1. Moles and Solutions 1.1.7 2008 SPECIFICATIONS

2. 1.1.7 Moles and solutions Calculate the amount of substance in moles using solution volume and concentration. Describe a solution’s concentration using the terms concentrated and dilute.

3. Procedure Fill your burette with : 0.100 mol dm –3 sulfuric acid, H 2 SO 4 : B2. Pipette 25.0 cm 3 of sodium hydroxide solution : A2 into a conical flask. Add a few drops of phenolphthalein indicator to the conical flask. Titrate the contents of the conical flask with sulfuric acid, H 2 SO 4 : B2. Repeat the titration until you have concordant results (within 0.10 cm 3 ). Record all titration results in a table showing initial and final burette readings. Equipment/materials A2: sodium hydroxide solution NaOH B2: 0.100 mol dm –3 sulfuric acid, H 2 SO 4 Phenolphthalein indicator 25.0 cm 3 pipette 50.0 cm 3 burette and stand Funnel Conical flask Wash bottle White tile – use white paper –Why ?

4. Data 2NaOH + H 2 SO 4 → Na 2 SO 4 + 2H 2 O H = 1.0, O = 16.0, Na = 23.0 Analysis of results Indicate which titres you use in your average. Calculate the average of your concordant results. Calculate the amount, in mol, of H 2 SO 4 reacting with the NaOH. Calculate the amount, in mol, of NaOH in the pipette. Calculate the amount, in mol, of NaOH in 1.00 dm 3 of A2. Calculate the mass of NaOH in 1.00 dm 3 of A2.

5. CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION UNITS concentrationmol dm -3 volumedm 3 BUT IF... concentrationmol dm -3 volumecm 3 THE MOLARITY MOLES CONC x VOLUME COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED MOLES = CONCENTRATION x VOLUME MOLES = CONCENTRATION (mol dm -3 ) x VOLUME (dm 3 ) MOLES = CONCENTRATION (mol dm -3 ) x VOLUME (cm 3 ) 1000 MOLES = CONCENTRATION (mol dm -3 ) x VOLUME (cm 3 ) 1000

7. From the examiner…  It is important to present your results (with units) in a table showing initial and final burette readings, using appropriate number of decimal places.  Show clearly how the average titre is obtained.  Explain each line of your calculation as shown in the Analysis of Results section above.  Consider how many significant figures to use within the calculation.  Consider how many significant figures to use in your final answer. Questions 1.Sulfuric acid is known as a dibasic acid. Explain what this means. 2.Explain why a titration reading of 23.58 cm 3 would generally be unacceptable. 3.Under what conditions would the acid salt, NaHSO 4, be produced? 4.Why is phenolphthalein such a good indicator to use in this case? 5.Explain why more than three significant figures for the final answer would not be appropriate.

8. All values in cm 3 rough1st2nd3rd4th initial0.0020.200.000.05 final22.1541.3022.2021.20 titre22.1521.1022.2021.15 Mean value cm 3 Mean value to 1 decimal place Concordancy is ± 0.10 cm 3 Mean value ( that the mean of concordant values )

9. All values in cm 3 rough1st2nd3rd4th initial0.0020.200.000.05 final22.1541.3022.2021.20 titre22.1521.1022.2021.15 Mean value cm 3 Mean value to 1 decimal place Concordancy is ± 0.10 cm 3 Mean value ( that the mean of concordant values )

10. All values in cm 3 rough1st2nd3rd4th initial0.0020.200.000.05 final22.1541.3022.2021.20 titre22.1521.1022.2021.15 Mean value 21.1 cm 3 Mean value to 1 decimal place Concordancy is ± 0.10 cm 3 Mean value ( that the mean of concordant values )

11. All values in cm 3 rough1st2nd3rd4th initial0.0020.200.000.050.00 final22.1541.3022.2021.20 titre22.1521.1022.2021.1521.20 Mean value cm 3 Mean value to 1 decimal place Concordancy is ± 0.10 cm 3 Mean value ( that the mean of concordant values )

12. All values in cm 3 rough1st2nd3rd4th initial0.0020.200.000.050.00 final22.1541.3022.2021.20 titre22.1521.1022.2021.1521.20 Mean value 21.2 cm 3 Mean value to 1 decimal place Concordancy is ± 0.10 cm 3 Mean value ( that the mean of concordant values )

13. CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION UNITS concentrationmol dm -3 volumedm 3 BUT IF... concentrationmol dm -3 volumecm 3 THE MOLARITY MOLES CONC x VOLUME COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED MOLES = CONCENTRATION x VOLUME MOLES = CONCENTRATION (mol dm -3 ) x VOLUME (dm 3 ) MOLES = CONCENTRATION (mol dm -3 ) x VOLUME (cm 3 ) 1000 MOLES = CONCENTRATION (mol dm -3 ) x VOLUME (cm 3 ) 1000

14. 1 Calculate the moles of sodium hydroxide in 25cm 3 of 2M NaOH MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME

15. 1 Calculate the moles of sodium hydroxide in 25cm 3 of 2M NaOH moles=conc x volume in cm 3 1000 = 2 mol dm -3 x 25cm 3 = 0.05 moles 1000 MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME

16. 1 Calculate the moles of sodium hydroxide in 25cm 3 of 2M NaOH moles=conc x volume in cm 3 1000 = 2 mol dm -3 x 25cm 3 = 0.05 moles 1000 2 What volume in cm 3 of 0.1M H 2 SO 4 contains 0.002 moles ? MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME

17. 1 Calculate the moles of sodium hydroxide in 25cm 3 of 2M NaOH moles=conc x volume in cm 3 1000 = 2 mol dm -3 x 25cm 3 = 0.05 moles 1000 2 What volume in cm 3 of 0.1M H 2 SO 4 contains 0.002 moles ? volume = 1000 x moles (re-arrangement of above) (in cm 3 ) conc = 1000 x 0.002 = 20 cm 3 0.1 mol dm -3 MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME

18. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm 3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out?

19. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm 3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? = 0.100 mol dm -3 How many moles will be in 1 dm 3 ? = 0.100 mol How many moles will be in 250cm 3 ? = 0.100/4 = 0.025 mol

20. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm 3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? = 0.100 mol dm -3 How many moles will be in 1 dm 3 ? = 0.100 mol How many moles will be in 250cm 3 ? = 0.100/4 = 0.025 mol What is the formula of anhydrous sodium carbonate?= Na 2 CO 3 What is the relative formula mass?= 106 What is the molar mass?= 106g mol -1

21. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm 3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? = 0.100 mol dm -3 How many moles will be in 1 dm 3 ? = 0.100 mol How many moles will be in 250cm 3 ? = 0.100/4 = 0.025 mol What is the formula of anhydrous sodium carbonate?= Na 2 CO 3 What is the relative formula mass?= 106 What is the molar mass?= 106g mol -1 What mass of Na 2 CO 3 is in 0.025 moles = 0.025 x 106 = 2.650g of Na 2 CO 3 ? (mass = moles x molar mass)

22. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm 3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? = 0.100 mol dm -3 How many moles will be in 1 dm 3 ? = 0.100 mol How many moles will be in 250cm 3 ? = 0.100/4 = 0.025 mol What is the formula of anhydrous sodium carbonate?= Na 2 CO 3 What is the relative formula mass?= 106 What is the molar mass?= 106g mol -1 What mass of Na 2 CO 3 is in 0.025 moles = 0.025 x 106 = 2.650g of Na 2 CO 3 ? (mass = moles x molar mass) ANS. The chemist will have to weigh out 2.650g, dissolve it in water and then make the solution up to 250cm 3 in a graduated flask.

23. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm 3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? = 0.100 mol dm -3 How many moles will be in 1 dm 3 ? = 0.100 mol How many moles will be in 250cm 3 ? = 0.100/4 = 0.025 mol What is the formula of anhydrous sodium carbonate?= Na 2 CO 3 What is the relative formula mass?= 106 What is the molar mass?= 106g mol -1 What mass of Na 2 CO 3 is in 0.025 moles = 0.025 x 106 = 2.650g of Na 2 CO 3 ? (mass = moles x molar mass) ANS. The chemist will have to weigh out 2.650g, dissolve it in water and then make the solution up to 250cm 3 in a graduated flask.

24. odds or evens... Up to 30 !! Now do all 31, 32, 33, 34, 35

25. odds or evens...

27. odds or evens... Up to 30 !! Now do all 31, 32, 33, 34, 35

29. 1.1.7 Tasks Write out these Key definitions The concentration of a solution is.. A standard solution is... Examiner tip is.... Worksheets Questions : 1, 2, 3 Pages 16-17

30. Volumetric solutions are made by dissolving a known amount of solute in a solvent (usually water) and then adding enough solvent to get the correct volume of solution. WRONG Dissolve 1g of solute in 250cm 3 of de-ionised water SOLUTIONS ‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’ 250cm 3 1g

31. Volumetric solutions are made by dissolving a known amount of solute in a solvent (usually water) and then adding enough solvent to get the correct volume of solution. WRONG Dissolve 1g of solute in 250cm 3 of de-ionised water RIGHT Dissolve 1g of solute in water and then add enough water to make 250cm 3 of solution SOLUTIONS ‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’ 250cm 3 1g 250cm 3 1g WATER

32. STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na 2 CO 3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm 3 graduated flask and made up to the mark with de-ionised (or distilled) water.

33. STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na 2 CO 3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm 3 graduated flask and made up to the mark with de-ionised (or distilled) water. What is the concentration of the solution in mol dm -3 ?

34. STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na 2 CO 3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm 3 graduated flask and made up to the mark with de-ionised (or distilled) water. What is the concentration of the solution in mol dm -3 ? mass of Na 2 CO 3 in a 250cm 3 solution = 4.240g molar mass of Na 2 CO 3 = 106g mol -1 no. of moles in a 250cm 3 solution= 4.240g / 106g mol -1 = 0.04 mol

35. STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na 2 CO 3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm 3 graduated flask and made up to the mark with de-ionised (or distilled) water. What is the concentration of the solution in mol dm -3 ? mass of Na 2 CO 3 in a 250cm 3 solution = 4.240g molar mass of Na 2 CO 3 = 106g mol -1 no. of moles in a 250cm 3 solution= 4.240g / 106g mol -1 = 0.04 mol Concentration is normally expressed as moles per dm 3 of solution Therefore, as it is in 250cm 3, the value is scaled up by a factor of 4 – that’s 1000/250

36. STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na 2 CO 3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm 3 graduated flask and made up to the mark with de-ionised (or distilled) water. What is the concentration of the solution in mol dm -3 ? mass of Na 2 CO 3 in a 250cm 3 solution = 4.240g molar mass of Na 2 CO 3 = 106g mol -1 no. of moles in a 250cm 3 solution= 4.240g / 106g mol -1 = 0.04 mol Concentration is normally expressed as moles per dm 3 of solution Therefore, as it is in 250cm 3, the value is scaled up by a factor of 4 no. of moles in 1000cm 3 (1dm 3 ) = 4 x 0.04 = 0.16 mol ANS. 0.16 mol dm -3

37. 250cm 3 25cm 3 250cm 3 CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask= 0.200 mol dm -3 volume pipetted out into the conical flask= 25.00 cm 3 THE MOLARITY

38. The original solution has a concentration of 0.200 mol dm -3 250cm 3 25cm 3 250cm 3 CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask= 0.200 mol dm -3 volume pipetted out into the conical flask= 25.00 cm 3 THE MOLE

39. The original solution has a concentration of 0.200 mol dm -3 This means that there are 0.200 mols of solute in every 1 dm 3 (1000 cm 3 ) of solution 250cm 3 25cm 3 250cm 3 CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask= 0.200 mol dm -3 volume pipetted out into the conical flask= 25.00 cm 3 THE MOLE

40. The original solution has a concentration of 0.200 mol dm -3 This means that there are 0.200 mols of solute in every 1 dm 3 (1000 cm 3 ) of solution Take out 25.00 cm 3 and you will take a fraction 25/1000 ( 1/40 ) of the number of moles 250cm 3 25cm 3 250cm 3 CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask= 0.200 mol dm -3 volume pipetted out into the conical flask= 25.00 cm 3 THE MOLE

41. The original solution has a concentration of 0.200 mol dm -3 This means that there are 0.200 mols of solute in every 1 dm 3 (1000 cm 3 ) of solution Take out 25.00 cm 3 and you will take a fraction 25/1000 or 1/40 of the number of moles moles in 1dm 3 (1000cm 3 )= 0.200 moles in 1cm 3 = 0.200/1000 moles in 25cm 3 = 25/1000 x 0.200 = 5.0 x 10 -3 mol 250cm 3 25cm 3 250cm 3 CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask= 0.200 mol dm -3 volume pipetted out into the conical flask= 25.00 cm 3 THE MOLE

42. The original solution has a concentration of 0.200 mol dm -3 This means that there are 0.200 mols of solute in every 1 dm 3 (1000 cm 3 ) of solution Take out 25.00 cm 3 and you will take a fraction 25/1000 or 1/40 of the number of moles moles in 1dm 3 (1000cm 3 )= 0.200 moles in 1cm 3 = 0.200/1000 moles in 25cm 3 = 25 x 0.200/1000 = 5.0 x 10 -3 mol 250cm 3 25cm 3 250cm 3 CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask= 0.200 mol dm -3 volume pipetted out into the conical flask= 25.00 cm 3 THE MOLE

43. Titrations1.1.13

44. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm 3 is neutralised by 25cm 3 of hydrochloric acid of concentration 0.100 mol dm -3.

45. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm 3 is neutralised by 25cm 3 of hydrochloric acid of concentration 0.100 mol dm -3. 1. Write out a BALANCED equationNaOH + HC l ——> NaC l + H 2 O

46. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm 3 is neutralised by 25cm 3 of hydrochloric acid of concentration 0.100 mol dm -3. 1. Write out a BALANCED equationNaOH + HC l ——> NaC l + H 2 O 2. Get a molar relationship between the reactants moles of NaOH = moles of HC l you need ONE NaOH for every ONE HC l

47. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm 3 is neutralised by 25cm 3 of hydrochloric acid of concentration 0.100 mol dm -3. 1. Write out a BALANCED equationNaOH + HC l ——> NaC l + H 2 O 2. Get a molar relationship between the reactants moles of NaOH = moles of HC l you need ONE NaOH for every ONE HC l 3. Calculate the number of moles of each substanceHC l 0.100 x 25/1000(i) M is the concentration in mol dm -3 NaOH M x 20/1000(ii)

48. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm 3 is neutralised by 25cm 3 of hydrochloric acid of concentration 0.100 mol dm -3. 1. Write out a BALANCED equationNaOH + HC l ——> NaC l + H 2 O 2. Get a molar relationship between the reactants moles of NaOH = moles of HC l you need ONE NaOH for every ONE HC l 3. Calculate the number of moles of each substanceHC l 0.100 x 25/1000(i) M is the concentration in mol dm -3 NaOH M x 20/1000(ii) 4. Look at the molar relationship and insert (i) and (ii) moles of NaOH = moles of HC l M x 20/1000 = 0.100 x 25/1000

49. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm 3 is neutralised by 25cm 3 of hydrochloric acid of concentration 0.100 mol dm -3. 1. Write out a BALANCED equationNaOH + HC l ——> NaC l + H 2 O 2. Get a molar relationship between the reactants moles of NaOH = moles of HC l you need ONE NaOH for every ONE HC l 3. Calculate the number of moles of each substanceHC l 0.100 x 25/1000(i) M is the concentration in mol dm -3 NaOH M x 20/1000(ii) 4. Look at the molar relationship and insert (i) and (ii) moles of NaOH = moles of HC l M x 20/1000 = 0.100 x 25/1000 5. Cancel the 1000’s M x 20 = 0.100 x 25 re-arrange the numbers to obtain M M = 0.100 x 25 20

50. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm 3 is neutralised by 25cm 3 of hydrochloric acid of concentration 0.100 mol dm -3. 1. Write out a BALANCED equationNaOH + HC l ——> NaC l + H 2 O 2. Get a molar relationship between the reactants moles of NaOH = moles of HC l you need ONE NaOH for every ONE HC l 3. Calculate the number of moles of each substanceHC l 0.100 x 25/1000(i) M is the concentration in mol dm -3 NaOH M x 20/1000(ii) 4. Look at the molar relationship and insert (i) and (ii) moles of NaOH = moles of HC l M x 20/1000 = 0.100 x 25/1000 5. Cancel the 1000’s M x 20 = 0.100 x 25 re-arrange the numbers to obtain M M = 0.100 x 25 20 6. Calculate the concentration of the NaOH= 0.125 mol dm -3

51. Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake. 2NaOH + H 2 SO 4 ——> Na 2 SO 4 + 2H 2 O you need 2 moles of NaOH to react with every 1 mole of H 2 SO 4 i.e moles of NaOH = 2 x moles of H 2 SO 4 or moles of H 2 SO 4 = moles of NaOH 2 VOLUMETRIC CALCULATIONS REMEMBER... IT IS NOT A MATHEMATICAL EQUATION 2 moles of NaOH DO NOT EQUAL 1 mole of H 2 SO 4 REMEMBER... IT IS NOT A MATHEMATICAL EQUATION 2 moles of NaOH DO NOT EQUAL 1 mole of H 2 SO 4 More examples follow

52. Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake. 2NaOH + H 2 SO 4 ——> Na 2 SO 4 + 2H 2 O you need 2 moles of NaOH to react with every 1 mole of H 2 SO 4 i.e moles of NaOH = 2 x moles of H 2 SO 4 or moles of H 2 SO 4 = moles of NaOH 2 VOLUMETRIC CALCULATIONS 2HC l + Na 2 CO 3 ——> 2NaC l + CO 2 + H 2 O you need 2moles of HC l to react with every 1 mole of Na 2 CO 3 i.e moles of HC l = 2 x moles of Na 2 CO 3 or moles of Na 2 CO 3 = moles of HC l 2

53. VOLUMETRIC CALCULATIONS Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake. MnO 4 ¯ + 8H + + 5Fe 2+ ——>Mn 2+ + 4H 2 O + 5Fe 3+ you need 5 moles of Fe 2+ to react with every 1 mole of MnO 4 ¯ i.e moles of Fe 2+ = 5 x moles of MnO 4 ¯ ormoles of MnO 4 ¯ = moles of Fe 2+ 5

54. VOLUMETRIC CALCULATIONS 2MnO 4 ¯ + 5H 2 O 2 + 6H + ——> 2Mn 2+ + 5O 2 + 8H 2 O you need 5 moles of H 2 O 2 to react with every 2 moles of MnO 4 ¯ i.e moles of H 2 O 2 = 5 x moles of MnO 4 ¯ 2 ormoles of MnO 4 ¯ = 2 x moles of H 2 O 2 5 Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake. MnO 4 ¯ + 8H + + 5Fe 2+ ——>Mn 2+ + 4H 2 O + 5Fe 3+ you need 5 moles of Fe 2+ to react with every 1 mole of MnO 4 ¯ i.e moles of Fe 2+ = 5 x moles of MnO 4 ¯ ormoles of MnO 4 ¯ = moles of Fe 2+ 5

55. Calculate the volume of sodium hydroxide (concentration 0.100 mol dm -3 ) required to neutralise 20cm 3 of sulphuric acid of concentration 0.120 mol dm -3. 2NaOH + H 2 SO 4 ——> Na 2 SO 4 + 2H 2 O you need 2 moles of NaOH to react with every 1 mole of H 2 SO 4 therefore moles of NaOH = 2 x moles of H 2 SO 4 moles of H 2 SO 4 = 0.120 x 20/1000(i) moles of NaOH= 0.100 x V/1000(ii) where V is the volume of alkali in cm 3 VOLUMETRIC CALCULATIONS

56. Calculate the volume of sodium hydroxide (concentration 0.100 mol dm -3 ) required to neutralise 20cm 3 of sulphuric acid of concentration 0.120 mol dm -3. 2NaOH + H 2 SO 4 ——> Na 2 SO 4 + 2H 2 O you need 2 moles of NaOH to react with every 1 mole of H 2 SO 4 therefore moles of NaOH = 2 x moles of H 2 SO 4 moles of H 2 SO 4 = 0.120 x 20/1000(i) moles of NaOH= 0.100 x V/1000(ii) where V is the volume of alkali in cm 3 substitute numbersmoles of NaOH = 2 x moles of H 2 SO 4 0.100 x V/1000 = 2 x 0.120 x 20/1000 cancel the 1000’s 0.100 x V = 2 x 0.120 x 20 re-arrange Volume of NaOH (V) = 2 x 0.120 x 20 = 48.00 cm 3 0.100 VOLUMETRIC CALCULATIONS

61. Exercise 1.1.13 – Titrations Simple volumetric calculations In this series of calculations you should start by writing the equation for the reaction taking place then generate the molarity/volume ratio. In some cases you will need to calculate the molarity of the solutions before you start the main part of the question.

62. 1 25.0 cm 3 of sodium hydroxide ( NaOH ) reacts with 21.0 cm 3 of 0.2 mol dm –3 hydrochloric acid (HCl) 2 25.0 cm 3 of sodium hydroxide ( NaOH ) reacts with 17.0 cm 3 of 0.1 mol dm –3 sulphuric acid (H2SO4) 3 20.0 cm 3 of hydrochloric acid (HCl) reacts with 23.6 cm 3 of 0.1 mol dm–3 NaOH 4 20.0 cm 3 of hydrochloric acid (HCl) reacts with 20.0 cm 3 of a solution of NaOH containing 40 g dm –3 of NaOH 5 25.0 cm3 of nitric acid ( HNO3)reacts with 15.0 cm 3 of a solution of 0.2 mol dm –3 Ammonium hydroxide (NH4OH) 6 25.0 cm 3 of a solution of barium chloride ( BaCl 2 ) reacts with 20.0 cm 3 of a solution of 0.05 mol dm –3 sulphuric acid (H2SO4) 7 25.0 cm 3 of a solution of sodium chloride (NaCl) reacts with 10.0 cm 3 of a 0.02 mol dm –3 silver nitrate ( AgNO 3 ) 8 10.0 cm 3 of a solution of Aluminium chloride (AlCl 3 ) reacts with 30.0 cm 3 of 0.01 mol dm -3 silver nitrate ( AgNO 3 ) 9 25.0 cm 3 of H x A reacts with 25.0 cm 3 of 0.2 mol dm –3 NaOH to give Na2A 10 25.0 cm 3 of H 3 PO 4 reacts with 100.0 cm 3 of 0.1 mol dm –3 NaOH to give NaH 2 PO 4

63. 1 25.0 cm 3 of sodium hydroxide ( NaOH ) reacts with 21.0 cm 3 of 0.2 mol dm –3 hydrochloric acid (HCl) 2 25.0 cm 3 of sodium hydroxide ( NaOH ) reacts with 17.0 cm 3 of 0.1 mol dm –3 sulphuric acid (H 2 SO 4 ) 3 20.0 cm 3 of hydrochloric acid (HCl) reacts with 23.6 cm 3 of 0.1 mol dm–3 NaOH 4 20.0 cm 3 of hydrochloric acid (HCl) reacts with 20.0 cm 3 of a solution of NaOH containing 40 g dm –3 of NaOH 0.136 mol dm –3 0.118 mol dm –3 1.0 mol dm –3 0.168 mol dm–3

64. 5 25.0 cm3 of nitric acid ( HNO 3 )reacts with 15.0 cm 3 of a solution of 0.2 mol dm –3 Ammonium hydroxide (NH 4 OH) 6 25.0 cm 3 of a solution of barium chloride ( BaCl 2 ) reacts with 20.0 cm 3 of a solution of 0.05 mol dm –3 sulphuric acid (H2SO4) 7 25.0 cm 3 of a solution of sodium chloride (NaCl) reacts with 10.0 cm 3 of a 0.02 mol dm –3 silver nitrate ( AgNO 3 ) 0.12 mol

65. 8 10.0 cm 3 of a solution of Aluminium chloride (AlCl 3 ) reacts with 30.0 cm 3 of 0.01 mol dm -3 silver nitrate ( AgNO 3 ) 9 25.0 cm 3 of H x A reacts with 25.0 cm 3 of 0.2 mol dm –3 NaOH to give Na2A 10 25.0 cm 3 of H 3 PO 4 reacts with 100.0 cm 3 of 0.1 mol dm –3 NaOH to give NaH 2 PO 4

72. +

81. A5 is the LIMEWATER C5 is the solution in the Volumetric Flask Its the diluted HYDROCHLORIC ACID (B5)

82. All values in cm 3 rough1st2nd3rd4th initial0.0020.200.000.05 final22.1541.3022.2021.20 titre22.1521.1022.2021.15 Mean value cm 3 Mean value to 1 decimal place Concordancy is ± 0.10 cm 3 Mean value ( that the mean of concordant values )

83.  Calculate the amount, in mol, of HCl in the volumetric flask.  Calculate the amount, in mol, of HCl used in titration.  Calculate the amount, in mol, of Ca(OH) 2 reacting.  Calculate the amount, in mol, of Ca(OH) 2 in 1 dm 3.  Calculate mass of Ca(OH) 2 in 1 dm 3. But first of all write the equation for the reaction 2HCl(aq) + Ca(OH)2 (aq)  CaCl2 (aq) + 2H 2 O(l)

84. 3.50 cm 3 HCl 2.00 Molar Moles = 3.50/1000 x 2.00 or 0.0035 x 2.00 = 0.007 moles  Calculate the amount, in mol, of HCl in the volumetric flask. Use a graduated pipette to add 3.50 cm 3 of 2.00 mol dm –3 hydrochloric acid to a 250 cm 3 volumetric flask. So whats the concentration of this dilute HCl solution ? Molarity = moles / volume = 0.007 /0.250 ( that’s 250/1000 ) = 0.028 moles /dm3

85. Molarity of HCl is 0.028 moles /dm3 Lets assume its was 40.0 cm 3 HCl needed to neutralise  Calculate the amount, in mol, of HCl in the volumetric flask.  Calculate the amount, in mol, of HCl used in titration.  Calculate the amount, in mol, of Ca(OH) 2 reacting 2HCl(aq) + Ca(OH)2 (aq)  CaCl2+ 2H 2 O(l) Moles of HCl Moles = 0.028 x 40.0/1000 = 0.00112 moles Moles of Ca(OH)2 = 0.00112 / 2 = 0.00056moles in 25 cm3 of lime water

86. Molar mass of Ca(OH) 2 40.1 16.0 x 2 1.0 x 2 74.1 grms  Calculate the amount, in mol, of HCl in the volumetric flask.  Calculate the amount, in mol, of HCl used in titration.  Calculate the amount, in mol, of Ca(OH) 2 in 1 dm 3.  Calculate mass of Ca(OH) 2 in 1 dm 3 to 3 significant figures. Moles of Ca(OH)2 in 25 cm = 0.00056moles So moles in 1000 cm3 must be = 0.00056moles x 40 = 0.0224 moles mass of Ca(OH) 2 = 74.1 x 0.0224 = 1.65984 1.66 grms

87. Question 23 What volume of carbon dioxide will be produced if 100 cm 3 of 0.2 mol dm –3 HNO 3 is added to excess sodium carbonate solution? Na 2 CO 3 + 2HNO 3  2NaNO 3 + CO 2 + H 2 O

89. Volume of CO 2 ? Moles ? 100.0 cm 3 HNO 3 0.20 Molar Moles = 0.100 x 0.20 = 0.020 Ratio : CO 2 : HNO 3 1 : 2 Moles CO 2 = 0.020/2 = 0.01 mole Volume = 0.01 x 24.0 dm 3 = 0.24 dm 3 Question 23 What volume of carbon dioxide will be produced if 100 cm3 of 0.2 mol dm –3 HNO 3 is added to excess sodium carbonate solution? Na 2 CO 3 + 2HNO 3  2NaNO 3 + CO 2 + H 2 O