example
If 400.0 mL of a solution contains 5.00 x 10^-3 moles of AgNO₃, what is the molarity of this solution?
    5.00 x 10^-3 mol AgNO₃ / 0.4000 L = 0.0125 M AgNO₃

Example
Calculate the molarity of an HCl solution containing 18.23 g of HCl in 355.0 mL of solution.
    A. Calculate the number of moles of HCl.
            (18.23 g HCl) (1 mol HCl/36.46 g HCl) = 0.5000 mol HCl

    B. Divide the number of moles of HCl by the total volume in liters.
            0.5000 mol HCl/0.3550 L solution = 1.408 M HCl

Example
How many moles of NaCl are contained in a 27.49 mL sample of a 0.350 M solution of NaCl?
    (0.350 mol/L)(0.02749 L) = 9.62 x 10^-3 mol NaCl

Example
What volume of 18.0 M H₂SO₄ must is required to make 100 mL of a 5.0 M solution of H₂SO₄? (28 mL)

Example
How many milliliters of water are required to dilute 6 M H₂SO₄ to 175 ml of 0.3 M H₂SO₄? (166 mL)

Molarity and Reactions in Aqueous Solutions

Example
How many mL of 6.0 M HCl are needed to digest 3.78 g of Mg? [52 mL]

Example
How many mL of 0.0487 M Ba(OH)₂ are needed to react with 35.67 mL of 0.0748 M HCl? [27.3 mL Ba(OH)₂]

Titrations
Titration is as an analytical procedure of determining the concentration of one substance in solution by reacting it with a solution of another substance whose concentration is known, called a titrant (or standard solution).

To carry out the process, we add the titrant, using a buret, to a known volume of the other solution until the reaction between the two substances is just complete, that is until chemically equivalent amounts of the two reactants are present. We often tell when we have reached this point, called the endpoint or equivalence point, by a change in color of an added chemical substance called an indicator.

The volume of the titrant is used to calculate the number of moles delivered from the buret. The moles of the other reactant are obtained from the coefficients in the balanced reaction equation, and this value is used with the volume of the solution of unknown concentration to calculate its molarity.

Volume of A ßà moles A ßà moles of B ßà Molarity of B

Example 1
In a titration 25.00 mL of sodium hydroxide solution was neutralized by 32.72 mL of hydrochloric acid. The HCl had a concentration of 0.129 M. Find the concentration of the NaOH solution.
    A.   Calculate the number of moles of HCl delivered.
            (0.03272 L HCl)(0.129 mol/L) = 4.22 x 10^-3 mol HCl

    B.    Calculate the number of moles of NaOH neutralized using the coefficients from the balanced neutralization reaction equation.
            HCl + NaOH ----> NaCl + H₂O
            (4.22 x 10^-3 mol HCl)(1 mol NaOH/1 mol HCl) = 4.22 x 10^-3 mol NaOH

    C.    Calculate the molarity of the NaOH using the volume of NaOH titrated and the number of moles of NaOH present in this volume.
            4.22 x 10^-3 mol NaOH / 0.02500 L NaOH solution = 0.169 M NaOH

Example 2
During a titration, a 20.00 mL portion of a 0.100 M H₂SO₄ solution was carefully measured into a flask and phenolphthalein was added to it. The solution was titrated with 18.47 mL of NaOH to reach the endpoint. What is the concentration of the NaOH?

    A.     (0.02000 L H₂SO₄)(0.100 mol H₂SO₄/L) = 2.00 x 10^-3 mol H₂SO₄
    B.     H₂SO₄ + 2 NaOH ---> Na₂SO₄ + 2 H₂O
            (2.00x10^-3 mol H₂SO₄)(2 mol NaOH/1 mol H₂SO₄) = 4.00 x 10^-3 mol NaOH
    C.     4.00 x 10^-3 mol NaOH / 0.01847 L NaOH solution = 0.217 M NaOH