7.4 NEUTRALISATION SLISS 2012

2. At the end of this lesson, students should be able to:
• Explain the meaning of neutralisation precisely.
• Explain the application of neutralisation in daily life.
• Write equations for neutralisation reactions
• Describe acid-base titration.
• Determine the end point of titration during
neutralisation.
• Solve numerical problems involving neutralisation
reactions to calculate either concentration or
volume of solutions.

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

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

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

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

7. a Acid b Base Salt Water
Molarity and
volume of acid M aVa a Mole of acid
Molarity and
volume of base M bVb b Mole of base

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

11. 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]

12. Solution:
Method 1
Step 1 : write down chemical equation
2NaOH + H2SO4  Na2SO4 + 2H2O
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

13. Step 3 : from the chemical reaction, the ratio of
number of moles of H 2 SO4 1
number of moles of NaOH 2
Step 4 : find the number of moles of H2SO4 reacted
2 mole of NaOH = 1 mole of H2SO4
0.0034 mole of NaOH = 0.0034 1
mol
2
= 0.0017 mol

14. 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

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

16. M a (0.025) 1
(0.1) (0.034) 2
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

17. Question 2:
What volume of 0.20 mol dm-3 nitric
acid is required to neutralise 0.14 g of
potassium hydroxide? [relative atomic
mass: O: 16, K:39, H:1]
Ans: 12.5 cm3/0.0125 dm3

18. Question 3:
15cm3 of an acid with the formula HaX of
0.1 mol dm-3 required 30 cm3 0f 0.15
mol dm-3 sodium hydroxide solution to
complete neutralisation. Calculate the
value of a.
Ans: 3