Chapter 11 Worked Solution

Worked Solutions Chapter 11 Question 5 What is the concentration in grams per litre of (a) a 0.1 M HCl solution (b) a 0.05 M Na2CO3 solution (c) a 0.05 M NaOH solution (d) a 0.01 M KOH solution (e) a 0.2 M HNO3 solution? Answer: (a) Molar mass of HCl = 36.5 g mol-1 0.1M HCl solution contains 0.1 moles = 0.1 X 36.5 g in 1 litre Concentration of HCl solution = 3.65 g l–1 (b) Molar mass of Na2CO3 = 106 g mol-1 0.05M Na2CO3 solution contains 0.05 moles = 0.05 X 106 g in 1 litre Concentration of Na2CO3 solution = 5.3 g l–1 (c) Molar mass of NaOH = 40 g mol-1 0.05M NaOH solution contains 0.05 moles = 0.05 X 40 g in 1 litre Concentration of NaOH solution = 2 g l–1 (d) Molar mass of KOH = 56 g mol-1 0.01M KOH solution contains 0.01 moles = 0.01 X 56 g in 1 litre Concentration of KOH solution = 0.56 g l–1 (e) Molar mass of HNO3 = 63 g mol-1 0.2M HNO3 solution contains 0.2 moles = 0.2 X 63 g in 1 litre Concentration of HNO3 solution = 12.6 g l–1 Question 6 A solution contains 25 g of potassium hydroxide (KOH) in 1 litre of solution. Calculate the concentration of the solution in moles per litre. Answer: 25 g KOH = 25/56 moles KOH = 0.45 moles KOH Concentration of the solution = 0.45 mol l-1 (=0.45 M) Question 7 The concentration of a potassium manganate(VII) solution (KMnO4) is 6.32 g l –1. What is the concentration of this solution in moles per litre? Answer: 6.32 g KMnO4 = 6.32/158 moles KMnO4 = 0.04 moles KMnO4 Concentration of the solution = 0.04 mol l-1 (= 0.04 M) Question 8 A solution contains 4 g of NaOH in 250 cm3 of solution. Calculate the concentration of the solution in moles per litre. Answer: 4 g of NaOH in 250 cm3 of solution is equivalent to 4 x 1000/250 = 16 g in 1 litre 16 g NaOH = 16/40 moles NaOH = 0.4 moles NaOH Concentration of the solution = 0.4 mol l-1 (= 0.4 M) Question 9 A solution contains 4.9 g H2SO4 in 100 cm3 of solution. Calculate the concentration of the solution in moles per litre. Answer: 4.9 g of H2SO4 in 100 cm3 of solution is equivalent to 4.9 x 1000/100 = 49 g in 1 litre 49 g H2SO4 = 49/98 moles H2SO4 = 0.5 moles H2SO4 Concentration of the solution = 0.5 mol l-1 (= 0.5 M) Question 10 A bottle of wine contains 49 g of ethanol (C2H5OH) in 700 cm3 of solution. Calculate the concentration of ethanol in moles per litre. Answer: 49 g of C2H5OH in 700 cm3 of solution is equivalent to 49 x 1000/700 = 70 g in 1 litre 70 g C2H5OH = 70/46 moles C2H5OH = 1.52 moles C2H5OH Concentration of the solution = 1.52 mol l-1 (= 1.52 M) Question 11 What mass of potassium hydroxide (KOH) is contained in 500 cm3 of 0.05 M potassium hydroxide solution? Answer: 1,000 cm3 of 0.05 M KOH solution contains 0.05 moles 500 cm3 of 0.05 M KOH solution contains 0.05 X 500 / 1,000 moles = 0.025 moles Molar mass of KOH = 56 g mol-1 Mass of KOH contained in 500 cm3 solution = 56 X 0.025 g = 1.4 g Question 12 What mass of sodium hydroxide (NaOH) is contained in 250 cm3 of 0.02 M sodium hydroxide solution? Answer: 1,000 cm3 of 0.02 M NaOH solution contains 0.02 moles 250 cm3 of 0.02 M NaOH solution contains 0.02 X 250 / 1,000 moles = 0.005 moles Molar mass of NaOH = 40 g mol-1 Mass of NaOH contained in 250 cm3 solution = 40 X 0.005 g = 0.2 g Question 13 How many moles of hydrochloric acid (HCl) are contained in 25 cm3 of 0.05 M hydrochloric acid solution? Answer: Number of moles = volume in cm3 X concentration in moles per litre / 1,000 = 25 X 0.05 / 1,000 = 0.00125 Question 14 How many moles of sulfuric acid (H2SO4) are contained in 15 cm3 of 0.02 M sulfuric acid solution? Answer: Number of moles = volume in cm3 X concentration in moles per litre / 1,000 = 15 X 0.02 / 1,000 = 0.0003 Question 15 How many moles of sodium carbonate are there in 100 cm3 of 0.05 M sodium carbonate solution? Answer: Number of moles = volume in cm3 X concentration in moles per litre / 1,000 = 100 X 0.05 / 1,000 = 0.005 Question 16 The concentration of dissolved oxygen in a water sample is 0.00002 M. Calculate the concentration of dissolved oxygen in the water sample in p.p.m. Answer: Molar mass of O2 = 32 g mol-1 There are 0.00002 moles of dissolved oxygen in 1 litre of the water sample. Mass of O2 in 1 litre = 0.00002 x 32 g = 0.00064 g = 0.00064 x 1,000 mg = 0.64 mg Concentration of dissolved oxygen in the water sample = 0.64 p.p.m. Question 17 A solution contains 25 g sodium hydroxide in 1 litre of solution. Express the concentration of the solution in % (w/v). Answer: The solution contains 25 g NaOH in 1 litre. 100 cm3 of solution contains 25 x 100 / 1,000 g NaOH = 2.5 g NaOH Concentration of NaOH in the solution = 2.5% (w/v) Question 18 A solution contains 8 g sodium hydroxide in 500 cm3 of solution. Express the concentration of the solution in % (w/v). Answer: The solution contains 8 g sodium hydroxide in 500 cm3 of solution. 100 cm3 of solution contains 8 x 500 / 1,000 g NaOH = 1.6 g NaOH Concentration of NaOH in the solution = 1.6% (w/v) Question 19 A bottle of vinegar contains 77 cm3 ethanoic acid in 700 cm3 of solution. Express the concentration of ethanoic acid in the solution in % (v/v). Answer: The solution contains 77 cm3 ethanoic acid in 700 cm3 of solution. 100 cm3 of solution contains 77 x 100 / 700 cm3 ethanooic acid = 11 cm3 ethanoic acid Concentration of ethanoic acid in the solution = 11% (v/v) Question 20 A solution contains 20 g potassium hydroxide in 250 g of solution. Express the concentration of the solution in % (w/w). Answer: The solution contains 20 g KOH in 250 g of solution. 100 g of solution contains 20 x 100 / 250 g KOH = 8 g KOH Concentration of KOH in the solution = 8% (w/w) Question 21 The label on a bottle of wine indicates that the concentration of alcohol in the wine is 9% (v/v). What volume of alcohol is there in 250 cm3 of the wine? Answer: 100 cm3 of the wine contains 9 cm3 alcohol. 250 cm3 of the wine contains 9 x 250 / 100 cm3 alcohol = 22.5 cm3 alcohol. Question 22 What volume of ethanol is there in 500 cm3 of a 5% solution of ethanol in water? Answer: 100 cm3 of the solution contains 5 cm3 ethanol. 500 cm3 of the solution contains 5 X 500 / 100 cm3 ethanol = 25 cm3 ethanol. Question 26 If 25 cm3 of a 1 M hydrochloric acid solution is diluted to a volume of 500 cm3 with water, what is the concentration of the diluted solution? Answer: Vdil X Mdil = Vconc X Mconc 500 X Mdil = 25 X 1 Mdil = 25 X 1 / 500 mol l-1 = 0.05 mol l-1 Concentration of the diluted solution = 0.05 mol l-1 Question 27 If 10 cm3 of a 0.1 M potassium hydroxide solution is diluted to a volume of 250 cm3 with water, what is the concentration of the diluted solution? Answer: Vdil X Mdil = Vconc X Mconc 250 X Mdil = 10 X 0.1 Mdil = 10 X 0.1 / 250 mol l-1 = 0.004 mol l-1 Concentration of the diluted solution = 0.004 mol l-1 Question 28 What volume of a 1 M hydrochloric acid solution is needed to make up 500 cm3 of a 0.1 M hydrochloric acid solution? Answer: Vdil X Mdil = Vconc X Mconc 500 X 0.1 = Vconc X 1 Vconc = 500 X 0.1 / 1 mol l-1 = 50 cm3 Volume of 1 M hydrochloric acid solution needed = 50 cm3 Question 29 What volume of a 0.5 M sulfuric acid solution is needed to make up 250 cm3 of a 0.1 M sulfuric acid solution? Answer: Vdil X Mdil = Vconc X Mconc 250 X 0.1 = Vconc X 0.5 Vconc = 250 X 0.1 / 0.5 mol l-1 = 50 cm3 Volume of 0.5 M sulfuric acid solution needed = 50 cm3 Question 40 In a titration, 25 cm3 of a 0.05 M sodium carbonate solution required 22 cm3 of a hydrochloric acid solution for complete neutralisation. Calculate the concentration of the hydrochloric acid solution. The equation for the reaction is 2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + H2O(l) + CO2(g) Answer: V1 x M1 x n2 = V2 x M2 x n1 22 x M1 x 1 = 25 x 0.05 x 2 M1 = 25 x 0.05 x 2 / (22 x 1) = 0.11 mol l-1 Concentration of hydrochloric acid solution = 0.11 mol l-1 Question 41 In a titration, 25 cm3 of a sodium hydroxide solution required 19.5 cm3 of a 0.1 M hydrochloric acid solution for complete neutralisation. Calculate the concentration of the sodium hydroxide solution. The equation for the reaction is HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Answer: V1 x M1 x n2 = V2 x M2 x n1 19.5 x 0.1 x 1 = 25 x M2 x 1 M2 = 19.5 x 0.1 x 1 / (25 x 1) = 0.078 mol l-1 Concentration of sodium hydroxide solution = 0.078 mol l-1 Question 42 In a titration, 25 cm3 of a potassium hydroxide solution required 22.5 cm3 of a 0.01 M hydrochloric acid solution for complete neutralisation. Calculate the concentration of the potassium hydroxide solution. The equation for the reaction is HCl(aq) + KOH(aq) KCl(aq) + H2O(l) Answer: V1 x M1 x n2 = V2 x M2 x n1 22.5 x 0.01 x 1 = 25 x M2 x 1 M2 = 22.5 x 0.01 x 1 / (25 x 1) = 0.009 mol l-1 Concentration of potassium hydroxide solution = 0.009 mol l-1 Question 43 In a titration, what volume of a 0.1 M hydrochloric acid is required to neutralise 25 cm3 of a 0.02 M sodium carbonate solution? The equation for the reaction is 2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + H2O(l) + CO2(g) Answer: V1 x M1 x n2 = V2 x M2 x n1 V1 x 0.1 x 1 = 25 x 0.02 x 2 V1 = 25 x 0.02 x 2 / (0.1 x 1) = 10 cm3 of hydrochloric acid Question 44 In a titration, what volume of a 0.05 M hydrochloric acid is required to neutralise 25 cm3 of a 0.04 M sodium hydroxide solution? The equation for the reaction is HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Answer: V1 x M1 x n2 = V2 x M2 x n1 V1 x 0.05 x 1 = 25 x 0.04 x 1 V1 = 25 x 0.04 x 1 / (0.05 x 1) = 20 cm3 of hydrochloric acid Question 45 What volume of 0.02 M H2SO4 is needed to neutralise 25 cm3 of a 0.05 M NaOH solution? The equation for the reaction is H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) Answer: V1 x M1 x n2 = V2 x M2 x n1 V1 x 0.02 x 2 = 25 x 0.05 x 1 V1 = 25 x 0.05 x 1 / (0.02 x 2) = 31.25 cm3 of sulfuric acid Question 46 In a titration, 25 cm3 of a 0.12 M sodium hydroxide solution required 24 cm3 of a sulfuric acid solution for complete neutralisation. Calculate (a) the number of moles of sodium hydroxide consumed (b) the number of moles of sulfuric acid consumed and (c) the concentration of the sulfuric acid solution. The equation for the reaction is H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) Answer: (a) In the titration, 25 cm3 of 0.12 M sodium hydroxide solution were used. 1,000 cm3 of 0.12 M sodium hydroxide solution contains 0.12 moles. 25 cm3 of 0.12 M sodium hydroxide solution contains 0.12 X 25 / 1,000 moles = 0.003 moles (b) The equation for the reaction is H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) 2 moles of NaOH react with 1 mole of H2SO4 1 mole of NaOH reacts with 0.5 moles of H2SO4 0.003 moles of NaOH reacts with 0.003 X 0.5 moles of H2SO4 = 0.0015 moles of H2SO4 (c) In 24 cm3 of the H2SO4 solution there are 0.0015 moles of H2SO4 In 1,000 cm3 of the H2SO4 solution there are 0.0015 X 1,000 / 24 moles = 0.0625 moles Concentration of the sulfuric acid solution = 0.0625 mol l-1 Question 47 20 cm3 of a 0.125 M HCl solution were neutralised by 25 cm3 of a KOH solution. The equation for the reaction is HCl(aq) + KOH(aq) KCl(aq) + H2O(l) (a) How many moles of HCl were consumed? (b) How many moles of KOH were neutralised? (c) How many moles of KOH were there in 25 cm3 of the KOH solution? (d) Calculate the concentration in mol l-1 of the KOH solution. solution. Answer: (a) In the titration, 20 cm3 of 0.125 M HCl solution were used. 1,000 cm3 of 0.125 M HCl solution contains 0.125 moles. 20 cm3 of 0.125 M HCl solution contains 0.125 X 20 / 1,000 moles = 0.0025 moles (b) The equation for the reaction is HCl(aq) + KOH(aq) KCl(aq) + H2O(l) One mole of HCl reacts with one mole of KOH. 0.0025 moles of HCl react with 0.0025 moles of KOH. (c) The 25 cm3 of KOH consumed contained 0.0025 moles. (d) In 25 cm3 of the KOH solution there are 0.0025 moles of KOH In 1,000 cm3 of the KOH solution there are 0.0025 X 1,000 / 25 moles = 0.1 moles Concentration of the KOH solution = 0.1 mol l-1 Question 48 What mass of magnesium will react with 20 cm3 of a 0.09 M hydrochloric acid solution? The equation for the reaction is 2HCl(aq) + Mg(s) MgCl2(aq) + H2(g) Answer: 1,000 cm3 of 0.09 M HCl solution contains 0.09 moles 20 cm3 of 0.09 M HCl solution contains 0.09 X 20 / 1,000 moles = 0.0018 moles From the equation, 1 mole of magnesium reacts with 2 mole of hydrochloric acid 0.0018 moles of hydrochloric acid react with 0.0018 / 2 = 0.0009 moles of magnesium Molar mass of magnesium = 24 g mol-1 0.0009 moles of magnesium = 0.0009 X 24 g = 0.0216 g magnesium Mass of magnesium = 0.022 g Question 49 What mass of magnesium will react with 15 cm3 of a 0.12 M sulfuric acid solution? The equation for the reaction is H2SO4(aq) + Mg (s) MgSO4(aq) + H2(g) Answer: 1,000 cm3 of 0.12 M H2SO4 solution contains 0.12 moles 15 cm3 of 0.12 M H2SO4 solution contains 0.12 X 15 / 1,000 moles = 0.0018 moles From the equation, 1 mole of magnesium reacts with 1 mole of sulfuric acid 0.0018 moles of magnesium react with 0.0018 moles of sulfuric acid Molar mass of magnesium = 24 g mol-1 0.0018 moles of magnesium = 0.0018 X 24 g = 0.0432 g magnesium Mass of magnesium = 0.043 g Question 50 A sample of vinegar was diluted from 25 cm3 to 250 cm3 with water. In a titration, 25 cm3 of a 0.1 M sodium hydroxide solution required 30 cm3 of the diluted vinegar for complete neutralisation. Calculate the concentration of ethanoic acid in the vinegar in (a) mol l-1 (b) % (w/v). The equation for the reaction is CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l) Answer: (a) The formula V1 x M1 x n2 = V2 x M2 x n1 is used first to find the concentration of ethanoic acid in the diluted vinegar. V1 x M1 x n2 = V2 x M2 x n1 30 x M1 x 1 = 25 x 0.1 x 1 M1 = 25 x 0.1 x 1 / (30 x 1) = 0.083 mol l-1 Concentration of ethanoic acid in the diluted vinegar = 0.083 mol l-1 The vinegar was originally diluted from 25 cm3 to 250 cm3 with water, i.e. a tenfold dilution. Concentration of ethanoic acid in the vinegar = 10 x 0.083 mol l-1 = 0.83 M (b) Molar mass of ethanoic acid (CH3COOH) = 60 g mol-1 Concentration of ethanoic acid in g l-1 in vinegar = 0.83 x 60 = 49.8 Percentage (w/v) of ethanoic acid in vinegar = 49.8 x 100 / 1000 = 4.98% Question 51 2.91 g of a monobasic acid, HX, were dissolved in water and made up to 250 cm3 with water. This solution was titrated with 0.108 M sodium hydroxide solution. 25 cm3 of the sodium hydroxide solution required 22.5 cm3 of the HX solution for complete neutralisation. The equation for the reaction is HX (aq) + NaOH(aq) NaX(aq) + H2O(l) (a) Calculate the concentration in (i) g l-1 (ii) mol l-1 of the acid. (b) Calculate the molar mass of HX. Answer: (i) 2.91 g of HX in 250 cm3 is equivalent to 2.91 X 4 g = 11.64 g in 1 litre. Thus the concentration is 11.64 g l-1 (ii) V1 x M1 x n2 = V2 x M2 x n1 22.5 x M1 x 1 = 25 x 0.108 x 1 M1 = 25 x 0.108 x 1 / (22.5 x 1) = 0.12 mol l-1 Concentration of HX solution = 0.12 mol l-1 0.12 moles has a mass of 11.64 g 1 mole has a mass of 11.64/0.12 g = 97 g The molar mass of HX is 97 g mol-1 Question 52 Crystals of hydrated sodium carbonate (Na2CO3.xH2O) of mass 3.15 g were dissolved in water and made up to 250 cm3 in a volumetric flask. 25 cm3 of this solution required 15 cm3 of a 0.15 M hydrochloric acid solution for complete neutralisation. The equation for the reaction is 2HCl (aq) + Na2CO3 (aq) 2NaCl (aq) + H2O(l) + CO2(g) Find (a) the concentration of the sodium carbonate solution (b) the value of x in the formula Na2CO3.xH2O (c) the percentage of water of crystallisation in the hydrated sodium carbonate. Answer: (a) V1 x M1 x n2 = V2 x M2 x n1 15 x 0.15 x 1 = 25 x M2 x 2 M2 = 15 x 0.15 x 1 / (25 x 2) = 0.045 mol l-1 Concentration of the sodium carbonate solution = 0.045 mol l-1 (b) Molar mass of Na2CO3.xH2O = 46 + 12 + 48 + 18x = 106 + 18x Concentration of sodium carbonate in mol l-1 = 0.045 Concentration of sodium carbonate in g l-1 = 3.15 X 4 = 12.6 Mass of 0.045 moles of Na2CO3.xH2O = 12.6 g Molar mass of Na2CO3.xH2O = 12.6 / 0.045 g mol1 = 280 g mol1 106 + 18x = 280 18x = 174 x = 9.7 (c) Molar mass of hydrated sodium carbonate = 280 g mol-1 Percentage of water of crystallisation in the compound = 174 x 100 / 280% = 62.14% Question 53 In a titration, 25 cm3 of a potassium hydroxide solution required 27 cm3 of a 0.12 M sulfuric acid solution for complete neutralisation. Calculate the concentration of the potassium hydroxide solution in (a) mol l-1 (b) g l-1. The equation for the reaction is H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l) Answer: (a) V1 x M1 x n2 = V2 x M2 x n1 27 x 0.12 x 2 = 25 x M2 x 1 M2 = 27 x 0.12 x 2 / (25 x 1) = 0.2592 mol l-1 Concentration of potassium hydroxide solution = 0.26 mol l-1 (b) Molar mass of potassium hydroxide = 56 g mol-1 Concentration of KOH = 0.26 x 56 g l-1 = 14.6 g l-1 Question 54 In a titration, 20 cm3 of 0.1 M potassium hydroxide solution required 25 cm3 of a hydrochloric acid solution for complete neutralisation. Calculate the concentration of the hydrochloric acid solution in (a) mol l-1 (b) g l-1. The equation for the reaction is HCl(aq) + KOH(aq) KCl(aq) + H2O(l) Answer: (a) V1 x M1 x n2 = V2 x M2 x n1 25 x M1 x 1 = 20 x 0.1 x 1 M1 = 20 x 0.1 x 1 / (25 x 1) = 0.08 mol l-1 Concentration of hydrochloric acid solution = 0.08 mol l-1 (b) Molar mass of hydrochloric acid = 36.5 g mol-1 Concentration of HCl = 0.08 X 36.5 g l-1 = 2.92 g l-1 Question 56 In a titration, 25 cm3 of a sodium carbonate solution required 22 cm3 of a 0.1 M hydrochloric acid solution for complete neutralisation. Calculate the concentration of the sodium carbonate solution in (a) mol l-1 (b) g l-1. The equation for the reaction is: 2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + H2O(l) + CO2(g) Answer: (a) V1 x M1 x n2 = V2 x M2 x n1 22 x 0.1 x 1 = 25 x M2 x 2 M2 = 22 x 0.1 x 1 / (25 x 2) = 0.044 mol l-1 Concentration of sodium carbonate solution = 0.044 mol l-1 (b) Molar mass of sodium carbonate = 106 g mol-1 Concentration of sodium carbonate solution = 0.044 x 106 g l-1 = 4.66 g l-1 Question 57 What volume of 0.01 M sodium hydroxide solution is required to neutralise 20 cm3 of a hydrochloric acid solution containing 0.73 g of hydrogen chloride per 250 cm3 of solution? The equation for the reaction is HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Answer: 0.73 g of hydrogen chloride per 250 cm3 of solution is equivalent to 0.73 x 4 g l-1 = 2.92 g l-1 = 2.92 / 36.5 mol l-1 = 0.08 mol l-1 V1 x M1 x n2 = V2 x M2 x n1 20 x 0.08 x 1 = V2 x 0.01 x 1 V2 = 20 x 0.08 x 1 / (0.01 x 1) = 160 cm3 of hydrochloric acid Question 58 What volume of a 0.01 M sodium carbonate solution is required to neutralise 25 cm3 of a solution containing 0.98 g l-1 of sulfuric acid? The equation for the reaction is H2SO4(aq) + Na2CO3(aq) Na2SO4(aq) + H2O(l) + CO2(g) Answer: Concentration of sulfuric acid solution = 0.98 g l-1 = 0.98 / 98 mol l-1 = 0.01 mol l-1 V1 x M1 x n2 = V2 x M2 x n1 25 x 0.01 x 1 = V2 x 0.01 x 1 V2 = 25 x 0.01 x 1 / (0.01 x 1) = 25 cm3 of sodium carbonate solution Question 59 In a titration, 25 cm3 of a 0.1 M sodium hydroxide solution required 28 cm3 of a sulfuric acid solution for complete neutralisation. Calculate (a) the number of moles of sodium hydroxide consumed (b) the number of moles of sulfuric acid consumed and (c) the concentration of the sulfuric acid solution in mol l-1. The equation for the reaction is H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) Answer: (a) In the titration, 25 cm3 of 0.1 M NaOH solution were used. 1,000 cm3 of 0.1 M NaOH solution contains 0.1 moles. 25 cm3 of 0.1 M NaOH solution contains 0.1 X 25 / 1,000 moles = 0.0025 moles (b) The equation for the reaction is H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) One mole of H2SO4 reacts with two moles of NaOH. 0.0025/2 moles = 0.00125 moles of H2SO4 react with 0.0025 moles of NaOH. The number of moles of sulfuric acid consumed is 0.00125 moles. (c) In 28 cm3 of the H2SO4 solution there are 0.00125 moles of H2SO4 In 1,000 cm3 of the H2SO4 solution there are 0.00125 X 1,000 / 28 moles = 0.045 moles Concentration of the H2SO4 solution = 0.045 mol l-1 Question 60 Sodium hydrogencarbonate solution reacts with hydrochloric acid solution according to the equation HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O(l) + CO2(g) It was found that 25 cm3 of a solution containing 8.4 g l-1 reacted with 27 cm3 of a solution of hydrochloric acid. Calculate the concentration of the hydrochloric acid solution in (a) mol l-1 (b) g l-1. Answer: (a) Concentration of sodium hydrogencarbonate solution = 8.4 g l-1 = 8.4 / 84 mol l-1 = 0.1 mol l-1 V1 x M1 x n2 = V2 x M2 x n1 27 x M1 x 1 = 25 x 0.1 x 1 M1 = 25 x 0.1 x 1 / (27 x 1) = 0.093 mol l-1 (b) Concentration of hydrochloric acid solution = 0.093 mol l-1 = 0.093 x 36.5 mol l-1 = 3.39 g l-1 1