2. Titration
Titration is a quantitative volumetric analysis to
determine the concentration of an acid or a
base.
The apparatus used in a titration includes the
following:
• Pipette
• Burette
• Retort stand
• Erlenmeyer flask (conical flask)
• Measuring cylinder
• Beaker
Beaker
Burette
Erlenmeyer flask
Hydrochloric acid
Sodium hydroxide solution
Retort stand
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3. A titration is a technique where a solution of known concentration is used to
determine the concentration of an unknown solution.
This process involves the gradual neutralisation of an acid by a base, or a
base by an acid.
The point where enough acid has been added to base (or base added to
acid) to bring about a colour change in the indicator is called the end point
of neutralisation.
TITRATIONS
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4. An indicator is used to determine the end point which becomes visible by
means of a colour change.
As we have already discovered, this will not necessarily be at a pH of 7, so
we must choose a suitable indicator for the titration that we do.
TITRATIONS
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5. If we are titrating a strong acid and a strong base or a weak acid and a weak
base, the end point would be at approximately a pH of 7, so we would use
bromothymol blue as the indicator.
If we are titrating a strong acid with a weak base, the end point would be at a
pH less than 7, so we would use methyl orange as the indicator.
If we are titrating a weak acid with a strong base, the end point would be at a
pH greater than 7, so we would use phenolphthalein as the indicator.
TITRATIONS - INDICATORS
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6. • Titrations are carried out to determine the unknown concentration of
an acid or a base.
• If one solution, either acid or base of known concentration is added
to a solution of unknown concentration until it is exactly neutralized,
the concentration of substance with the unknown concentration can
be calculated.
• This is known as volumetric analysis and the titration is the technique
used to perform this.
• Note: A solution of known concentration which is used in the titration
is called a standard solution.
TITRATIONS
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7. VolumetricAnalysis
Acid
solution
Base
solution
Titration
◦ A procedure for determining the amount of substanceA by adding a carefully measured volume
(burette) of a solution with a known concentration of B until the reaction ofA and B is just complete.
◦ Volumetric analysis is a method of analysis based on titration
◦ An indicator is a substance that changes color when the reaction of interest approaches completion
e.g.
phenolphthalein.
Equivalence point
Colourles
s
(acidic)
Equivalen
ce
(neutral)
Pink
(basic
)
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8. 1. 82 cm3 of hydrochloric acid of concentration 0,2 mol.dm–3 is required to
neutralize 60 cm3 of Ba(OH)2, of unknown concentration. Calculate the
concentration of Ba(OH)2.
HCl + Ba(OH)2→ BaCl2 + H2O
The number of moles of acid and base are the ratio of acid to base in the equation
and are the same as the number in front of the acid and base in the balanced
equation.
Therefore, it is always necessary to be working from a balanced equation when
doing titration calculations.
acid base
2HCl + 1Ba(OH)2→ BaCl2 + 2H2O
Mole ratio: 2 : 1
EXAMPLES OF TITRATION QUESTIONS
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9. • 2HCl + 1Ba(OH)2→ BaCl2 + 2H2O
• ca = 0,2 mol.dm-3 Va = 82 cm3 cb = ? Vb = 60 cm3
• Moles acid (concentration and volume), moles base (ratio), concentration base (moles and volume
• Or
• STEP 1:
• Determine the mole ratio between the two reactants, using a balanced
equation.
• Acid : Base
• HCl : NaOH
• 2 : 1
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10. • 2HCl + 1Ba(OH)2→ BaCl2 + 2H2O
• STEP 2:
• Using the equation n = c.V, formulate an equation that equates the number of moles of
acid to the number of moles of base, using the mole ratio from step 1.
• na / nb = 2 / 1
• We know that n = c.V, therefore:
• ca Va / cb Vb = 2 / 1
• STEP 3:
• Solve for the unknown.
• ca Va / cb Vb = 2 / 1
• (0,2)(82) / cb (60) = 2 / 1
• cb = 0,14 mol.d m-3
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11. 1. The concentration of a solution of sodium carbonate is 0,5 mol.dm–
3. 20 cm3 of this solution neutralises exactly 24,5 cm3 of sulphuric
acid. Calculate the concentration of the sulphuric acid.
• Na2CO3 + H2SO4→ Na2SO4 + H2O + CO2
• SOLUTION:
• base acid
• 1Na2CO3 + 1H2SO4→ Na2SO4 + H2O + CO2
• Mole ratio: 1 : 1
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12. • 1Na2CO3 + 1H2SO4→ Na2SO4 + H2O + CO2
• ca = ?
• Va = 24,5 cm3
• cb = 0,5 mol.dm-3
• Vb = 20 cm3
• STEP 1:
• Determine the mole ratio between the two reactants, using a balanced
equation.
• Acid : Base
• Na2CO3 : H2SO4
• 1 : 1
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13. • 1Na2CO3 + 1H2SO4→ Na2SO4 + H2O + CO2
• STEP 2:
• Using the equation n = c.V, formulate an equation that equates the number of moles of
acid to the number of moles of base, using the mole ratio from step 1.
• na / nb = 1 / 1
• We know that n = c.V, therefore:
• ca Va / cb Vb = 1 / 1
• STEP 3:
• Solve for the unknown.
• ca Va / cb Vb = 1 / 1
• ca (24,5) / (0,5)(20) = 1 / 1
• ca = 0,41 mol.d m-3
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14. 4/25/2023 Acids and bases - Titrations
PERCENTAGE PURITY
Reaction of a sea shell with hydrochloric acid
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
15. Acids and bases - Titrations
PERCENTAGE PURITY
Aim: To prepare a standard solution of oxalic acid.
OXALIC ACID – SODIUM HYDROXIDE – ETHANOIC ACID %ge purity
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16. The substances that we use in titrations are not always completely pure.
They may contain impurities and sometimes we have to calculate how
pure the sample is or what percentage by mass there exists in a sample.
PERCENTAGE PURITY
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17. • 3,68 g of washing soda crystals (Na2CO3.10H2O) are dissolved in water
and made up to a volume of 275 cm3. 25 cm3 of this sample is
neutralised by 23,5 cm3 of HCl of concentration 0,11 mol.dm–3.
Calculate the percentage Na2CO3 in commercial washing soda.
• Na2CO3 + 2HCl → CO2 + 2NaCl + H2O
PERCENTAGE PURITY
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18. • It is important to note that 3,68 g is not the mass of Na2CO3 – it is the mass of
washing soda crystals, which is comprised of some Na2CO3 as well as water of
crystallization and some impurities as well.
• The aim of our calculations is to determine how much of the 3,68 g of washing
soda is actually Na2CO3.
• The HCl will only react with the Na2CO3 in the washing soda, so we will use the
titration information to determine the mass of Na2CO3 in the washing soda.
PERCENTAGE PURITY
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19. STEP 1:
Calculate the number of moles of the pure acid used in the titration.
c = n / V
n = c.V
n = (0,11)(0,0235)
n = 2,59 × 10−3 mol
PERCENTAGE PURITY
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20. STEP 2:
Using the mole ratio, find the number of moles of Na2CO3 used in the titration.
We divide the number of moles of HCl by 2 to get the number of moles of Na2CO3
because the mole ration of HCl to Na2CO3 is 2:1 in the balanced equation.
n = (2,59 × 10−3 ) ÷ 2
∴n = 1,29 × 10−3 mol
PERCENTAGE PURITY
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21. • STEP 3:
• Find the number of moles of Na2CO3 in the original 275 cm3 sample.
• 1,29 × 10–3 mol is the number of moles in 25 cm3 of the Na2CO3 solution and we need to know the
number of moles in the original 275 cm3 sample.
• n = (1,29 x 10−3 / 25) × 275
• ∴n = 1,42 × 10 −2 mol Na2CO3 in 275 cm3 sample
PERCENTAGE PURITY
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22. STEP 4:
Calculate the mass of Na2CO3 in the original 275 cm3 sample.
n = m / M
m = n.M
m = (1,42 × 10−2 )(106)
m = 1,51 g Na2CO3
PERCENTAGE PURITY
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23. STEP 5:
Using the original mass used, calculate the percentage of Na2CO3 the original sample.
Only 1,51 g of the original sample is pure Na2CO3.
Therefore, the percentage purity of the Na2CO3 is:
% purity = (1,51 / 3,68) × 100 = 41 %
PERCENTAGE PURITY
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24. THE APPLICATION OF ACIDS AND BASES
• ACIDS AND BASES IN THE CHLOR-ALKALI INDUSTRY
• ACIDS AND BASES IN THE CHEMISTRY OF HAIR
• Hair relaxers
• Permanent waves
• Hair colouring
4/25/2023 Acids and bases - Titrations