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# C06 concentration of solutions and volumetric analysis

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### C06 concentration of solutions and volumetric analysis

1. 1. Chapter 6 Concentration of Solutions and Volumetric Analysis LEARNING OUTCOMES  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 concentration and mass concentration of reactants
2. 2. Chapter 6 Concentration of Solutions and Volumetric Analysis 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. 
3. 3. Chapter 6 Concentration of Solutions and Volumetric Analysis Concentrated or dilute? Do you like “strong” or “weak” tea? Instead of using the words “strong” and “weak” to describe tea, 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. 
4. 4. Chapter 6 Concentration of Solutions and Volumetric Analysis Concentration of solutions  In order to standardise the volume of the solution, chemists use 1 dm3 as the unit for measurement. 1 dm3 = 1000 cm3 The concentration of a solution is the mass of solute dissolved in 1 dm3 of the solution.
5. 5. Chapter 6 Concentration of Solutions and Volumetric Analysis Concentration of solutions  Concentrations can be expressed in two ways as:   grams/dm3 or g/dm3 moles/dm3 or mol/dm3 5
6. 6. Chapter 6 Concentration of Solutions and Volumetric Analysis Concentration of solution in g/dm3 ► Suppose a solution of sodium chloride is made by dissolving 58.5g of the salt in 1 dm3 of the solution. The concentration of the sodium chloride solution is equal to: 58.5 g /dm3
7. 7. Chapter 6 Concentration of Solutions and Volumetric Analysis Concentration of solution in mol/dm3 ► 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/dm3 (or 1 M).  The number of moles per dm3 of a solution is also called the molarity of the solution.
8. 8. Chapter 6 Concentration of Solutions and Volumetric Analysis Formulae Concentration = Mass of solute in grams in g/dm3 Volume of solution in dm3 Concentration = No. of moles of solute in mol/dm3 Volume of solution in dm3 Mass of solute = Volume of solution in dm3 x Concentration in g/dm3
9. 9. Chapter 6 Concentration of Solutions and Volumetric Analysis Concentration of solutions Worked example 1 A solution of sodium chloride is made by dissolving 11.7 g of sodium chloride in 500 cm3 of the solution. Find the concentration of the solution in (a) g/dm3, (b) mol/dm3. Solution Volume of solution = 500 cm3 = 500 = 0.5 dm3 1000 (a) Concentration = Mass in grams Volume in dm3 = 11.7 g = 23.4 g/dm3 0.5 dm3 (b) No. of moles = 11.7 g = 11.7 = 0.2 mol Mr of NaCl 58.5 Concentration = No. of moles Volume in dm3 = 0.2 mol = 0.4 mol/dm3 0.5 dm3
10. 10. Chapter 6 Concentration of Solutions and Volumetric Analysis Concentration of solutions Worked example 2 A solution of magnesium chloride has a concentration of 23.75 g/dm3. (a) What is the concentration of the solution in mol/dm3? (b) If 200 cm3 of the solution is evaporated to dryness, what mass of salt can be obtained? Solution (a) Number of moles of MgCl2 in 1 dm3 = 23 g/dm3 = 23.75 = 0.25 mol Mr of MgCl2 95 Concentration = 0.25 mol = 0.25 mol/dm3 1 dm3 (b) Mass of solute = Concentration x Volume of solution = 23.75 g/dm3 x 200 dm3 1000 = 4.75 g
11. 11. Chapter 6 Concentration of Solutions and Volumetric Analysis Concentration of solutions Worked example 3 A solution of sulphuric acid has a concentration of 0.25 mol/dm3 (a) What is the concentration of the solution in g/dm3 ? (b) What mass of acid will be contained in 250 cm3 of the solution? Solution Mass of H2SO4 = 0.25 mol x Mr = 0.25 x 98 g = 24.5 g (a) Concentration = Mass in grams Volume in dm3 = 24.5 g = 24.5 g/dm3 1 dm3 (b) Mass of acid = Concentration x Volume of solution = 24.5 g/dm3 x 250 dm3 1000 = 6.125 g
12. 12. Chapter 6 Concentration of Solutions and Volumetric Analysis Concentration of solutions Worked example 4 25 cm3 of a solution of sulphuric acid of concentration 0.400 mol/dm3 is neutralised with a solution of sodium hydroxide of concentration 0.625 mol/dm 3. What is the volume of sodium hydroxide solution required? Solution Equation of reaction: H2SO4 + 2NaOH  Na2SO4 + 2H2O From the equation, No. of moles of H2SO4 = 1 No. of moles of NaOH 2 Vol. of H2SO4 x Conc. of H2SO4 = 1 Vol. of NaOH x Conc. of NaOH 2 0.025 dm3 x 0.400 mol/dm3 = 1 Vol. of NaOH x 0.625 mol/dm3 2 Vol. of NaOH = 2 x 0.025 x 0.400 = 0.032 dm3 0.625 3
13. 13. Chapter 6 Concentration of Solutions and Volumetric Analysis Quick check 1. 2. 3. 4. A solution of calcium chloride (CaCl2) contains 37 g of the salt in 250 cm3 of the solution. Find the concentration of the solution in (a) g/dm3, (b) mol/dm3. 500 cm3 of a solution of sodium nitrate contains 14.7 g of the salt. (a) Find the concentration of the solution in mol/dm3. (b) If 100 cm3 of the solution is evaporated, how much salt can be obtained? A solution of magnesium sulphate has a concentration of 0.25 mol/dm3. (a) What is the concentration of the solution in g/dm3? (b) What mass of magnesium sulphate is contained in 250 cm3 of the solution? A solution of nitric acid has an unknown concentration. 25.0 cm3 of the acid is completely neutralised by 22.5 cm3 of potassium hydroxide solution of concentration 0.485 mol/dm3. What is the concentration of the nitric acid? Solution
14. 14. Chapter 6 Concentration of Solutions and Volumetric Analysis Solution to Quick check 1. 2. (a) Concentration = 37 g = 148 g/ dm3 0.25 dm3 (b) No. of moles = 37 = 0.333 mol 111 Concentration = 0.333 = 1.33 mol/dm3 0.25 dm3 (a) No. of moles = 14.7 = 0.173 mol 85 Concentration = 0.173 = 0.346 mol/dm3 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 dm3 = 30 g/dm3 (b) Mass of magnesium sulphate = 0.250 x 30 = 7.5 g 4. Equation: HNO3 + KOH  KNO3 + H2O No. of moles of nitric acid = 1 No. of moles of KOH 1 3 25.0 cm x Conc. of acid = 1 22.5 cm3 x 0.485 mol/dm3 1 Conc. of nitric acid = 0.437 mol/dm3 Return
15. 15. Chapter 6 Concentration of Solutions and Volumetric Analysis To Learn more about Concentrations of Solutions, click on the links below! http://www.ausetute.com.au/concsols.html 2. http://dl.clackamas.edu/ch105-04/tableof.htm 3. http://en.wikipedia.org/wiki/Concentration 1.
16. 16. Chapter 6 Concentration of Solutions and Volumetric Analysis 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. 16
17. 17. Chapter 6 Concentration of Solutions and Volumetric Analysis Using a pipette        A pipette is used to deliver an exact volume, usually 25.0 cm3 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. 17
18. 18. Chapter 6 Concentration of Solutions and Volumetric Analysis 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. 18
19. 19. Chapter 6 Concentration of Solutions and Volumetric Analysis 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 cm3. E.g. 24.0 cm3, not 24 cm3. 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. 19
20. 20. Chapter 6 Concentration of Solutions and Volumetric Analysis 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). 20
21. 21. Chapter 6 Concentration of Solutions and Volumetric Analysis 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. 21
22. 22. Chapter 6 Concentration of Solutions and Volumetric Analysis Use of Indicators Indicator Colour in acids Colour at end point Colour in alkalis Methyl orange red orange yellow Screened methyl orange red grey green Litmus red purple blue Phenolphthalein colourless pink red 22
23. 23. Chapter 6 Concentration of Solutions and Volumetric Analysis Titration readings 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.  Titration number 1 2 3 4 Final burette reading /cm3 25.2 24.8 33.3 24.9 Initial burette reading /cm3 0.0 0.0 7.4 0.1 Volume of NaOH used /cm3 25.2 24.8 25.9 24.8 Best titration results (√) √ √ 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.  23
24. 24. Chapter 6 Concentration of Solutions and Volumetric Analysis Titration of a known acid with an alkali Suppose that in an experiment, you are asked to find the concentration of a solution of sulphuric acid by titrating 25.0 cm3 of the acid against a standard solution of sodium hydroxide of concentration 0.100 mol/dm3, 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.  24
25. 25. Chapter 6 Concentration of Solutions and Volumetric Analysis Results Suppose the following readings are obtained: Titration number 1 2 3 4 Final burette reading /cm3 25.2 24.8 33.3 24.9 Initial burette reading /cm3 0.0 0.0 7.4 0.1 Volume of NaOH used /cm3 25.2 24.8 25.9 24.8 Best titration results (√) √ √ Mean volume of sodium hydroxide used = 24.8 cm3 25
26. 26. Chapter 6 Concentration of Solutions and Volumetric Analysis 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: H2SO4 + 2NaOH  Na2SO4 + 2H2O From the equation, No. of moles of H2SO4 = 1 No. of moles of NaOH 2 Vol. of H2SO4 x Conc. of H2SO4 = 1 Vol. of NaOH x Conc. of NaOH 2 25.0 x Conc. of H2SO4 = 1 24.8 x 0.100 mol/dm3 2 Therefore, Conc. of H2SO4 = 1 x 24.8 x 0.100 mol/dm3 2 x 25.0 = 0.0496 mol/dm3 26
27. 27. Chapter 6 Concentration of Solutions and Volumetric Analysis Acid-base titration 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 : Va x Ma = x Vb x Mb y where Ma, Mb are the concentrations of the acid and base and Va, Vb are the volumes of the acid and base used in the titration. 27
28. 28. Chapter 6 Concentration of Solutions and Volumetric Analysis Titration of an unknown acid with an alkali Aim: You are provided with a solution containing 5.00 g/dm3 of the acid H3XO4. You are to find the relative molecular mass of the acid by titrating 25.0 cm3 portions of the acid with the standard (0.100 mol/dm3) sodium hydroxide solution, and hence find the relative atomic mass of element X. The equation for the reaction is: H3XO4 + 2NaOH  Na2HXO4 + 2H2O Results: Titration No. 1 2 3 4 Final reading/ cm3 25.4 25.5 25.6 35.8 Initial reading/ cm3 0.0 0.0 0.0 10.0 Volume of NaOH/ cm3 25.4 25.5 25.6 25.8  Average volume of NaOH used = 25.5 cm3 28
29. 29. Chapter 6 Concentration of Solutions and Volumetric Analysis Titration of unknown acid with an alkali  From the equation, No. of moles of H3XO4 = 1 No. of moles of NaOH 2 Vol. of H3XO4 x Conc. of H3XO4 = 1 Vol. of NaOH x Conc. of NaOH 2 25.0 x Conc. of H3XO4 = 1 25.5 x 0.100 mol/dm3 2 Therefore, conc. of H3XO4 = 1 x 25.5 x 0.100 2 x 25.0 = 0.0510 mol/dm3  Since 1 dm3 of the acid contains 5.00 g of the acid, therefore 0.0510 x Mr of H3XO4 = 5.00 g Mr of H3XO4 = 5.00 = 98.0 0.0510  Calculate the relative atomic mass of X: 1x3 + X + 16x4 = 98 X = 98 – 67 = 31 29
30. 30. Chapter 6 Concentration of Solutions and Volumetric Analysis Titration of hydrogen peroxide with potassium manganate(VII) solution 25.0 cm3 portions of hydrogen peroxide solution (H2O2) was titrated with standard (0.020 mol/dm3) 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. 30
31. 31. Chapter 6 Concentration of Solutions and Volumetric Analysis Titration of hydrogen peroxide with potassium manganate(VII) solution Results: Titration No. 1 2 3 Final reading/ cm3 25.1 25.2 25.2 Initial reading/ cm3 0.0 0.0 0.0 Volume of KMnO4/ cm3 25.1 25.2 25.2 Volume of KMnO4 used = 25.2 cm3 A. Calculate the number of moles of KMnO4 used. No. of moles of KMnO4 = Volume in dm3 x Conc. = 25.2 dm3 x 0.020 mol/dm3 1000 = 0.000504 mol  31
32. 32. Chapter 6 Concentration of Solutions and Volumetric Analysis Titration of hydrogen peroxide with potassium manganate(VII) solution B. C. If 1 mole of KMnO4 reacts with 2.5 moles of H2O2, (a) Calculate the number of moles of H2O2 that react with the KMnO4. (b) Find the concentration of H2O2 solution. (a) No. of moles of H2O2 = 2.5 x 0.000504 mol = 0.00126 mol (b) Concentration = 0.00126 mol = 0.0504 mol/dm3 0.025 dm3 If 2 moles of H2O2 decompose during the reaction to give 1 mole of oxygen, calculate the volume of oxygen given off during the titration. No. of moles of O2 given off = 1 x 0.00126 mol 2 = 0.00063 mol Therefore, Volume of O2 = 0.00063 x 24000 cm3 32 = 15.1 cm3
33. 33. Chapter 6 Concentration of Solutions and Volumetric Analysis Quick Check 1. 2. 3. 4. After washing the pipette, it should be rinsed with ________. (A) distilled water (B) the titrate (C) the titrant (D) tap water After washing the titration flask, it should be rinsed with ________. (A) distilled water (B) the titrate (C) the titrant (D) tap water After washing the burette, it should be rinsed with ________. (A) distilled water (B) the titrate (C) the titrant (D) tap water A titration flask contains 25.0 cm3 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 33
34. 34. Chapter 6 Concentration of Solutions and Volumetric Analysis Quick Check 25.0 cm3 samples of sodium hydroxide solution are titrated against hydrochloric acid which has a concentration of 0.225 mol/dm3. The results obtained are shown in the table below. Titration No. 1 2 3 4 Final burette reading/ cm3 24.4 48.9 23.6 48.0 Initial burette reading/ cm3 0.0 24.4 0.0 23.6 Volume of HCl/ cm3 Best titration result (√) (a) Complete the table above. (b) Calculate the concentration of the sodium hydroxide solution. Solution 34
35. 35. Chapter 6 Concentration of Solutions and Volumetric Analysis To learn more about titration, click on the links below! 1. http://www.tele.ed.nom.br/buret.html 2. http://www.chem.ubc.ca/courseware/154/tutorials/exp6A/ 35
36. 36. Chapter 6 Concentration of Solutions and Volumetric Analysis Solution to Quick check 1. 2. 3. 4. After washing the pipette, it should be rinsed with (A) distilled water (B) the titrate (C) the titrant (D) tap water After washing the titration flask, it should be rinsed with (A) distilled water (B) the titrate (C) the titrant (D) tap water After washing the burette, it should be rinsed with (A) distilled water (B) the titrate (C) the titrant (D) tap water A titration flask contains 25.0 cm3 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 36
37. 37. Chapter 6 Concentration of Solutions and Volumetric Analysis Solution to Quick check 5. Titration No. 1 2 3 4 Final burette reading/ cm3 24.4 48.9 23.6 48.0 Initial burette reading/ cm3 0.0 24.4 0.0 23.6 Volume of HCl/ cm3 24.4 24.5 23.6 24.4 Best titration result (√) √ √ Average volume of HCl used = 24.4 cm3 Equation: NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) 25.0 x Conc. of NaOH = 1 24.4 x 0.225 mol/dm3 1 Conc. of NaOH = 24.4 x 0.225 mol/dm3 25.0 = 0.220 mol/dm3 Return 37