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.
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)
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.
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
Examples of Indicators Indicator Colour Acid Neutral AlkaliLitmus Solution Red Purple BluePhenolphthalein Colourless Colourless PinkMethyl orange Red Orange Yellow Universal Red Green Purple indicator
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]
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
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
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
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
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
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