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Moles and Solutions 1.1.7

Moles and Solutions 1.1.7. 2008 SPECIFICATIONS. 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. Procedure

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Moles and Solutions 1.1.7

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  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 : 0.100moldm–3 sulfuric acid, H2SO4 : B2. • Pipette 25.0cm3 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, H2SO4 : B2. • Repeat the titration until you have concordant results (within 0.10cm3). • Record all titration results in a table showing initial and final burette readings. Equipment/materials A2: sodium hydroxide solution NaOH B2: 0.100moldm–3 sulfuric acid, H2SO4 Phenolphthalein indicator 25.0cm3 pipette 50.0cm3 burette and stand Funnel Conical flask Wash bottle White tile – use white paper –Why ?

  4. Data 2NaOH + H2SO4→ Na2SO4 + 2H2O 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 H2SO4 reacting with the NaOH. • Calculate the amount, in mol, of NaOH in the pipette. • Calculate the amount, in mol, of NaOHin 1.00dm3 of A2. • Calculate the mass of NaOH in 1.00dm3 of A2.

  5. MOLES CONC x VOLUME MOLES = CONCENTRATION (mol dm-3) x VOLUME (cm3) 1000 COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED THE MOLARITY CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION UNITSconcentration mol dm-3 volume dm3 BUT IF... concentration mol dm-3 volume cm3 MOLES = CONCENTRATION x VOLUME MOLES = CONCENTRATION (mol dm-3) x VOLUME (dm3)

  6. 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 Sulfuric acid is known as a dibasic acid. Explain what this means. Explain why a titration reading of 23.58cm3 would generally be unacceptable. Under what conditions would the acid salt, NaHSO4, be produced? Why is phenolphthalein such a good indicator to use in this case? Explain why more than three significant figures for the final answer would not be appropriate.

  7. Mean value to 1 decimal place Concordancy is ± 0.10 cm3 Mean value ( that the mean of concordant values )

  8. Mean value to 1 decimal place Concordancy is ± 0.10 cm3 Mean value ( that the mean of concordant values )

  9. Mean value to 1 decimal place Concordancy is ± 0.10 cm3 Mean value ( that the mean of concordant values )

  10. Mean value to 1 decimal place Concordancy is ± 0.10 cm3 Mean value ( that the mean of concordant values )

  11. Mean value to 1 decimal place Concordancy is ± 0.10 cm3 Mean value ( that the mean of concordant values )

  12. MOLES CONC x VOLUME MOLES = CONCENTRATION (mol dm-3) x VOLUME (cm3) 1000 COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED THE MOLARITY CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION UNITSconcentration mol dm-3 volume dm3 BUT IF... concentration mol dm-3 volume cm3 MOLES = CONCENTRATION x VOLUME MOLES = CONCENTRATION (mol dm-3) x VOLUME (dm3)

  13. MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME 1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH

  14. MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME 1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH moles = conc x volume in cm3 1000 = 2 mol dm-3 x 25cm3= 0.05 moles 1000

  15. MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME 1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH moles = conc x volume in cm3 1000 = 2 mol dm-3 x 25cm3 = 0.05 moles 1000 2 What volume in cm3of 0.1M H2SO4 contains 0.002 moles ?

  16. MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME 1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH moles = conc x volume in cm3 1000 = 2 mol dm-3 x 25cm3 = 0.05 moles 1000 2 What volume in cm3of 0.1M H2SO4 contains 0.002 moles ? volume = 1000 x moles (re-arrangement of above) (in cm3) conc = 1000 x 0.002 = 20 cm3 0.1 mol dm-3

  17. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out?

  18. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 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 dm3 ? = 0.100 mol How many moles will be in 250cm3 ? = 0.100/4 = 0.025 mol

  19. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 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 dm3 ? = 0.100 mol How many moles will be in 250cm3 ? = 0.100/4 = 0.025 mol What is the formula of anhydrous sodium carbonate? = Na2CO3 What is the relative formula mass? = 106 What is the molar mass? = 106g mol -1

  20. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 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 dm3 ? = 0.100 mol How many moles will be in 250cm3 ? = 0.100/4 = 0.025 mol What is the formula of anhydrous sodium carbonate? = Na2CO3 What is the relative formula mass? = 106 What is the molar mass? = 106g mol -1 What mass of Na2CO3 is in 0.025 moles = 0.025 x 106 = 2.650g of Na2CO3 ? (mass = moles x molar mass)

  21. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 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 dm3 ? = 0.100 mol How many moles will be in 250cm3 ? = 0.100/4 = 0.025 mol What is the formula of anhydrous sodium carbonate? = Na2CO3 What is the relative formula mass? = 106 What is the molar mass? = 106g mol -1 What mass of Na2CO3 is in 0.025 moles = 0.025 x 106 = 2.650g of Na2CO3 ? (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 250cm3 in a graduated flask.

  22. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 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 dm3 ? = 0.100 mol How many moles will be in 250cm3 ? = 0.100/4 = 0.025 mol What is the formula of anhydrous sodium carbonate? = Na2CO3 What is the relative formula mass? = 106 What is the molar mass? = 106g mol -1 What mass of Na2CO3 is in 0.025 moles = 0.025 x 106 = 2.650g of Na2CO3 ? (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 250cm3 in a graduated flask.

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

  24. odds or evens . . .

  25. odds or evens . . .

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

  27. 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

  28. 1g 250cm3 SOLUTIONS ‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’ 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 250cm3 of de-ionised water

  29. 1g WATER WATER 1g 250cm3 250cm3 SOLUTIONS ‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’ 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 250cm3 of de-ionised water RIGHT Dissolve 1g of solute in water and then add enough water to make 250cm3 of solution

  30. STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 graduated flask and made up to the mark with de-ionised (or distilled) water.

  31. STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 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 ?

  32. STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 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 Na2CO3 in a 250cm3 solution = 4.240g molar mass of Na2CO3 = 106g mol -1 no. of moles in a 250cm3 solution = 4.240g / 106g mol -1 = 0.04 mol

  33. STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 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 Na2CO3 in a 250cm3 solution = 4.240g molar mass of Na2CO3 = 106g mol -1 no. of moles in a 250cm3 solution = 4.240g / 106g mol -1 = 0.04 mol Concentration is normally expressed as moles per dm3 of solution Therefore, as it is in 250cm3, the value is scaled up by a factor of 4 – that’s 1000/250

  34. STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 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 Na2CO3 in a 250cm3 solution = 4.240g molar mass of Na2CO3 = 106g mol -1 no. of moles in a 250cm3 solution = 4.240g / 106g mol -1 = 0.04 mol Concentration is normally expressed as moles per dm3 of solution Therefore, as it is in 250cm3, the value is scaled up by a factor of 4 no. of moles in 1000cm3 (1dm3) = 4 x 0.04 = 0.16 molANS.0.16 mol dm-3

  35. THE MOLARITY 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 cm3 25cm3 250cm3 250cm3

  36. THE MOLE 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 cm3 25cm3 250cm3 250cm3 The original solution has a concentration of0.200 mol dm-3

  37. THE MOLE 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 cm3 25cm3 250cm3 250cm3 The original solution has a concentration of0.200 mol dm-3 This means that there are 0.200 mols of solute in every 1 dm3 (1000 cm3) of solution

  38. THE MOLE 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 cm3 25cm3 250cm3 250cm3 The original solution has a concentration of0.200 mol dm-3 This means that there are 0.200 mols of solute in every 1 dm3 (1000 cm3) of solution Take out 25.00 cm3 and you will take a fraction 25/1000 ( 1/40 ) of the number of moles

  39. THE MOLE 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 cm3 25cm3 250cm3 250cm3 The original solution has a concentration of0.200 mol dm-3 This means that there are 0.200 mols of solute in every 1 dm3 (1000 cm3) of solution Take out 25.00 cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles moles in 1dm3 (1000cm3) = 0.200 moles in 1cm3 = 0.200/1000 moles in 25cm3 = 25/1000 x 0.200 = 5.0 x 10-3 mol

  40. THE MOLE 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 cm3 25cm3 250cm3 250cm3 The original solution has a concentration of0.200 mol dm-3 This means that there are 0.200 mols of solute in every 1 dm3 (1000 cm3) of solution Take out 25.00 cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles moles in 1dm3 (1000cm3) = 0.200 moles in 1cm3 = 0.200/1000 moles in 25cm3 = 25 x 0.200/1000 = 5.0 x 10-3 mol

  41. Titrations 1.1.13

  42. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.

  43. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3. 1. Write out a BALANCED equationNaOH + HCl ——> NaCl + H2O

  44. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3. 1. Write out a BALANCED equationNaOH + HCl ——> NaCl + H2O 2. Get a molar relationship between the reactantsmoles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl

  45. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3. 1. Write out a BALANCED equationNaOH + HCl ——> NaCl + H2O 2. Get a molar relationship between the reactantsmoles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl 3. Calculate the number of moles of each substance HCl0.100 x 25/1000 (i) M is the concentration in mol dm-3NaOHM x 20/1000 (ii)

  46. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3. 1. Write out a BALANCED equationNaOH + HCl ——> NaCl + H2O 2. Get a molar relationship between the reactantsmoles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl 3. Calculate the number of moles of each substance HCl0.100 x 25/1000 (i) M is the concentration in mol dm-3NaOHM x 20/1000 (ii) 4. Look at the molar relationship and insert (i) and (ii) moles of NaOH = moles of HCl M x 20/1000 = 0.100 x 25/1000

  47. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3. 1. Write out a BALANCED equationNaOH + HCl ——> NaCl + H2O 2. Get a molar relationship between the reactantsmoles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl 3. Calculate the number of moles of each substance HCl0.100 x 25/1000 (i) M is the concentration in mol dm-3NaOHM x 20/1000 (ii) 4. Look at the molar relationship and insert (i) and (ii) moles of NaOH = moles of HCl 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

  48. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3. 1. Write out a BALANCED equationNaOH + HCl ——> NaCl + H2O 2. Get a molar relationship between the reactantsmoles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl 3. Calculate the number of moles of each substance HCl0.100 x 25/1000 (i) M is the concentration in mol dm-3NaOHM x 20/1000 (ii) 4. Look at the molar relationship and insert (i) and (ii) moles of NaOH = moles of HCl M x 20/1000 = 0.100 x 25/1000 5. Cancel the 1000’sM x 20 = 0.100 x 25 re-arrange the numbers to obtain MM = 0.100 x 25 20 6. Calculate the concentration of the NaOH = 0.125 mol dm-3

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