chapter 7 acid & bases part 4
DESCRIPTION
7.4 NEUTRALISATION SLISS 2012TRANSCRIPT
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7.4 NEUTRALISATION
NURUL ASHIKIN BT ABD RAHMAN
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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.
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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)
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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)
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APPLICATION OF NEUTRALISATION
Soil treatment
Treat gastric
Treat wasp stings
Prevent coogulation latex
Baking powder
Manufacture detergent
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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.
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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
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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
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ACID-BASE TITRATION
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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]
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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
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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
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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
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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
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(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
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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
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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
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THE END...
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