LC-Learn. Leaving Cert Chemistry Notes Higher Level Volumetric Analysis

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1 Lving Cert Chemistry Notes Higher Level Volumetric Analysis

2 Powered By: Volumetric Analysis Essential Theory A standard solution is a solution whose concentration is accurately known. A primary standard is a compound that can be obtained in a pure, soluble, stable and solid form so that it can be weighed out and dissolved in deionised water to make a solution of accurately known concentration. A secondary standard is obtained when a primary standard is used to standardise (i.e. to accurately determine the concentration of) another solution The following are unsuitable as primary standards: Iodine decomposes in the presence of sunlight (is not stable) Sodium hydroxide and potassium hydroxide they absorbed water vapour from the air (not stable/pure) Sodium thiosulphate and potassium permanganate they cannot be obtained in a state of high purity. Only a small amount of indicator should be used in a titration. This is because indicators are wk acids and bases themselves which may ld to an inaccuracy in the titration figures if too much is used. Molarity refers to the number of moles of a substance in 1 litre of solution. Titration calculations The equation used to calculate the concentration of either the acid or base used in the titration is given by : Molarity of acid Volume of base used Molarity of base Volume of acid used Number of moles of the base Number of moles of the acid (From the balanced equation) This equation can be used to solve for the unknown concentration/molarity of the acid or base used. 1

3 Powered By: (1) Preparation of a standard solution of anhydrous sodium carbonate Method Weigh out required mass of pure anhydrous sodium carbonate on a clock glass using a balance. Empty the contents of the clock glass into a cln, dry bker. Wash down the clock glass with deionised water and transfer these washings into the bker. This ensures that all of the sodium carbonate is transferred to the bker. Add sufficient deionised water to the bker, whilst stirring, until the sodium carbonate is completely dissolved. Transfer solution to a volumetric flask using a funnel with washings from the stirring rod, funnel and bker. Make up to the mark in the volumetric flask using deionised water. Add the deionised water drop by drop when the solution is within 1cm3 of the mark, ensuring to rd from the bottom of the meniscus. Stopper and invert flask 20 times to make solution homogeneous. (2) To standardise a solution of hydrochloric acid using a standard solution of sodium carbonate Na2CO3 + 2HCl 2NaCl +H2O + CO2 Method Rinse the conical flask with deionised water To remove any impurities. Rinse the pipette with deionised water and then with a sample of the standard sodium carbonate solution made in the previous experiment. Rinsing the pipette with deionised water removes any impurities Rinsing the pipette with the solution it is to contain removes the deionised water. Transfer 25cm3 of the standard solution of sodium carbonate into the conical flask using the pipette. Add 2-3 drops of methyl orange indicatior to the conical flask. 2

4 Powered By: Rinse the burette with deionised water and then with the hydrochloric acid solution. (to remove impurities and then to remove the deionised water as above) Fill the burette with the hydrochloric acid solution above the zero mark. Open the tap and allow some of the HCl to escape into a spare bker until the bottom of the meniscus of the HCl solution rests on the zero mark. This ensures that the portion below the tap in the burette is filled. Begin titration; slowly add the acid in the burette to the sodium carbonate in the conical flask, whilst swirling, until a colour change of orange pink is observed in the flask. Note rding on the burette. Rept titration until results agree to within 0.1cm3 HCl solution Sodium carbonate solution + Methyl orange indicator To calculate the molarity (M) of the hydrochloric acid Use titration equation and solve for Ma Molarity of acid Volume of base used Molarity of base Volume of acid used Number of moles of the acid Number of moles of the base 3

5 Powered By: (3) To determine the concentration of ethanoic acid in vinegar The concentration of ethanoic acid in a sample of vinegar can be determined by titrating the vinegar against a standard solution of sodium hydroxide solution, which was standardised in the previous experiment. CH3COOH + NaOH CH3COONa + H2O The indicator used in the titration is phenolphthalein which has a pink colour. Phenolphthalein is the suitable indicator for the titration as we are titrating a wk acid against a strong base. The colour change for the titration is pink colourless. The vinegar must be diluted. This is done to avoid having to use a large volume of sodium hydroxide solution to neutralise it. Also, since the vinegar is much more concentrated than the sodium hydroxide solution, a very small amount of vinegar would contribute to a higher percentage error in the experiment. Carry out titration as usual and calculate the molarity of the acid using the titration equation. Ethanoic acid Dilute NaOH + phenolphthalein indicator 4

6 Powered By: (4) To determine the amount of water of crystallisation in hydrated sodium carbonate (washing soda) A solution of washing soda (sodium carbonate Na2CO3) is prepared and titrated against a standard solution of HCl. The titration is identical to the rlier titration in which a solution of hydrochloric acid was standardised using a standard solution of sodium carbonate. Hence the steps are exactly the same and the indicator used is methyl orange. Again the colour change for the titration is orange pink. This time however, the concentration of the HCl is accurately known and it is the molarity of the sodium carbonate we wish to determine. Once the titration has been carried out, we can use the titration equation to solve for the concentration of sodium carbonate in the washing soda. This will allow us to calculate the amount of water of crystallisation in the washing soda. 5

7 Powered By: Sample Exam Questions 2016 (Vinegar) a) i) Pour some of the vinegar into a cln, dry bker. (Ensure there is more than 25cm3 of vinegar in the bker) ii) Using a cln, dry pipette and a pipette filler, transfer 25cm3 of the vinegar to a cln, dry 250cm3 volumetric flask using a funnel with washings from the pipette and funnel. iii) Make up to the mark in the volumetric flask using deionised water, adding drop wise using a dropping pipette when solution is with 1cm3 of the mark. Ensure to rd from the bottom of the meniscus. iv) Stopper and invert flask 20 times to make solution homogeneous. b) Phenolphthalein Pink colourless 6

8 Powered By: c) i) Since we are not given the molarity of the sodium hydroxide we must calculate it ourselves. We know that 1.2g of sodium hydroxide was used in a 500cm3 solution This implies that there would have been 2.4 g of the NaOH in a litre of solution. Since the relative molecular mass of NaOH is ( ) 40, (i.e. there are 40 g in one mole of NaOH) we can calculate the molarity of the NaOH used here to be: = 0.06M (this is the molarity of the NaOH) Hence in 25cm3 of the NaOH x 0.06 = 1.5x10-3 moles of NOH in ch 25cm3 portion ii) Now we can use the titration equation to solve for the molarity of the vinegar Ma - x Va na 1 Mb 0.06 Vb 25 nb - 1 Solving for x gives x =0.08 M (this is the number of moles of vinegar per litre, i.e. per 1000cm3) Since the question requires us to calculate the number of moles per cm3 we must divide this answer by 1000 = 8x10-5 moles (this is the number of moles of vinegar per cm3) d) i) We found that the molarity of the vinegar was 0.08 moles Since the original vinegar was diluted by a factor of 10 we must multiply the molarity of the vinegar by 10 to calculate the molarity of the original vinegar sample. (0.08)(10) = 0.8M 7

9 Powered By: ii) Weight/volume mns grams/cm3 Mr of ethanoic acid = 60 (i.e. there are 60g in a mole of ethanoic acid) Since we have 0.8 moles of ethanoic acid in one litre, 60 x 0.8 = 48g (We have 48 g of ethanoic acid in one litre) Since there are 1000cm3 in one litre dividing 48 by 1000 will give us the amount of grams of ethanoic acid in 1cm3 = g/cm3 Since we wish to calculate the percentage weight per volume, we essentially want to calculate the number of g in 100cm3. Therefore by multiplying our answer by 100 we will get a value for the % w/v of ethanoic acid in the original vinegar. (0.048)(100) = 4.8% (w/v) 8

10 Powered By: 2012 (HCl/sodium carbonate titration) a) A primary standard is a compound that can be obtained in a pure, soluble, stable and solid form so that it can be weighed out and dissolved in deionised water to make a solution of accurately known concentration. b) 1. Empty the contents of the clock glass into a cln, dry bker. Rinse down the clock glass with deionised water into the bker. 2. Add sufficient deionised water to the bker and stir using a stirring rod until the sodium carbonate is fully dissolved. 3. Transfer the solution to a 500cm3 volumetric flask using a funnel with washings from the funnel, stirring rod and bker. 4. Make up to the mark with deionised water in the volumetric flask, add drop wise using a dropping pipette nr the end point. Ensure to rd from the bottom of the meniscus. 5. Stopper and invert flask 20 times to make solution homogeneous. c) i) The burette was filled above the zero mark. The tap of the burette was then opened until the bottom of the meniscus rested on the zero mark. 9

11 Powered By: ii) The solution may be corrosive so using your mouth to suck up solution into pipette would be dangerous. d) Methyl orange. orange pink e)i) Ma - x Va 20.8 na 2 Mb 0.05 Vb 25 nb - 1 x = 0.12 M ii) Mr of HCl = (1) + (35.5) = 36.5g This is the number of grams of HCl in one mole Hence the concentration of HCl in grams/litre in this titration was (36.5)(0.12) = 4.38g/L 10

Powered By: 2014 (Washing soda crystals) a) A standardised solution is one whose concentration has been accurately determined by another titration. b) 1. Transfer the crystals to a cln, dry bker. 2. Rinse down the clock glass with deionised water and empty these washings into the bker.

12 Powered By: 2014 (Washing soda crystals) a) A standardised solution is one whose concentration has been accurately determined by another titration. b) 1. Transfer the crystals to a cln, dry bker. 2. Rinse down the clock glass with deionised water and empty these washings into the bker. 3. Add deionised water to the bker whilst stirring with a stirring rod until the crystals are fully dissolved. 4. Transfer the solution to a 250cm3 volumetric flask using a funnel with washings from the bker, stirring rod and funnel. 5. Make up to the mark in the volumetric flask using deionised water, adding drop wise using a dropping pipette nr the end point. Ensure to rd from the bottom of the meniscus. 6. Stopper and invert the flask 20 times to make the solution homogeneous. c) Rinse down the side of the conical flask with deionised water at regular intervals to ensure thorough mixing of the rctants. - Any of the solution from the burette which may have fallen onto the side of the conical flask and did not mix with the solution in the conical flask will be washed into the solution to ensure the volume rding from the burette is accurate. d) methyl orange orange pink 11

13 Powered By: e) i) Ma 0.1 Va 21.6 na 2 Mb x Vb 25 nb - 1 x = M ii) Mr of Na2CO3 = 106 (0.0432)(106) = g/L of Na2CO3 iii) Since there are g of Na2CO3 in one litre There are g of Na2CO3 in 250cm3 Since we used 2.5 g of the crystals in 250cm3, the difference must be the mass of the water in the crystals = g of H2O So to calculate the percentage water of crystallisation x 100 = % To calculate the value of x = moles/250cm3 of H2O Mr of H2O And we know that there are moles of Na2CO3 in 250cm3 =

14 Powered By: (1) To prepare a standard solution of ammonia iron (II) sulfate, to use this solution to standardise a solution of potassium permanganate by titration and hence be able to determine the amount of iron in an iron tablet MnO4- + 8H+ + 5Fe2+ Mn2+ + 5Fe3+ + 4H2O Ammonia Iron II sulphate Reducing agent Ammonia Iron II sulphate is a primary standard as it is available in a highly pure state and is stable. Iron II sulphate can t be used as a primary standard as the crystals tend to lose their water of crystallisation to the air, i.e. they tend to be oxidised slightly by the air When making up the ammonia iron II sulphate solution sulfuric acid (H2SO4) is added to prevent the Fe2+ ions being oxidised to Fe3+ ions by oxygen in the air. Potassium permanganate (KMnO4) Oxidising agent It is not a primary standard as it cannot be obtained in a state of high purity and it decomposes in the presence of sunlight. KMnO4 is placed in the burette and is rd from the top of the meniscus due to the intense purple colour (meniscus is not visible). Dilute sulfuric acid is added in excess to the conical flask before the titration to ensure that all of the MnO4- ions are reduced to Mn2+. Otherwise, if none or insufficient acid is used, a brown precipitate is formed due to the presence of the intermediate compound MnO2 MnO4- acts as its own indicator This is because when all the Fe2+ ions have rcted, the next drop of KMnO4 solution gives a pale permanent pink colour. Hence, the colour change is colourless pale permanent pink colour 2+ Mn is an autocatalyst an autocatalyst is a product of a rction which catalyses the rction 13

15 Powered By: The titration between the potassium permanganate and the ammonia iron II sulfate is carried out in the usual titration manner. Potassium Permanganate Ammonia Iron (II) Sulphate The titration formula can again be used to calculate the concentration of the potassium permanganate Molarity of reducing agent Volume of reducing agent Volume of oxidising agent Number of moles of the reducing agent (from the balanced equation) Molarity of oxidising agent Number of moles of the oxidising agent To calculate the amount of iron in an iron tablet: The titration between the potassium permanganate and the iron (II) sulfate (which is obtained from the iron tablets) is carried out in an identical manner. This time however we use the potassium permanganate solution we have just standardised to be able to find the concentration of the iron (II) sulfate solution. i.e. the concentration of iron (II) sulfate in the iron tablets. Since potassium permanganate decomposes in the presence of sunlight, the standardised solution must be prepared immediately before it is used to determine the amount of iron in iron tablets. 14

16 Powered By: 2003 a) To prevent atmospheric oxygen from oxidising the Fe2+ ions to Fe3+ and to also help to dissolve the tablets. b) 1. Crush the tablets using a pestle and mortar in dilute sulfuric acid. 2. Transfer the crushed tablets to a bker. Wash the pestle and mortar with deionised water into the bker. 3. Add sufficient deionised water to the bker and stir using a stirring rod to dissolve the powder. 4. Transfer the solution using a funnel to a 250cm3 volumetric flask with washings from the funnel, bker and stirring rod. 5. Make up to the mark in the volumetric flask using deionised water, adding drop wise nr the end point using a dropping pipette. Ensure to rd from the bottom of the meniscus. 6. Stopper and invert flask 20 times to make solution homogeneous. c) To ensure sufficient H+ ions were present to ensure the complete reduction of the MnO4- ions to Mn2+ ions. Otherwise a brown precipitate would have been formed in the conical flask due to the presence of the intermediate compound MnO2. d) The colour change of colourless point of the titration. first permanent pale pink colour was the end 15

17 Powered By: e) i) Mo 0.01 Vo 13.9 no 1 Mred x Vr 25 nr - 5 x = M ii) Mr of Fe is 56 (56)(0.0278) = g/l of Fe g/250 cm3 of Fe Since there were 4 tablets in the 250 cm3 solution, = g/tablet of Fe x 100 = 27% iii) 16

18 Powered By: To prepare a solution of sodium thiosulfate, to standardise it by titration against a solution of iodine and hence to be able to determine the percentage (w/v) of sodium hypochlorite in blch Iodine oxidising agent A standard solution of iodine cannot be made up by directly weighing as iodine does not dissolve in water and it vaporises at room temperature. A standard solution of iodine is obtained by rcting a standard solution of potassium permanganate with excess potassium iodide solution 2MnO I- + 16H+ 2Mn2+ + 5I2 + 8H2O The I- ions are kept in excess This is done to help keep the iodine ions that are produced in solution. The mns that amount of iodine formed depends on the amount of KMnO4 present. Sulfuric acid is added in excess to the KMnO4 solution in the conical flask to supply sufficient H+ ions for the complete reduction of the MnO4- ions to Mn2+ ions. Sodium Thiosulphate reducing agent Sodium thiosulphate is used for extracting gold and to fix the image when developing photographic film Sodium Thiosulphate is not a primary standard as it cannot be obtained in a sufficiently pure state and the crystals are efflorescent (they spontaneously lose their water of crystallisation to the air.) Sodium Thiosulphate is a reducing agent and it rcts with iodine molecules (I2) to convert them into iodide ions I- I2 + 2S2O32- S4O I- 17

19 Powered By: Deionised water should be used as tap water contains small amounts of chlorine. Chlorine displaces iodine from iodide solutions, which would incrse the amount of iodine present, lding to an inaccurate result. Method 1. A solution of sodium thiosulphate is prepared and placed in the burette. 2. Pipette a standardised solution of potassium permanganate into the conical flask. 3. Add some dilute sulfuric acid to the conical flask immediately To supply sufficient H+ ions for the complete reduction of the MnO4ions to Mn2+ ions. 4. Add potassium iodide in excess to the conical flask, The solution in the conical flask tus a red brown colour. This is the colour of the liberated iodine, which was liberated according to the equation 2MnO I- + 16H+ 2Mn2+ + 5I2 + 8H2O 5. Carry out titration as usual. I2 + 2S2O32- S4O I- The red/brown colour becomes less and less intense as the iodine is used up and fades to a pale yellow colour. The solution will go colourless when all the iodine is used up but since the solution fades slowly, the endpoint is not clr. 6. Add a few drops of freshly prepared starch solution. The starch solution rcts with the small amount of iodine left in the conical flask and will tu blue/black in colour. 7. The titration is continued until the blue/black colour disapprs. i.e. the colour change is blue/black colourless The blue/black colour disapprs when there is no more iodine left in the conical flask, i.e. all of the iodine has rcted with the sodium thiosulphate. Hence, the starch is simply used to make the end point of the titration obvious, it does not become involved in the rction. If the starch is added too rly the iodine is strongly absorbed onto it and the accuracy of the value for the endpoint is reduced. 8. Note the titration figure from the burette. 18

20 Powered By: To determine the percentage (w/v) of sodium hypochlorite (NaClO) in household blch Blches contain oxidising agents, the most common one being sodium hypochlorite, NaClO. In acidic solution hypochlorite ions oxidise iodide ions I- to form I2. The iodine that is produced is then titrated against a standard solution of sodium thiosulphate. 2H+ + ClO- + 2I- Cl- + I2 + H2O The titration carried out in the same manner as the previous experiment. This time however the concentration of the sodium thiosulphate solution is known and it is the concentration of the iodine ions that we are trying to determine. From that we will be able to calculate the concentration of NaClO in the sample of the blch. Household blch is highly concentrated so it must first be diluted. Otherwise a large amount of sodium thiosulphate would be required to rct with the very large amount of iodine liberated. The potassium iodide acts to keep the iodine in solution and so it is added in excess. As before, sulfuric acid is added in excess to the conical flask to supply sufficient H+ ions so that all of the hypochlorite ions (ClO-) are completely reduced to Chlorine ions (Cl-) 19

Powered By: Sample Exam Question 2007 a) A primary standard is a substance that can be obtained in a soluble, solid, pure and stable form so that it can be weighed out and dissolved in deionised

21 Powered By: Sample Exam Question 2007 a) A primary standard is a substance that can be obtained in a soluble, solid, pure and stable form so that it can be weighed out and dissolved in deionised water to give a solution of accurately know concentration. b) 1. Transfer crystals to a cln, dry bker. Wash down the clock glass with deionised water into the bker. 2. Add sufficient deionised water to the bker and stir using a stirring rod to dissolve crystals. 3. Transfer solution to a 500cm3 volumetric flask using a funnel with washings from the funnel, stirring rod and bker. 4. Make up to the mark in the volumetric flask with deionised water, adding drop wise nr the endpoint using a dropping pipette. Ensure to rd from the bottom of the meniscus. 5. Stopper and invert the flask 20 times to make solution homogeneous. c) Potassium iodide. d) red/brown (iodine) pale yellow (fading as free iodine is used up) blue/black (starch forms a blue/black compound with remaining iodine) colourless (no iodine left, end point) 20

22 Powered By: e) Mo 0.05 Vo 25 no 1 Mred x Vr 20 nr - 2 x = M Mr of Na2S2O3 = 248 (248)(0.125) = 31g/L 21

Volumetric Analysis: Redox

Volumetric Analysis: Redox Name: Volumetric Analysis Objectives 3. Volumetric Analysis carry out a potassium manganate(vii)/ammonium iron(ii) sulfate titration determine the amount of iron in an iron tablet carry out an iodine/thiosulfate