Topic 9 - Reactions of acids
What Type of Substance Can Neutralise an Acid ?
There are four categories of substance which react with acids and bring their pH towards 7.
We will call these substances neutralisers. They can be recalled in an exam by remembering
the letters CAMO !!!
C
A
M
O
stands for metal or ammonium
stands for
stands for reactive
stands for metal or ammonium
Carbonate
Alkali
Metal
Oxide
Be careful - not all metals react, only those
which we consider to be reactive (i.e. above
Hydrogen on the electrochemical series)
Everyday examples of Neutralisation
Farmers add lime, calcium oxide, to soil help neutralise the soil as most plants and vegetables
grow best in neutral or slightly alkaline soil. The lime can neutralise acidity of soil caused by
acid rain. Lime is also used to neutralise lakes made acidic by acid rain.
Milk of magnesia, magnesium oxide/hydroxide, is used to neutralise excess stomach acid.
Dental Care - bacteria react with the carbohydrates in the mouth to form acids capable of
dissolving the enamel, causing tooth decay. Toothpaste should be alkali to neutralise the acid
produced by bacteria which helps to decay teeth.
Insect Stings - A wasp sting is alkaline and a weak acid such as vinegar can be rubbed onto
it to alleviate pain. In contrast the sting of a bee is acidic and an alkali such as bicarbonate
of soda can be used to treat it.
Effect of neutralisation on pH
The pH of acids rise toward 7 when being neutralised, the pH of alkalis fall towards 7 when
being neutralised. Neutralisation moves the pH of a solution closer to 7 (neutral).
Base versus Alkali
We have looked at four substances which will neutralise an acid ie. CAMO substances. Each
of these react with an acid by accepting the H+ ion from the acid. Every substance which
reacts with an acid in this way is called a base. So a base by definition is a hydrogen ion
acceptor. Bases can be soluble or insoluble in water. Alkalis are those bases which are soluble
in water. This can be visualised using the diagram below :
Bases
insoluble
alkalis
soluble
Making a Salt
If we remove the hydrogen from each acid and replace it with a metal or ammonium ion from
one of the 4 neutralisers in the previous page this would form a salt. The last part in the
name of the salt formed would be determined by the acid used in the reaction. Each acid
gives rise to a family of salts
Acid
Hydrochloric
Sulphuric
Nitric
e.g.
Salt
Chloride
Sulphate
Nitrate
Hydrochloric acid + Potassium hydroxide  Potassium chloride + water
sulphuric acid + sodium hydroxide  sodium sulphate + water
Ammonia + nitric acid  ammonium nitrate + water
There are many different methods used for making a salt but the method you use depends on
two factors:
 the solubility of the base used and
 the solubility of the salt to be made.
The following table summarises information found on page 5 of your data booklet regarding
the solubility of various compounds which you may find useful:
compound
metal compound soluble in water
metal compound insoluble in water
hydroxide
sodium, potassium, ammonium, calcium all others
oxide
sodium, potassium, calcium
all others
carbonate
sodium, potassium, ammonium
all others
nitrate
all possible
none
sulphate
all others
lead, barium
chloride
all others
silver, lead
If the new salt is insoluble a precipitate is obtained.
Salts by Precipitation
Insoluble salts are best made by precipitation, in this method two soluble salts one containing
the positive ion and the other containing the negative ion are mixed together, an insoluble
precipitate containing the desired salt is formed. e.g. barium sulphate
barium nitrate + sodium sulphate  barium sulphate (precipitate) + sodium nitrate
Writing Ionic Equations
These equations are carried out in the same way as before.
Assuming that you will want to go through all of the various steps then they are outlined
below :
Write a word equation from an account of the reaction.
Turn your word equation into a formula equation.
Put in state symbols. (these are important in this case)
Balance your equation.
Turn it into an ionic equation by putting in charges on any ions which exist.
Separate ions which are free to move away from each other.
Example 1
Write an ionic equation for the reaction between calcium hydroxide and nitric acid.
Step 1 Word Equation
calcium hydroxide +
nitric acid
calcium nitrate + water
Step 2 Formula Equation
Ca (OH)2
+
HNO3
Ca ( NO3 )2
+
H2O
Step 3 Put in state symbols ( look them up in the solubility table on page 5 of your data
booklet)
Ca (OH)2 (aq) + HNO3(aq)
Ca ( NO3 )2 (aq) +
H2O(l)
Step 4 Balance the Equation (see rules in topic 5)
Ca (OH)2 (aq)
+ 2HNO3(aq)
Ca ( NO3 )2
Step 5 Put in ionic charges
Ca 2+ (OH-)2 (aq) + 2H+ NO3-(aq)
Ca
Step 6 Separate ions which are free to move
Ca 2+(aq) + 2OH- (aq) + 2H+(aq) + 2NO3-(aq)
2+
(aq)
2H2O(l)
+
( NO3 -)2
Ca
2+
(aq)
(aq)
2H2O(l)
+
+ 2NO3 -(aq) + 2H2O(l)
If we examine the final ionic equation above we notice that there are two ions which appear
as both reactants and products, namely Ca 2+(aq) and NO3 -(aq) . They appear to be doing
nothing to take part in the reaction. For this reason we call ions which appear on both sides
of an equation and in the same state spectator ions. We can rewrite the equation with
spectator ions omitted as follows:
Ca
2+
(aq)
+ 2OH- (aq)
+ 2H+(aq)
i.e. 2OH- (aq)
+
2NO3-(aq)
Ca
+ 2H+(aq)
2+
(aq)
+ 2NO3 -(aq) + 2H2O(l)
2H2O(l)
Without including the spectator ions, the reaction appears to be simply the reaction between
H+ and OH- ions to form H2O. This is true for all neutralisation reactions between acids
and alkalis.
Example 2
Write a balanced ionic equation, omitting spectator ions, for the reaction between aluminium
carbonate and hydrochloric acid.
Jumping straight to Step 4
Al2(CO3)3 (s) + HCl
Step 5
(aq)
Balance the equation
Al2(CO3)3 (s) + 6HCl
(aq)
AlCl3 (aq)
+
CO
2AlCl3 (aq)
+
3CO 2(g) +
2(g)
+
H2O(l)
3H2O(l)
Step 6
Put in Ionic Charges
(Al 3+ )2(CO32-)3 (s) + 6H+ Cl-
(aq)
Step 7 Separate ions which are free to move
(Al 3+ )2(CO32-)3 (s) + 6H+ (aq) + 6Cl- (aq)
Step 8 Score out spectator ions
(Al 3+ )2(CO32-)3 (s) + 6H+ (aq) + 6ClStep 9
(aq)
2Al 3+ (Cl-)3 (aq) + 3CO 2(g) +
3H2O(l)
2Al 3+(aq) + 6Cl- (aq) + 3CO 2(g) + 3H2O(l)
2Al 3+(aq) + 6Cl- (aq) + 3CO 2(g) + 3H2O(l)
Rewrite the equation without the spectator ions
(Al 3+ )2(CO32-)3 (s) + 6H+ (aq)
2Al 3+(aq)
+
3CO 2(g) + 3H2O(l)
Volumetric Analysis
Using volumetric analysis the molarity or the volume of an alkali can be determined if the
molarity and volume of the acid are known along with either the molarity or volume of the
alkali. The results of a titration are used.
Acid Rain
For causes and effects of acid rain see topic 8 notes.