Chapter 7 acid & bases part 4

7.4 NEUTRALISATION

NURUL ASHIKIN BT ABD RAHMAN

LEARNING OUTCOMES

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.

NEUTRALISATION

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)

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)

APPLICATION OF NEUTRALISATION

Soil treatment

Treat gastric

Treat wasp stings

Prevent coogulation latex

Baking powder

Manufacture detergent

ACID-BASE TITRATION

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.

ACID-BASE TITRATION

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

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

ACID-BASE TITRATION

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]

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

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

2mol

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

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

(0.025) 1

(0.1) (0.034) 2aM

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

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

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

THE END...

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