JK-TK/Newdata huggingface dataset:
Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 24.5 24.5 24.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.5 24.5 24.5 Sample calculations (a)Calculate the average volume of solution L used(1mk) 24.5 + 24.5 + 24.5 = 24.5cm3 3 (b)How many moles of sodium hydroxide are present in the average volume of solution L used? Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles
42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K? Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 25cm3(volume pipetted) =0.49 /4.9 x 10-1moles (d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used Original moles = Original molarity x pipetted volume => 1000cm3 1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles 1000 (e)How many moles of hydrochloric acid were used to react with calcium carbonate present?
Sample Table 2 1 2 3 Final burette reading (cm3) 24.5 24.5 24.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.5 24.5 24.5 Sample calculations (a)Calculate the average volume of solution L used(1mk) 24.5 + 24.5 + 24.5 = 24.5cm3 3 (b)How many moles of sodium hydroxide are present in the average volume of solution L used? Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles
42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K? Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 25cm3(volume pipetted) =0.49 /4.9 x 10-1moles (d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used Original moles = Original molarity x pipetted volume => 1000cm3 1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles 1000 (e)How many moles of hydrochloric acid were used to react with calcium carbonate present? Moles that reacted = original moles –moles in average titre => 0.04807/4.807 x 10-2moles - 0.01225 /1.225 x 10-2moles = 0.03582/3.582 x 10 -2 moles (f)Write the equation for the reaction between calcium carbonate and hydrochloric acid.
Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles
42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K? Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 25cm3(volume pipetted) =0.49 /4.9 x 10-1moles (d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used Original moles = Original molarity x pipetted volume => 1000cm3 1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles 1000 (e)How many moles of hydrochloric acid were used to react with calcium carbonate present? Moles that reacted = original moles –moles in average titre => 0.04807/4.807 x 10-2moles - 0.01225 /1.225 x 10-2moles = 0.03582/3.582 x 10 -2 moles (f)Write the equation for the reaction between calcium carbonate and hydrochloric acid. CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l) (g)Calculate the number of moles of calcium carbonate that reacted with hydrochloric acid.
Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 25cm3(volume pipetted) =0.49 /4.9 x 10-1moles (d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used Original moles = Original molarity x pipetted volume => 1000cm3 1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles 1000 (e)How many moles of hydrochloric acid were used to react with calcium carbonate present? Moles that reacted = original moles –moles in average titre => 0.04807/4.807 x 10-2moles - 0.01225 /1.225 x 10-2moles = 0.03582/3.582 x 10 -2 moles (f)Write the equation for the reaction between calcium carbonate and hydrochloric acid. CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l) (g)Calculate the number of moles of calcium carbonate that reacted with hydrochloric acid. From the equation CaCO3(s):2HCl(aq) = 1:2 => Moles CaCO3(s) = 1/2moles HCl = 1/2 x 0.03582/3.582 x 10 -2 moles = 0.01791 /1.791 x 10-2moles
43 43 (h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0) Molar mass CaCO3 = 100g Mass CaCO3 = moles x molar mass => 0.01791 /1.791 x 10-2moles x 100g = 1.791g (i)Determine the % of calcium carbonate present in the mixture % CaCO3 = mass of pure x 100% => 1.791g x 100% = 44.775% Mass of impure 4.0 (d)Sample titration practice 4 (Multiple titration) You are provided with: (i)sodium L containing 5.0g per litre of a dibasic organic acid H2X.2H2O.
Moles that reacted = original moles –moles in average titre => 0.04807/4.807 x 10-2moles - 0.01225 /1.225 x 10-2moles = 0.03582/3.582 x 10 -2 moles (f)Write the equation for the reaction between calcium carbonate and hydrochloric acid. CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l) (g)Calculate the number of moles of calcium carbonate that reacted with hydrochloric acid. From the equation CaCO3(s):2HCl(aq) = 1:2 => Moles CaCO3(s) = 1/2moles HCl = 1/2 x 0.03582/3.582 x 10 -2 moles = 0.01791 /1.791 x 10-2moles
43 43 (h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0) Molar mass CaCO3 = 100g Mass CaCO3 = moles x molar mass => 0.01791 /1.791 x 10-2moles x 100g = 1.791g (i)Determine the % of calcium carbonate present in the mixture % CaCO3 = mass of pure x 100% => 1.791g x 100% = 44.775% Mass of impure 4.0 (d)Sample titration practice 4 (Multiple titration) You are provided with: (i)sodium L containing 5.0g per litre of a dibasic organic acid H2X.2H2O. (ii)solution M which is acidified potassium manganate(VII) (iii)solution N a mixture of sodium ethanedioate and ethanedioic acid (iv)0.1M sodium hydroxide solution P (v)1.0M sulphuric(VI) You are required to: (i)standardize solution M using solution L (ii)use standardized solution M and solution P to determine the % of sodium ethanedioate in the mixture. Procedure 1 Fill the burette with solution M.
From the equation CaCO3(s):2HCl(aq) = 1:2 => Moles CaCO3(s) = 1/2moles HCl = 1/2 x 0.03582/3.582 x 10 -2 moles = 0.01791 /1.791 x 10-2moles
43 43 (h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0) Molar mass CaCO3 = 100g Mass CaCO3 = moles x molar mass => 0.01791 /1.791 x 10-2moles x 100g = 1.791g (i)Determine the % of calcium carbonate present in the mixture % CaCO3 = mass of pure x 100% => 1.791g x 100% = 44.775% Mass of impure 4.0 (d)Sample titration practice 4 (Multiple titration) You are provided with: (i)sodium L containing 5.0g per litre of a dibasic organic acid H2X.2H2O. (ii)solution M which is acidified potassium manganate(VII) (iii)solution N a mixture of sodium ethanedioate and ethanedioic acid (iv)0.1M sodium hydroxide solution P (v)1.0M sulphuric(VI) You are required to: (i)standardize solution M using solution L (ii)use standardized solution M and solution P to determine the % of sodium ethanedioate in the mixture. Procedure 1 Fill the burette with solution M. Pipette 25.0cm3 of solution L into a conical flask. Heat this solution to about 70oC(but not to boil).Titrate the hot solution L with solution M until a permanent pink colour just appears .Shake thoroughly during the titration. Repeat this procedure to complete table 1.
Pipette 25.0cm3 of solution L into a conical flask. Heat this solution to about 70oC(but not to boil).Titrate the hot solution L with solution M until a permanent pink colour just appears .Shake thoroughly during the titration. Repeat this procedure to complete table 1. Sample Table 1 1 2 3 Final burette reading (cm3) 24.0 24.0 24.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.0 24.0 24.0 Sample calculations
44 44 (a)Calculate the average volume of solution L used (1mk) 24.0 + 24.0 + 24.0 = 24.0cm3 3 (b)Given that the concentration of the dibasic acid is 0.05molesdm-3.determine the value of x in the formula H2X.2H2O (H=1.0,O=16.0) Molar mass H2X.2H2O= mass per litre => 5.0g/litre = 100g Moles/litre 0.05molesdm-3 H2X.2H2O =100 X = 100 – ((2 x1) + 2 x (2 x1) + (2 x 16) => 100 – 34 = 66 (c) Calculate the number of moles of the dibasic acid H2X.2H2O. Moles = molarity x pipette volume => 0.5 x 25 = 0.0125/1.25 x10 -2 moles 1000 1000 (d)Given the mole ratio manganate(VII)(MnO4-): acid H2X is 2:5, calculate the number of moles of manganate(VII) (MnO4-) in the average titre. Moles H2X = 2/5 moles of MnO4- => 2/5 x 0.0125/1.25 x10 -2 moles = 0.005/5.0 x 10 -3moles (e)Calculate the concentration of the manganate(VII)(MnO4-) in moles per litre.
Moles = molarity x pipette volume => 0.5 x 25 = 0.0125/1.25 x10 -2 moles 1000 1000 (d)Given the mole ratio manganate(VII)(MnO4-): acid H2X is 2:5, calculate the number of moles of manganate(VII) (MnO4-) in the average titre. Moles H2X = 2/5 moles of MnO4- => 2/5 x 0.0125/1.25 x10 -2 moles = 0.005/5.0 x 10 -3moles (e)Calculate the concentration of the manganate(VII)(MnO4-) in moles per litre. Moles per litre/molarity = moles x 1000 average burette volume =>0.005/5.0 x 10 -3moles x 1000 = 0.2083 molesl-1/M 24.0 Procedure 2 With solution M still in the burette ,pipette 25.0cm3 of solution N into a conical flask. Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 12.5 12.5 12.5
45 45 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 12.5 12.5 12.5 Sample calculations (a)Calculate the average volume of solution L used (1mk) 12.5 + 12.5 + 12.5 =12.5cm3 3 (b)Calculations: (i)How many moles of manganate(VII)ions are contained in the average volume of solution M used?
Moles per litre/molarity = moles x 1000 average burette volume =>0.005/5.0 x 10 -3moles x 1000 = 0.2083 molesl-1/M 24.0 Procedure 2 With solution M still in the burette ,pipette 25.0cm3 of solution N into a conical flask. Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 12.5 12.5 12.5
45 45 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 12.5 12.5 12.5 Sample calculations (a)Calculate the average volume of solution L used (1mk) 12.5 + 12.5 + 12.5 =12.5cm3 3 (b)Calculations: (i)How many moles of manganate(VII)ions are contained in the average volume of solution M used? Moles = molarity of solution M x average burette volume 1000 => 0.2083 molesl-1/ M x 12.5 = 0.0026 / 2.5 x 10-3 moles 1000 (ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation: 2MnO4- (aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M.
Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 12.5 12.5 12.5
45 45 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 12.5 12.5 12.5 Sample calculations (a)Calculate the average volume of solution L used (1mk) 12.5 + 12.5 + 12.5 =12.5cm3 3 (b)Calculations: (i)How many moles of manganate(VII)ions are contained in the average volume of solution M used? Moles = molarity of solution M x average burette volume 1000 => 0.2083 molesl-1/ M x 12.5 = 0.0026 / 2.5 x 10-3 moles 1000 (ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation: 2MnO4- (aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M. From the stoichiometric equation,mole ratio MnO4- (aq): C2O42- (aq) = 2:5 => moles C2O42- = 5/2 moles MnO4- => 5/2 x 0.0026 / 2.5 x 10-3 moles = 0.0065 /6.5 x10-3 moles (iii)Calculate the number of moles of ethanedioate ions contained in 250cm3 solution N.
Moles = molarity of solution M x average burette volume 1000 => 0.2083 molesl-1/ M x 12.5 = 0.0026 / 2.5 x 10-3 moles 1000 (ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation: 2MnO4- (aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M. From the stoichiometric equation,mole ratio MnO4- (aq): C2O42- (aq) = 2:5 => moles C2O42- = 5/2 moles MnO4- => 5/2 x 0.0026 / 2.5 x 10-3 moles = 0.0065 /6.5 x10-3 moles (iii)Calculate the number of moles of ethanedioate ions contained in 250cm3 solution N. 25cm3 pipette volume -> 0.0065 /6.5 x10-3 moles 250cm3 -> 0.0065 /6.5 x10-3 moles x 250 = 0.065 / 6.5 x10-2 moles 25 Procedure 3
46 46 Remove solution M from the burette and rinse it with distilled water. Fill the burette with sodium hydroxide solution P. Pipette 25cm3 of solution N into a conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this solution N with solution P from the burette. Repeat the procedure to complete table 3.
Pipette 25cm3 of solution N into a conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this solution N with solution P from the burette. Repeat the procedure to complete table 3. Sample Table 2 1 2 3 Final burette reading (cm3) 24.9 24.9 24.9 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.9 24.9 24.9 Sample calculations (a)Calculate the average volume of solution L used (1mk) 24.9 + 24.9 + 24.9 = 24.9 cm3 3 (b)Calculations: (i)How many moles of sodium hydroxide solution P were contained in the average volume? Moles = molarity of solution P x average burette volume 1000 => 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles 1000 (ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is: 2NaOH (aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l) Calculate the number of moles of ethanedioic acid that were used in the reaction From the stoichiometric equation,mole ratio NaOH(aq): H2C2O4 (aq) = 2:1 => moles H2C2O4 = 1/2 moles NaOH => 1/2 x 0.00249 / 2.49 x 10-3 moles = 0.001245/1.245 x10-3 moles. (iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N?
Sample Table 2 1 2 3 Final burette reading (cm3) 24.9 24.9 24.9 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.9 24.9 24.9 Sample calculations (a)Calculate the average volume of solution L used (1mk) 24.9 + 24.9 + 24.9 = 24.9 cm3 3 (b)Calculations: (i)How many moles of sodium hydroxide solution P were contained in the average volume? Moles = molarity of solution P x average burette volume 1000 => 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles 1000 (ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is: 2NaOH (aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l) Calculate the number of moles of ethanedioic acid that were used in the reaction From the stoichiometric equation,mole ratio NaOH(aq): H2C2O4 (aq) = 2:1 => moles H2C2O4 = 1/2 moles NaOH => 1/2 x 0.00249 / 2.49 x 10-3 moles = 0.001245/1.245 x10-3 moles. (iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N? 47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) Molar mass H2C2O4 = 90.0g Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4 =>0.01245/1.245 x10-2 moles x 90.0 = 1.1205g % by mass of sodium ethanedioate =(Mass of mixture - mass of H2C2O4) x 100% Mass of mixture => 2.0 - 1.1205 g = 43.975% 2.0 Note (i) L is 0.05M Oxalic acid (ii) M is 0.01M KMnO4 (iii) N is 0.03M oxalic acid(without sodium oxalate) Practice example 5.(Determining equation for a reaction) You are provided with -0.1M hydrochloric acid solution A -0.5M sodium hydroxide solution B You are to determine the equation for thereaction between solution A and B Procedure Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1 Table 1(Sample results) Titration 1 2 3 Final volume(cm3) 12.5 25.0 37.5 Initial volume(cm3) 0.0 12.5 25.0 Volume of solution A used(cm3) 12.5 12.5 12.5
48 48 Sample questions Calculate the average volume of solution A used.
Moles = molarity of solution P x average burette volume 1000 => 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles 1000 (ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is: 2NaOH (aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l) Calculate the number of moles of ethanedioic acid that were used in the reaction From the stoichiometric equation,mole ratio NaOH(aq): H2C2O4 (aq) = 2:1 => moles H2C2O4 = 1/2 moles NaOH => 1/2 x 0.00249 / 2.49 x 10-3 moles = 0.001245/1.245 x10-3 moles. (iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N? 47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) Molar mass H2C2O4 = 90.0g Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4 =>0.01245/1.245 x10-2 moles x 90.0 = 1.1205g % by mass of sodium ethanedioate =(Mass of mixture - mass of H2C2O4) x 100% Mass of mixture => 2.0 - 1.1205 g = 43.975% 2.0 Note (i) L is 0.05M Oxalic acid (ii) M is 0.01M KMnO4 (iii) N is 0.03M oxalic acid(without sodium oxalate) Practice example 5.(Determining equation for a reaction) You are provided with -0.1M hydrochloric acid solution A -0.5M sodium hydroxide solution B You are to determine the equation for thereaction between solution A and B Procedure Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1 Table 1(Sample results) Titration 1 2 3 Final volume(cm3) 12.5 25.0 37.5 Initial volume(cm3) 0.0 12.5 25.0 Volume of solution A used(cm3) 12.5 12.5 12.5
48 48 Sample questions Calculate the average volume of solution A used. 12.5+12.5+12.5 = 12.5cm3 3 Theoretical Practice examples 1.
(iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N? 47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) Molar mass H2C2O4 = 90.0g Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4 =>0.01245/1.245 x10-2 moles x 90.0 = 1.1205g % by mass of sodium ethanedioate =(Mass of mixture - mass of H2C2O4) x 100% Mass of mixture => 2.0 - 1.1205 g = 43.975% 2.0 Note (i) L is 0.05M Oxalic acid (ii) M is 0.01M KMnO4 (iii) N is 0.03M oxalic acid(without sodium oxalate) Practice example 5.(Determining equation for a reaction) You are provided with -0.1M hydrochloric acid solution A -0.5M sodium hydroxide solution B You are to determine the equation for thereaction between solution A and B Procedure Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1 Table 1(Sample results) Titration 1 2 3 Final volume(cm3) 12.5 25.0 37.5 Initial volume(cm3) 0.0 12.5 25.0 Volume of solution A used(cm3) 12.5 12.5 12.5
48 48 Sample questions Calculate the average volume of solution A used. 12.5+12.5+12.5 = 12.5cm3 3 Theoretical Practice examples 1. 1.0g of dibasic acid HOOC(CH2)xCOOH was dissolved in 250cm3 solution.
47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) Molar mass H2C2O4 = 90.0g Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4 =>0.01245/1.245 x10-2 moles x 90.0 = 1.1205g % by mass of sodium ethanedioate =(Mass of mixture - mass of H2C2O4) x 100% Mass of mixture => 2.0 - 1.1205 g = 43.975% 2.0 Note (i) L is 0.05M Oxalic acid (ii) M is 0.01M KMnO4 (iii) N is 0.03M oxalic acid(without sodium oxalate) Practice example 5.(Determining equation for a reaction) You are provided with -0.1M hydrochloric acid solution A -0.5M sodium hydroxide solution B You are to determine the equation for thereaction between solution A and B Procedure Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1 Table 1(Sample results) Titration 1 2 3 Final volume(cm3) 12.5 25.0 37.5 Initial volume(cm3) 0.0 12.5 25.0 Volume of solution A used(cm3) 12.5 12.5 12.5
48 48 Sample questions Calculate the average volume of solution A used. 12.5+12.5+12.5 = 12.5cm3 3 Theoretical Practice examples 1. 1.0g of dibasic acid HOOC(CH2)xCOOH was dissolved in 250cm3 solution. 25.0 cm3 of this solution reacted with 30.0cm3 of 0.06M sodium hydroxide solution.
12.5+12.5+12.5 = 12.5cm3 3 Theoretical Practice examples 1. 1.0g of dibasic acid HOOC(CH2)xCOOH was dissolved in 250cm3 solution. 25.0 cm3 of this solution reacted with 30.0cm3 of 0.06M sodium hydroxide solution. Calculate the value of x in HOOC(CH2)xCOOH.
46.0g of a metal carbonate MCO3 was dissolved 160cm3 of 0.1M excess hydrochloric acid and the resultant solution diluted to one litre.25.0cm3 of this solution required 20.0cm3 of 0.1M sodium hydroxide solution for complete neutralization. Calculate the atomic mass of ‘M’ Equation Chemical equation NaOH(aq) + HCl(aq) -> KCl(aq) + H2O(l) Moles of NaOH = Molarity x volume=> 0.1 x20 = 0.002 moles 1000 1000 Mole ratio HCl; NaOH = 1:1 Excess moles of HCl = 0.002 moles 25cm3 -> 0.002 moles 1000cm3 -> 1000 x 0.002 = 0.08moles 25cm3 Original moles of HCl = Molarity x volume => 1M x 1litre = 1.0 moles Moles of HCl reacted with MCO3 = 1.0 - 0.08 moles = 0.92moles Chemical equation MCO3(s) + 2HCl(aq) -> MCl2 (aq) + CO2 (g) + H2O(l) Mole ratio MCO3(s) : HCl(aq) =1:2 Moles of MCO3 = 0.92moles => 0.46moles 2 Molar mass of MCO3= mass => 46g = 100 g moles 0.46moles M= MCO3 - CO3 =>100g – (12+ 16 x3 = 60) = 40
52 52 6. 25.0cm3 of a mixture of Fe2+ and Fe3+ ions in an aqueous salt was acidified with sulphuric(VI)acid then titrated against potassium manganate(VI).The salt required 15cm3 ofe0.02M potassium manganate(VI) for complete reaction. A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against the same concentration of potassium manganate(VI).19.0cm3 of potassium manganate(VI)solution was used for complete reaction.