Chapter 7:

Chapter 7: Neutralization Reactions

Neutralization Reaction

A reaction between an acid and a base to produce a salt and water. (Remember you will need to be able to write the ionic and net ionic equations for these reactions.)

To begin with we need to be able to list some acids and bases that were are familiar with...here are a few I think everyone should know...already knows from our discussion in Chapter 6 on nomenclature.

Important Acids

Name	Formula
Sulfuric acid	H₂SO₄
Sulfurous acid	H₂SO₃
Nitric acid	HNO₃
Nitrous acid	HNO₂
Phosphoric acid	H₃PO₄
Phosphorus acid	H₃PO₃
Carbonic acid	H₂CO₃
Perchloric acid	HClO₄
Acetic acid	HC₂H₃O₂

Formula	Name
HF(aq)	Hydrofluoric acid
HCl(aq)	Hydrochloric acid
HBr(aq)	Hydrobromic acid
HI(aq)	Hydroiodic acid
H₂S(aq)	Hydrosulfuric acid
HCN(aq)	Hydrocyanic acid

Important Bases (All of the Group IA and IIA hydroxides)

Name of Base	Formula of Base
Sodium hydroxide	NaOH
Potassium hydroxide	KOH
Barium hydroxide	Ba(OH)₂
Ammonia	NH₃
Calcium hydroxide	Ca(OH)₂
Aluminum hydroxide	Al(OH)₃

Let's consider a few neutralization reactions and how we write the equations.

Consider the reaction between hydrochloric acid and sodium hydroxide;

HCl_(aq) + NaOH_(aq) --->

To write the products we combine the anion of the acid with the cation of the base and write the correct formula following the principle of electroneutrality. The other product is water.

Molecular equation: HCl_(aq) + NaOH_(aq) ---> NaCl_(aq) + H₂O_(l)

So the molecular form of the equation is shown above. To write the ionic equation we must separate all aqueous species into their ions and leave any solid, liquid or gaseous substance in its molecular form. So in this case HCl_(aq), NaOH_(aq), and NaCl_(aq)must be written as aqueous ions and H₂O_(l)remains in its molecualr form.

ionic equation: H⁺_(aq) + Cl^-_(aq) + Na⁺_(aq) + OH^-_(aq) ---> Na⁺_(aq) + Cl^-_(aq) + H₂O_(l)

Finally to write the net ionic equation we must cancel all species common to both sides of the equation

ionic equation: H⁺_(aq) + Cl^-_(aq) + Na⁺_(aq) + OH^-_(aq) ---> Na⁺_(aq) + Cl^-_(aq) + H₂O_(l)

net ionic equation: H⁺_(aq) + OH^-_(aq) ---> H₂O_(l)

Consider the reaction between nitric acid and calcium hydroxide;

HNO_3(aq) + Ca(OH)_2(aq) --->

To write the products we combine the anion of the acid with the cation of the base and write the correct formula following the principle of electroneutrality. The other product is water. Be sure to balance this equation.

Molecular equation: 2HNO_3(aq) + Ca(OH)_2(aq) ---> Ca(NO₃)_2(aq) + 2H₂O_(l)

So the molecular form of the equation is shown above. To write the ionic equation we must separate all aqueous species into their ions and leave any solid, liquid or gaseous substance in its molecular form. So in this case HNO_3(aq), Ca(OH)_2(aq), and Ca(NO₃)_2(aq)must be written as aqueous ions and H₂O_(l)remains in its molecualr form.

ionic equation: 2H⁺_(aq) + 2NO₃^-_(aq) + Ca²⁺_(aq) + 2OH^-_(aq) ---> Ca²⁺_(aq) + 2NO₃^-_(aq) + 2H₂O_(l)

Finally to write the net ionic equation we must cancel all species common to both sides of the equation

ionic equation: 2H⁺_(aq) + 2NO₃^-_(aq) + Ca²⁺_(aq) + 2OH^-_(aq) ---> Ca²⁺_(aq) + 2NO₃^-_(aq) + 2H₂O_(l)

net ionic equation: 2H⁺_(aq) + 2OH^-_(aq) ---> 2H₂O_(l)

or

net ionic equation: H⁺_(aq) + OH^-_(aq) ---> H₂O_(l)

Consider the reaction between hydrobromic acid and ammonia;

HBr_(aq) + NH_3(aq) --->

To write the products we combine the anion of the acid with the cation of the base and write the correct formula following the principle of electroneutrality. The other product is water. Be sure to balance this equation.

Molecular equation: HBr_(aq) + NH_3(aq) ---> NH₄Br_(aq)

Notice in this particular neutralization equation no water is formed. Since there is no hydroxide ions we can not write water as a product. So when ammonia is one of the reactants we do not include water as a product. Water is present since the reaction occurs in aqueous solution, we just do not write it as a product.

So the molecular form of the equation is shown above. To write the ionic equation we must separate all aqueous species into their ions and leave any solid, liquid or gaseous substance in its molecular form. So in this case HBr_(aq) and NH₄Br_(aq)must be written as aqueous ions and NH_3(aq)remains in its molecualr form.

ionic equation: H⁺_(aq) + Br^-_(aq) + NH_3(aq) ---> NH₄⁺_(aq) + Br^-_(aq)

Finally to write the net ionic equation we must cancel all species common to both sides of the equation

ionic equation: H⁺_(aq) + Br^-_(aq) + NH_3(aq) ---> NH₄⁺_(aq) + Br^-_(aq)

net ionic equation: H⁺_(aq) + NH_3(aq) ---> NH₄⁺_(aq)

Consider ONE more reaction, between sulfuric acid and barium hydroxide;

H₂SO_4(aq) + Ba(OH)_2(aq) --->

To write the products we combine the anion of the acid with the cation of the base and write the correct formula following the principle of electroneutrality. The other product is water. Be sure to balance this equation.

Molecular equation: H₂SO_4(aq) + Ba(OH)_2(aq) ---> BaSO_4(s) + 2H₂O_(l)

So the molecular form of the equation is shown above. To write the ionic equation we must separate all aqueous species into their ions and leave any solid, liquid or gaseous substance in its molecular form. So in this case H₂SO_4(aq) and Ba(OH)_2(aq)must be written as aqueous ions and BaSO_4(s) and 2H₂O_(l)remains in their molecualr form.

ionic equation: 2H⁺_(aq) + SO₄^2-_(aq) + Ba²⁺_(aq) + 2OH^-_(aq) ---> BaSO_4(s) + 2H₂O_(l)

Finally to write the net ionic equation we must cancel all species common to both sides of the equation. But there are no species common to both sides of the equation!

net ionic equation: 2H⁺_(aq) + SO₄^2-_(aq) + Ba²⁺_(aq) + 2OH^-_(aq) ---> BaSO_4(s) + 2H₂O_(l)

So the net ionic equation and the ionic equaton are the same.

Looking at the list of acids and bases at the top of the page you can imagine ALL the possibilities. So practice a few on your own until you get comfortable with writing neutralization equations.

Single replacement Reactions

A reaction in which an element in a reactant compound is replaced by a second reacting element producing a new compound and an element. (Remember you will need to be able to write the ionic and net ionic equations.)

2Li_(s) + 2H₂O_(l) ---> 2LiOH_(aq) + H_2(g)

2Na_(s) + 2H₂O_(l) ---> 2NaOH_(aq) + H_2(g)

2K_(s) + 2H₂O_(l) ---> 2KOH_(aq) + H_2(g)

Mg_(s) + 2HCl_(aq) ---> MgCl_2(aq) + H_2(g)

We looked at the reactions of lithium with water, sodium with water, potassium with water and barium with water. All of these are examples of single replacement reactions and following the equation as show about.

We also looked at the reaction of magnesium with hydrochloric acid. In general only the most reactive metal (Group IA) react with water. Group IIA element are not as reactive with water. To get magnesium to react with water requires the water to be hot. Calcium and barium both react with water, but the reactions are not as impressive at the Group IA elements. To get the Group IIA elements to liberate hydrogen we need to react the metals with an acid like hydrochloric acid. Zinc will also react with hydrochloric acid like magnesium.

Decomposition Reactions

The only example of a decomposition reaction I showed was that of potassium chlorate decomposing when heated.

2KClO_3(s) ---> 2KCl_(s) + 3O_2(g)