1. Chapter 6 L EARNING O UTCOMES Define the term standard solution Use results from volumetric analysis to calculate the number of moles reacting, the mole ratio in which the reactants combine and the concentrate and mass concentration of reactants Concentration of Solutions and Volumetric Analysis

2. Solute and Solvent A solution is made up of two parts: solute + solvent = solution The solute is the substance dissolved in a solution. The solvent is the substance in which the solute has dissolved. For example, in a beaker of sugar solution, the sugar is the solute and the water is the solvent. Chapter 6 Concentration of Solutions and Volumetric Analysis

3. Concentrated or dilute? Instead of using the words “strong” and “weak” to describe coffee, we can use the terms concentrated and dilute. A concentrated solution will contain more solute dissolved in a certain volume of solution. A dilute solution will contain less solute dissolved in the same volume of solution. Do you like “ strong ” or “weak” coffee? Chapter 6 Concentration of Solutions and Volumetric Analysis

4. Concentration of solutions In order to standardise the volume of the solution, chemists use 1 dm 3 as the unit for measurement. The concentration of a solution is the mass of solute dissolved in 1 dm 3 of the solution. 1 dm 3 = 1000 cm 3 Chapter 6 Concentration of Solutions and Volumetric Analysis

5. 5 Chapter 6 Concentration of solutions Concentrations can be expressed in two ways as:  grams/dm 3 or g/dm 3  moles/dm 3 or mol/dm 3 Concentration of Solutions and Volumetric Analysis

6. Concentration of solution in g/dm 3 ► Suppose a solution of sodium chloride is made by dissolving 58.5g of the salt in 1 dm 3 of the solution. The concentration of the sodium chloride solution is equal to: 58.5 g /dm 3 Chapter 6 Concentration of Solutions and Volumetric Analysis

7. Concentration of solution in mol/dm 3 ► Since 58.5 g of sodium chloride is equal to 1 mole of the salt, The concentration of the solution is also equal to: 1 mol/dm 3 (or 1 M). The number of moles per dm 3 of a solution is also called the molarity of the solution. Chapter 6 Concentration of Solutions and Volumetric Analysis

8. Formulae Concentration = Mass of solute in grams in g/dm 3 Volume of solution in dm 3 Concentration = No. of moles of solute in mol/dm 3 Volume of solution in dm 3 Mass of solute = Volume of solution in dm 3 x Concentration in g/dm 3 Chapter 6 Concentration of Solutions and Volumetric Analysis

9. Worked example 1 A solution of sodium chloride is made by dissolving 11.7 g of sodium chloride in 500 cm 3 of the solution. Find the concentration of the solution in (a) g/dm 3, (b) mol/dm 3. Solution Volume of solution = 500 cm 3 = 500 = 0.5 dm 3 1000 (a) Concentration = Mass in grams Volume in dm 3 = 11.7 g = 23.4 g/dm 3 0.5 dm 3 (b) No. of moles = 11.7 g = 11.7 = 0.2 mol M r of NaCl 58.5 Concentration = No. of moles Volume in dm 3 = 0.2 mol = 0.4 mol/dm 3 0.5 dm 3 Chapter 6 Concentration of solutions Concentration of Solutions and Volumetric Analysis

10. A solution of magnesium chloride has a concentration of 23.75 g/dm 3. (a) What is the concentration of the solution in mol/dm 3 ? (b) If 200 cm 3 of the solution is evaporated to dryness, what mass of salt can be obtained? Solution (a) Number of moles of MgCl 2 in 1 dm 3 = 23 g/dm 3 = 23.75 = 0.25 mol M r of MgCl 2 95 Concentration = 0.25 mol = 0.25 mol/dm 3 1 dm 3 (b) Mass of solute = Concentration x Volume of solution = 23.75 g/dm 3 x 200 dm 3 1000 = 4.75 g Chapter 6 Worked example 2 Concentration of solutions Concentration of Solutions and Volumetric Analysis

11. A solution of sulphuric acid has a concentration of 0.25 mol/dm 3 (a) What is the concentration of the solution in g/dm 3 ? (b) What mass of acid will be contained in 250 cm 3 of the solution? Solution Mass of H 2 SO 4 = 0.25 mol x M r = 0.25 x 98 g = 24.5 g (a) Concentration = Mass in grams Volume in dm 3 = 24.5 g = 24.5 g/dm 3 1 dm 3 (b) Mass of acid = Concentration x Volume of solution = 24.5 g/dm 3 x 250 dm 3 1000 = 6.125 g Chapter 6 Worked example 3 Concentration of solutions Concentration of Solutions and Volumetric Analysis

12. 25 cm 3 of a solution of sulphuric acid of concentration 0.400 mol/dm 3 is neutralised with a solution of sodium hydroxide of concentration 0.625 mol/dm 3. What is the volume of sodium hydroxide solution required? Equation of reaction: H 2 SO 4 + 2NaOH  Na 2 SO 4 + 2H 2 O From the equation, No. of moles of H 2 SO 4 = 1 No. of moles of NaOH 2 Vol. of H 2 SO 4 x Conc. of H 2 SO 4 = 1 Vol. of NaOH x Conc. of NaOH 2 0.025 dm 3 x 0.400 mol/dm 3 = 1 Vol. of NaOH x 0.625 mol/dm 3 2 Vol. of NaOH = 2 x 0.025 x 0.400 = 0.032 dm 3 0.625 = 32 cm 3 Solution Chapter 6 Worked example 4 Concentration of solutions Concentration of Solutions and Volumetric Analysis

13. Quick check 1. 1. A solution of calcium chloride (CaCl 2 ) contains 37 g of the salt in 250 cm 3 of the solution. Find the concentration of the solution in (a) g/dm 3, (b) mol/dm 3. 2. 2. 500 cm 3 of a solution of sodium nitrate contains 14.7 g of the salt. (a) Find the concentration of the solution in mol/dm 3. (b) If 100 cm 3 of the solution is evaporated, how much salt can be obtained? 3. 3. A solution of magnesium sulphate has a concentration of 0.25 mol/dm 3. (a) What is the concentration of the solution in g/dm 3 ? (b) What mass of magnesium sulphate is contained in 250 cm 3 of the solution? 4. 4. A solution of nitric acid has an unknown concentration. 25.0 cm 3 of the acid is completely neutralised by 22.5 cm 3 of potassium hydroxide solution of concentration 0.485 mol/dm 3. What is the concentration of the nitric acid? Solution Chapter 6 Concentration of Solutions and Volumetric Analysis

14. Solution to Quick check 1. 1. (a) Concentration = 37 g = 148 g/ dm 3 0.25 dm 3 (b)No. of moles = 37 = 0.333 mol 111 Concentration = 0.333 = 1.33 mol/dm 3 0.25 dm 3 2.(a) No. of moles = 14.7 = 0.173 mol 85 Concentration = 0.173 = 0.346 mol/dm 3 0.5 (b) Mass of salt = 0.1 x 0.346 x 85 = 2.94 g 3.(a) Concentration = (0.25 x 120) mol x 1 dm 3 = 30 g/dm 3 (b) Mass of magnesium sulphate = 0.250 x 30 = 7.5 g 4. Equation:HNO 3 + KOH  KNO 3 + H 2 O No. of moles of nitric acid = 1 No. of moles of KOH 1 25.0 cm 3 x Conc. of acid = 1 22.5 cm 3 x 0.485 mol/dm 3 1 Conc. of nitric acid = 0.437 mol/dm 3 Chapter 6 Concentration of Solutions and Volumetric Analysis Return

15. 1. 1. http://www.ausetute.com.au/concsols.html http://www.ausetute.com.au/concsols.html 2. 2. http://dl.clackamas.edu/ch105-04/tableof.htm http://dl.clackamas.edu/ch105-04/tableof.htm 3. 3. http://en.wikipedia.org/wiki/Concentration http://en.wikipedia.org/wiki/Concentration To Learn more about Concentrations of Solutions, click on the links below! Chapter 6 Concentration of Solutions and Volumetric Analysis

16. 16 Introduction Volumetric Analysis or VA is a method of finding out the quantity of substance present in a solid or solution. It usually involves titrating a known solution, called a standard solution, with an unknown solution. Based on the equation of reaction, calculations are then made to find out the concentration of the unknown solution. Chapter 6 Concentration of Solutions and Volumetric Analysis

17. 17 Using a pipette A pipette is used to deliver an exact volume, usually 25.0 cm 3 of solution into a conical flask. The solution in the titrating flask is called the titrate. Before using a pipette, it should be washed with tap water, then rinsed with distilled water and finally with the liquid it is to be filled. For safety reasons, a pipette filler is used to suck up the solution. To use the pipette filler, first fit it to the top of the pipette, as shown in the diagram. Squeeze valve 1 with right index finger and thumb and squeeze the bulb with the left palm to expel all the air in the bulb. Then place the tip of the pipette below the surface of the liquid to be sucked up, and squeeze valve 2 to suck up the liquid. Chapter 6 Concentration of Solutions and Volumetric Analysis

18. 18 Using a pipette When the liquid rises to a level higher than the mark, remove the tip of the pipette from the liquid. Gently squeeze valve 3 to release the liquid slowly until the meniscus of the liquid is exactly at the mark of the pipette. Now place the tip of the pipette into the titration flask, and squeeze valve 3 to release all the liquid into the flask. When all the liquid in the pipette has run out, touch the tip of the pipette on the inside of the flask so that only a drop of liquid is left inside the tip of the pipette. Chapter 6 Concentration of Solutions and Volumetric Analysis

19. 19 Using a burette A burette is used to contain and measure the volume of the liquid, called the titrant used in the titration. Before using a burette, it should be washed first with tap water, then rinsed with distilled water and finally with the liquid (titrant) it is to be filled. The liquid (titrant) in the burette must be released slowly, a few drops at a time, into the titration flask. The readings must be taken accurate to 0.1 cm 3. E.g. 24.0 cm 3, not 24 cm 3. Make sure that the clip of the burette is tight and the liquid is not leaking. Also make sure that the burette jet is filled with liquid, it must not contain any air bubbles. Chapter 6 Concentration of Solutions and Volumetric Analysis

20. 20 Using a burette The burette should be clamped to the retort stand in a vertical position so that the reading will be accurate. When reading the burette, the eye must be horizontal to the bottom of meniscus to avoid parallax error. (See diagram). Chapter 6 Concentration of Solutions and Volumetric Analysis

21. 21 Other tips on safety and accuracy When filling or reading the burette, it should be lowered to a suitable height. Do not attempt to read it by climbing onto a stool. Make sure that the tip of the pipette is always kept below the surface of the liquid when it is being filled, otherwise air bubbles will get into the pipette. After filling a burette, the small funnel should be removed from the top of the burette, otherwise drops of liquid may run down into the burette during a titration and affect the reading. The titration flask should be placed on a white tile or paper so that the colour of the indicator can be seen easily. Use the wash bottle to wash down the insides of the conical flask towards the end of the titration. Chapter 6 Concentration of Solutions and Volumetric Analysis

22. 22 Use of Indicators Indicator Colour in acids Colour at end point Colour in alkalis Methyl orange redorangeyellow Screened methyl orange redgreygreen Litmusredpurpleblue Phenolphthaleincolourlesspinkred Chapter 6 Concentration of Solutions and Volumetric Analysis

23. 23 In a normal titration, candidates are usually advised to carry out at least one rough and two accurate titrations. You should record your readings in a table like this. Chapter 6 Titration readings In general, you should carry out as many titrations as needed to obtain two or more consistent volumes. If no consistent volumes are obtained, the average value should be calculated. Titration number 1234 Final burette reading /cm 3 25.224.833.324.9 Initial burette reading /cm 3 0.00.07.40.1 Volume of NaOH used /cm 3 25.224.825.924.8 Best titration results (√) Concentration of Solutions and Volumetric Analysis √√

24. 24 Chapter 6 Suppose that in an experiment, you are asked to find the concentration of a solution of sulphuric acid by titrating 25.0 cm 3 of the acid against a standard solution of sodium hydroxide of concentration 0.100 mol/dm 3, using phenolphthalein as an indicator. First set up the apparatus as shown in the diagram and then carry out the titration, repeating it as many times as necessary to obtain a set of consistent results. Titration of a known acid with an alkali Concentration of Solutions and Volumetric Analysis

25. 25 Mean volume of sodium hydroxide used = 24.8 cm 3 Chapter 6 Titration number 1234 Final burette reading /cm 3 25.224.833.324.9 Initial burette reading /cm 3 0.00.07.40.1 Volume of NaOH used /cm 3 25.224.825.924.8 Best titration results (√) √√ Suppose the following readings are obtained: Results Concentration of Solutions and Volumetric Analysis

26. 26 Titration of a known acid with an alkali You are then asked to calculate the concentration of the sulphuric acid from your results. The equation for the reaction is: H 2 SO 4 + 2NaOH  Na 2 SO 4 + 2H 2 O From the equation, No. of moles of H 2 SO 4 = 1 No. of moles of NaOH 2 Vol. of H 2 SO 4 x Conc. of H 2 SO 4 = 1 Vol. of NaOH x Conc. of NaOH 2 25.0 x Conc. of H 2 SO 4 = 1 24.8 x 0.100 mol/dm 3 2 Therefore, Conc. of H 2 SO 4 = 1 x 24.8 x 0.100 mol/dm 3 2 x 25.0 = 0.0496 mol/dm 3 Chapter 6 Concentration of Solutions and Volumetric Analysis

27. 27 V a x M a = x V b x M b y where M a, M b are the concentrations of the acid and base and V a, V b are the volumes of the acid and base used in the titration. Chapter 6 In general if x moles of an acid reacts with y moles of a base, then No. of moles of acid = x No. of moles of base y Vol. of acid x Conc. of acid = x Vol. of base x Conc. of base y Hence, it can be shown that : Acid-base titration Concentration of Solutions and Volumetric Analysis

28. 28 Aim: You are provided with a solution containing 5.00 g/dm 3 of the acid H 3 XO 4. You are to find the relative molecular mass of the acid by titrating 25.0 cm 3 portions of the acid with the standard (0.100 mol/dm 3 ) sodium hydroxide solution, and hence find the relative atomic mass of element X. The equation for the reaction is: H 3 XO 4 + 2NaOH  Na 2 HXO 4 + 2H 2 O Titration No.1234 Final reading/ cm 3 25.425.525.635.8 Initial reading/ cm 3 0.0 10.0 Volume of NaOH/ cm 3 25.425.525.625.8 Average volume of NaOH used = 25.5 cm 3 Results: Chapter 6 Titration of an unknown acid with an alkali Concentration of Solutions and Volumetric Analysis

29. 29 From the equation, No. of moles of H 3 XO 4 = 1 No. of moles of NaOH 2 Vol. of H 3 XO 4 x Conc. of H 3 XO 4 = 1 Vol. of NaOH x Conc. of NaOH 2 25.0 x Conc. of H 3 XO 4 = 1 25.5 x 0.100 mol/dm 3 2 Therefore, conc. of H 3 XO 4 = 1 x 25.5 x 0.100 2 x 25.0 = 0.0510 mol/dm 3 Since 1 dm 3 of the acid contains 5.00 g of the acid, therefore 0.0510 x M r of H 3 XO 4 = 5.00 g M r of H 3 XO 4 = 5.00 = 98.0 0.0510 Calculate the relative atomic mass of X: 1x3 + X + 16x4 = 98 X = 98 – 67 = 31 Chapter 6 Titration of unknown acid with an alkali Concentration of Solutions and Volumetric Analysis

30. 30 25.0 cm 3 portions of hydrogen peroxide solution (H 2 O 2 ) was titrated with standard (0.020 mol/dm 3 ) potassium manganate(VII) solution. Introduction: Oxidising agents can be titrated with reducing agents. Hydrogen peroxide is a reducing agent and can be titrated against acidified potassium manganate(VII), an oxidising agent. No indicator is required for this titration as potassium manganate(VII) solution is purple in colour and is decolourised by the hydrogen peroxide solution when the reaction is complete. Chapter 6 Titration of hydrogen peroxide with potassium manganate(VII) solution Concentration of Solutions and Volumetric Analysis

31. 31 Volume of KMnO 4 used = 25.2 cm 3 A. Calculate the number of moles of KMnO 4 used. No. of moles of KMnO 4 = Volume in dm 3 x Conc. = 25.2 dm 3 x 0.020 mol/dm 3 1000 = 0.000504 mol Chapter 6 Results: Titration No.123 Final reading/ cm 3 25.125.2 Initial reading/ cm 3 0.0 Volume of KMnO 4 / cm 3 25.1 25.2 Titration of hydrogen peroxide with potassium manganate(VII) solution Concentration of Solutions and Volumetric Analysis

32. 32 B. If 1 mole of KMnO 4 reacts with 2.5 moles of H 2 O 2, (a) Calculate the number of moles of H 2 O 2 that react with the KMnO 4. (b) Find the concentration of H 2 O 2 solution. (a) No. of moles of H 2 O 2 = 2.5 x 0.000504 mol = 0.00126 mol (b) Concentration = 0.00126 mol = 0.0504 mol/dm 3 0.025 dm 3 C. If 2 moles of H 2 O 2 decompose during the reaction to give 1 mole of oxygen, calculate the volume of oxygen given off during the titration. No. of moles of O 2 given off = 1 x 0.00126 mol 2 = 0.00063 mol Therefore, Volume of O 2 = 0.00063 x 24000 cm 3 = 15.1 cm 3 Chapter 6 Titration of hydrogen peroxide with potassium manganate(VII) solution Concentration of Solutions and Volumetric Analysis

33. 33 1. 1. After washing the pipette, it should be rinsed with ________. (A) distilled water(B) the titrate (C) the titrant (D) tap water 2. 2. After washing the titration flask, it should be rinsed with ________. (A) distilled water(B) the titrate (C) the titrant (D) tap water 3. 3. After washing the burette, it should be rinsed with ________. (A) distilled water(B) the titrate (C) the titrant (D) tap water 4. 4. A titration flask contains 25.0 cm 3 of sodium hydroxide and a few drops of phenolphthalein as indicator. It is titrated against hydrochloric acid contained in a burette. What colour change would you observe when the end point is reached? (A) colourless to light pink(B) light pink to colourless (C) red to colourless(D) blue to pink Solution Chapter 6 Quick Check Concentration of Solutions and Volumetric Analysis

34. 34 25.0 cm 3 samples of sodium hydroxide solution are titrated against hydrochloric acid which has a concentration of 0.225 mol/dm 3. The results obtained are shown in the table below. Titration No.1234 Final burette reading/ cm 3 24.448.923.648.0 Initial burette reading/ cm 3 0.024.40.023.6 Volume of HCl/ cm 3 Best titration result (√) Chapter 6 Solution Quick Check Concentration of Solutions and Volumetric Analysis (a) Complete the table above. (b) Calculate the concentration of the sodium hydroxide solution.

35. 35 1.http://www.tele.ed.nom.br/buret.htmlhttp://www.tele.ed.nom.br/buret.html 2.http://www.chem.ubc.ca/courseware/154/tutorials/exp6A/http://www.chem.ubc.ca/courseware/154/tutorials/exp6A/ To learn more about titration, click on the links below! Chapter 6 Concentration of Solutions and Volumetric Analysis

36. 36 1. 1. After washing the pipette, it should be rinsed with (A) distilled water(B) the titrate (C) the titrant (D) tap water 2. 2. After washing the titration flask, it should be rinsed with (A) distilled water(B) the titrate (C) the titrant (D) tap water 3. 3. After washing the burette, it should be rinsed with (A) distilled water(B) the titrate (C) the titrant (D) tap water 4. 4. A titration flask contains 25.0 cm 3 of sodium hydroxide and a few drops of phenolphthalein as indicator. It is titrated against hydrochloric acid contained in a burette. What colour change would you observe when the end point is reached? (A) colourless to light pink(B) light pink to colourless (C) red to colourless(D) blue to pink Return Chapter 6 Solution to Quick check Concentration of Solutions and Volumetric Analysis

37. 37 5. Average volume of HCl used = 24.4 cm 3 Equation:NaOH(aq) + HCl(aq)  NaCl(aq) + H 2 O(l) 25.0 x Conc. of NaOH = 1 24.4 x 0.225 mol/dm 3 1 Conc. of NaOH= 24.4 x 0.225 mol/dm 3 25.0 = 0.220 mol/dm 3 Chapter 6 Return Solution to Quick check Titration No.1234 Final burette reading/ cm 3 24.448.923.648.0 Initial burette reading/ cm 3 0.024.40.023.6 Volume of HCl/ cm 3 24.424.523.624.4 Best titration result (√)√√ Concentration of Solutions and Volumetric Analysis

38. References Chemistry for CSEC Examinations by Mike Taylor and Tania Chung Longman Chemistry for CSEC by Jim Clark and Ray Oliver 38