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At the end of this lesson, students should be able to:  • Explain the meaning of neutralisation precisely.  • Explain the ...
Acid        Base              Salt           WaterNeutralisation is a reaction between acid and base            to produce...
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)ChemicalEquation           H+Cl-(aq) + Na+OH- (aq)  Na+Cl- (aq) + H2O (l)  Ioni...
Soil treatment                                      Treat gastricBaking powder  Manufacture                         Treat ...
Quantitative analysis that involves the            gradual addition of a chemicalTitration   solution from a burette to an...
a Acid          b Base           Salt      Water Molarity andvolume of acid            M aVa   a      Mole of acid Molarit...
Examples of Indicators   Indicator                   Colour                     Acid       Neutral     AlkaliLitmus Soluti...
Question 1:25.0 cm3 of sulphuric acid is neutralised by 34.0cm3 of 0.1 mol of dm-3 NaOH. Calculate theconcentration of sul...
Solution:Method 1 Step 1 : write down chemical equation2NaOH + H2SO4  Na2SO4 + 2H2OStep 2 : find the number of mole NaOHn...
Step 3 : from the chemical reaction, the ratio of  number of moles of H 2 SO4            1  number of moles of NaOH       ...
Step 5 : find the concentration of H2SO4 in mol dm-3   Concentration         = mol/volume                         = 0.0017...
Method 2: Step 1 : write down chemical equation2NaOH + H2SO4  Na2SO4 + 2H2OStep 2 : find the concentration of H2SO4 in mo...
M a (0.025)      1    (0.1) (0.034)     2               Ma = 0.068 mol dm-3Step 3 : find the concentration of H2SO4 in g d...
Question 2:  What volume of 0.20 mol dm-3 nitric  acid is required to neutralise 0.14 g of  potassium hydroxide? [relative...
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 sol...
Chapter 7 acid & bases part 4
Chapter 7 acid & bases part 4
Chapter 7 acid & bases part 4
Chapter 7 acid & bases part 4
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Chapter 7 acid & bases part 4

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7.4 NEUTRALISATION
SLISS 2012

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Transcript of "Chapter 7 acid & bases part 4"

  1. 1. 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.
  2. 2. Acid Base Salt WaterNeutralisation 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)
  3. 3. HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)ChemicalEquation H+Cl-(aq) + Na+OH- (aq)  Na+Cl- (aq) + H2O (l) Ionic H+(aq) + OH- (aq)  H2O (l)Equation
  4. 4. Soil treatment Treat gastricBaking powder Manufacture Treat wasp detergent stings Prevent coogulation latex
  5. 5. Quantitative analysis that involves the gradual addition of a chemicalTitration 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.
  6. 6. a Acid b Base Salt Water Molarity andvolume of acid M aVa a Mole of acid Molarity andvolume of base M bVb b Mole of base
  7. 7. Examples of Indicators Indicator Colour Acid Neutral AlkaliLitmus Solution Red Purple BluePhenolphthalein Colourless Colourless PinkMethyl orange Red Orange Yellow Universal Red Green Purple indicator
  8. 8. Question 1:25.0 cm3 of sulphuric acid is neutralised by 34.0cm3 of 0.1 mol of dm-3 NaOH. Calculate theconcentration of sulphuric acid in: (a) mol dm -3 (b) g dm-3[relative atomic mass; H:1, S:32, O:16]
  9. 9. Solution:Method 1 Step 1 : write down chemical equation2NaOH + H2SO4  Na2SO4 + 2H2OStep 2 : find the number of mole NaOHn=MVMoles of NaOH= molarity X Volume (dm3)= 0.1 X 0.034= 0.0034 mol
  10. 10. Step 3 : from the chemical reaction, the ratio of number of moles of H 2 SO4 1 number of moles of NaOH 2Step 4 : find the number of moles of H2SO4 reacted2 mole of NaOH = 1 mole of H2SO40.0034 mole of NaOH = 0.0034 1 mol 2 = 0.0017 mol
  11. 11. 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-3Molar 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
  12. 12. Method 2: Step 1 : write down chemical equation2NaOH + H2SO4  Na2SO4 + 2H2OStep 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
  13. 13. M a (0.025) 1 (0.1) (0.034) 2 Ma = 0.068 mol dm-3Step 3 : find the concentration of H2SO4 in g dm-3Molar 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
  14. 14. 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
  15. 15. 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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