Neutralization is a reaction between acids and bases that produces salt and water. It involves the reaction of hydrogen ions (H+) from an acid with hydroxide ions (OH-) from a base to form water. Common examples provided neutralize hydrochloric acid with sodium hydroxide and sulfuric acid with copper oxide to form the respective salts and water. Titration can be used to quantitatively analyze reactions by gradually adding one solution from a burette to another in a flask and tracking the endpoint when an indicator changes color, allowing calculation of molar concentrations.

2. Neutralisation is a reaction between acid and base
to produce salt and water.
Acid Base Salt Water
Examples:
HCl (aq) + NaOH (aq)  NaCl (aq)+ H2O (l)
H2SO4 (aq) + CuO (aq)  CuSO4 (aq) + H2O (l)

3. HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
H+Cl-
(aq) + Na+OH-
(aq)  Na+Cl-
(aq) + H2O (l)
Chemical
Equation
Ionic
Equation
H+
(aq) + OH-
(aq)  H2O (l)

4. Soil treatment
Treat gastric
Treat wasp
stings
Prevent coogulation
latex
Baking powder
Manufacture
detergent

5. Titration
End point Is the point in the titration at which
the indicatorchanges colour.
Quantitative analysis that involves the
gradual addition of a chemical
solution from a burette to another
chemical solution of known quantity
in a conical flask.

6. a a
b b
M V a
M V b

Molarity and
volume of acid Mole of acid
Molarity and
volume of base
Mole of base
a Acid b Base Salt Water

7. Examples of Indicators
Indicator Colour
Acid Neutral Alkali
Litmus Solution Red Purple Blue
Phenolphthalein Colourless Colourless Pink
Methyl orange Red Orange Yellow
Universal
indicator
Red Green Purple

10. Question 1:
25.0 cm3 of sulphuric acid is neutralised by 34.0
cm3 of 0.1 mol of dm-3 NaOH. Calculate the
concentration of sulphuric acid in:
(a) mol dm -3
(b) g dm-3
[relative atomic mass; H:1, S:32, O:16]

11. 2NaOH + H2SO4  Na2SO4 + 2H2O
Step 1 : write down chemical equation
Step 2 : find the number of mole NaOH
n=MV
Moles of NaOH
= molarity X Volume (dm3)
= 0.1 X 0.034
= 0.0034 mol
Solution:
Method 1

12. Step 3 : from the chemical reaction, the ratio of
2 4 1
2
number of moles of H SO
number of moles of NaOH

Step 4 : find the number of moles of H2SO4 reacted
2 mole of NaOH = 1 mole of H2SO4
0.0034 mole of NaOH =
= 0.0017 mol
0.0034 1
2
mol


13. Step 5 : find the concentration of H2SO4 in mol dm-3
Concentration = mol/volume
= 0.0017/ 0.025
= 0.068 mol dm-3
Step 6 : find the concentration of H2SO4 in g dm-3
Molar mass H2SO4 = 1(2) + 32+ 16(4)
= 98 g mol-
Concentration = concentration in mol dm-3 x molar mass H2SO4
= 0.068 x 98
= 6.664 g dm-3

14. Method 2:
a a
b b
M V a
M V b

2NaOH + H2SO4  Na2SO4 + 2H2O
Step 1 : write down chemical equation
Step 2 : find the concentration of H2SO4 in mol dm-3
Ma = ? Va = 25 cm3
Mb = 0.1 mol dm-3 Vb = 34 cm3

15. (0.025) 1
(0.1) (0.034) 2
aM

Ma = 0.068 mol dm-3
Step 3 : find the concentration of H2SO4 in g dm-3
Molar mass H2SO4 = 1(2) + 32+ 16(4)
= 98 g mol-
Concentration = concentration in mol dm-3 x molar mass H2SO4
= 0.068 x 98
= 6.664 g dm-3