Chemistry Module Form 4 (7)

mohd faisol mansor/chemistry form 4/chapter 7

130

CHAPTER 7 ACIDS AND BASES

The role of water

In the presence of water an acid will ionise to form hydrogen ion and

alkaline will dissociate into hydroxide ions, OH-.

Therefore, water is essential for the formation of hydrogen ions, H+

that cause acidity and hydroxide ions, OH- that cause alkalinity.

Arrhenius Theory

An acid is a chemical

compound that

produces hydrogen ions,

H+ or hydroxonium ions

H3O+ when dissolve in

water.

A base defined as a

chemical substance

that can neutralise an

acid to produce a

salt and water.

An alkali is defined as

a chemical

compound that

dissolve in water to

produce hydroxide

ions, OH- .

Example

HCl (g) H2O H+ (aq) + Cl- (aq)

NaOH(s) H2O Na+(aq) + OH-(aq)


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BASICITY OF AN ACID

Is the number of ionisable hydrogen atoms per molecule of an

acid.

Example: a) CH3COOH? (ethanoic acid) =

b) H2SO4? (sulphuric acid ) =

c) HNO3? (Nitric acid) =

d) H3PO4? (Phosphoric acid) =

STRENGTH OF ACID & ALKALI

Monoprotic Acid

Acid which produces

1 hydrogen ion when

one molecule of an

acid ionises in H2O.

Diprotic Acid

Triprotic Acid


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STRONG ACID

Strong acid will dissociate or ionize

completely in water to produce

hydrogen, H+ ions.

Degree of dissociation is higher.

Thus, higher concentration of

hydrogen ions in aqueous acid

solution.

Therefore, low pH value of the acid

solution.

WEAK ACID

STRONG ALKALI

WEAK ALKALI


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PHYSICAL PROPERTIES OF ACID & ALKALI

ACID

ALKALI pH Value Litmus paper

Taste Corrosive

pH Value Litmus paper

Corrosive Taste

Conduct

electricity

Conduct

electricity


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CHEMICAL PROPERTIES OF ACID & ALKALI

Acids can react with,

i) bases to produce salts and water

eg :

ii) metal to produce salts and hydrogen gas

eg :

iii) metal carbonates to produce salts, carbon dioxide

and water

eg :

iv) alkali to produce salts and water (neutralization)

eg :

Alkali can react with,

i) acid to produce salts and water (neutralization)

eg :

ii) ammonium salt to produce salts, water and ammonia

gas

eg :


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The concentration of acid and alkali

Concentration are measurement of the quantity of solutes dissolved

in a quantity of solvent.

1. The molarity of a bottle of nitric acid, HNO3 solution is 2.0 mol dm-3.

What is the concentration of the solution in g dm-3?

[RAM: H, 1 ; N, 14 ; O , 16]

Grams per dm3

( g dm-3)

Moles per dm3

( mol dm-3)

*known as Molarity Concentration

Grams per

dm3 ( g dm-3)

Moles per dm3

( mol dm-3)

Unit Conversion

molar mass

molar mass

Example


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2. Calculate the molarity of a sodium sulphate, Na2SO4 solution with

a concentration of 28.4 g dm-3. [RAM: O, 16 ; Na, 23 ; S, 32]

Calculating Involving Concentration and Molarity

1. 5.00 g of copper (II) sulphate is dissolved in water to form 500

cm3 solution. Calculate the concentration of copper (II) sulphate

in g dm-3.

2. A 250 cm3 nitric acid solution contains 0.4 moles. Calculate the

molarity of the nitric acid.

No of mole = Molarity x Volume ( cm3)

1000

n = MV

1000


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3. What is the mass of sodium carbonate required to dissolve in

water to prepare 200 cm3 solution contains 50 g dm-3.

4. Calculate the number of moles of ammonia in 150 cm3 of 2 mol

dm-3 aqueous ammonia.

5. Calculate the volume in dm3 of a 0.8 mol dm-3 sulphuric acid

that contains 0.2 mol.

6. 4.0 g sodium carbonate powder, Na2CO3 is dissolved in water

and made up to 250 cm3. What is the molarity of the sodium

carbonate solution. [RAM: C,12;O,16;Na,23]


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7. Dilute hydrochloric acid used in the school laboratories usually

has a concentration of 2.0mol dm-3. Calculate the mass of

hydrogen chloride that found in 250 cm3 of the hydrochloric

acid? [RAM : H,1; Cl,35.5]

8. The concentration of a potassium hydroxide solution is

84.0 g dm-3. Calculate the number of moles of potassium

hydroxide present in 300 cm3 of the solution. [RAM: K,39,H,1 O,16]


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9. Calculate the number of moles of hydrogen ions present in

200 cm3 of 0.5 mol dm-3 sulphuric acid.

Preparation of Standard Solution

A solution in which its concentration is accurately known is a

standard solution.

Preparation of a solution by dilution method

Adding water to a concentrated solution changes the concentration

of the solution but does not change the amount solutes ( number of

moles)of solution present in solution.

Moles of stock solution = moles of dilute solution

n1 = n2

M1V1 = M2V2

Add water Stock solution

Dilute solution


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1. Find the volume of 2.0 mol dm-3 sulphuric acid, H2SO4 needed to

prepare 100 cm3 of 1.0 mol dm-3 sulphuric acid, H2SO4.

2. Calculate the volume of a concentrated solution needed to

prepare each of the following dilute solution:

a) 50 cm3 of 0.1 mol dm-3 sodium hydroxide, NaOH solution

from 2.0 mol dm-3 sodium hydroxide, NaOH solution.

b) 100 cm3 of 0.5 mol dm-3 potassium manganate(VII),

KMnO4 solution from 1.0 mol dm-3 potassium

manganate(VII), KMnO4 solution.

Exercise


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3. Calculate the volume of 2.0 mol dm-3 sulphuric acid, H2SO4

needed to prepare 2.5 dm3 of 0.5 mol dm-3 of the same acid

solution.

4. Calculate the molarity of potassium hydroxide, KOH, solution if

200 cm3 of 2.0 mol dm-3 potassium hydroxide, KOH, solution is

added to 200 cm3 of water.

5. 60 cm3 of 0.5 mol dm-3 sodium hydroxide, NaOH, solution is

diluted with 30 cm3 of water. Calculate the molarity of the

solution produced.


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The pH Values and Molarity

As the molarity of an acid increases, the pH value of the acid

decreases, however the pH value of an alkali increases when the

molarity of the alkali increases.

Neutralisation

Reaction of an acid

and a base that

produce salt and

water.

Acid-base titration

Titration is a very useful

laboratory technique in

which one solution is

used to analyse another

solution.

An acid of known

concentration is

carefully delivered from

burette to completely

neutralise a known

volume of an alkali in a

conical flask.

Acid-base indicator is

used to detect the end

of titration.

Eg: methyl orange,

phenolphthalein and

litmus.

The point at which the

colour of the solution

change is called the

end point.

Titration using

phenolphthalein


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ACID-BASE

TITRATION

NEUTRALISATION


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1. Write a balanced equation for the neutralization of each of the

following:

a) Sulphuric acid, H2SO4 and barium hydroxide, Ba(OH)2

solution.

b) Nitric acid, HNO3 and calcium hydroxide, Ca(OH)2 solution.

c) Ethanoic acid, CH3COOH and potassium hydroxide, KOH

solution.

Acid-base Indicator

Indicator Colour in

alkalis

Colour in neutral

solution

Colour in

acids

Methyl orange

phenolphthalein

litmus

The end-point of neutralisation also can determined by another two

method:-

i) Measurement of pH values by computer

ii) Measurement of electrical conductivity during titration

Example


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Numerical Problem involving Neutralisation

Say the balance equation is

aA + bB product

which,

A = acid

a = no of mole of acid

B = base

b = no of mole of base

MAVA = a and MBVB = b

Therefore MAVA a

MBVB b

1. In an experiment, 25.0 cm3 of a sodium hydroxide solution of

unknown concentration required 26.50 cm3 of 1.0 mol dm-3

sulphuric acid to complete a reaction in titration. Calculate the

molarity of sodium hydroxide.

=

Exercise


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2. What is the volume of 0.5 mol dm-3 sulphuric acid, H2SO4 needed

to neutralize 25.0 cm3 of 0.8 mol dm-3 ammonia, NH3 solution?

3. A sample of copper(II) oxide, CuO was found to completely

neutralize 100 cm3 of 0.5 mol dm-3 hydrochloric acid, HCl.

Calculate the mass of the sample. [RAM: O, 16 ; Cu, 64]


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4. The volume of 0.15 mol dm-3 sulphuric acid, H2SO4 required to

completely neutralize 25.0 cm3 of potassium hydroxide, KOH

solution is 30.5 cm3. Calculate the molarity of the potassium

hydroxide, KOH solution.

5. A student dissolved 3.65 g of hydrogen chloride gas, HCl in water

to make 1.0 dm3 of solution. Calculate the volume of a 0.1 mol

dm-3 barium hydroxide, Ba(OH)2 solution required to completely

neutralize 25.0 cm3 of the acid solution. [RAM : H, 1 ; Cl, 35.5]

Chemistry Module Form 4 (7)

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