ࡱ>     '` mbjbjLULU .r.?.?WQq````jfzfzfzfz |% ~6egggggg$hl`9``ĭ `leer T, ~ fzU0a9,ڭ0 o̲&̲ ̲͢lHDŽm %%%<"aD%%%"a f`````` F6664.7 44.7 4.74Acid/base equilibria 4.7a The meaning of acidity 4.7a demonstrate an understanding that the theory about acidity developed in the 19th and 20th centuries from a substance with a sour taste to a substance which produces an excess of hydrogen ions in solution (Arrhenius theory) to the Brnsted-Lowry theory  INCLUDEPICTURE "http://crystalcraven.files.wordpress.com/2009/07/lemon.jpg" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.asaltandbattery.com/assets/images/products/copies/lg_vinegar.jpg" \* MERGEFORMATINET  Early ideas about acids linked to their sour taste and their use in foods. Citric acid in lemons and ethanoic acid in vinegar are examples. By 1777 Antoine Lavoisier was defining acids as compounds containing oxygen, but then someone discovered HCl. By 1810 Humphrey Davy was saying acids contained hydrogen. By 1838 Justin von Leibig was defining acids as producing hydrogen with metals. By 1884 Svante Arrhenius had proposed that aqueous acid solutions contained hydrogen ions. Acids dissociate in water: e.g. HNO3(aq) ( H+(aq) + NO3-(aq) In 1923 Bronsted and Lowry side stepped the need for water by defining acids as proton donors. So conc. sulfuric acid behaves like an acid when mixed with conc nitric acid even with no water present. H2SO4(l) + HNO3(l) ( HSO4-(l) + H2NO3+(l) Task4.7a State whether the following are acids or bases and why: orange juice, CH3COOH(aq), OH-(aq), HCl(aq), HCl(g), HNO3(l), NH3(g), H2O(l), HW 4.7a Explain what is meant by the correct definition of an acid 4.7b BrnstedLowry acid-base equilibria 4.7b demonstrate an understanding that a BrnstedLowry acid is a proton donor and a base a proton acceptor and that acid-base equilibria involve transfer of protons If an acid is a proton donor then an ammonium ion NH4+ can be an acid. NH4+ (aq) ( NH3(aq) + H+(aq) A base can have a similar definition. A base is a proton (hydrogen ion) acceptor. This means that water can act as a base. H2O(l) + H+(aq) ( H3O+(aq) Many acid base reaction are better written as equilibria e.g. H2SO4(l) + HNO3(l) ( HSO4-(l) + H2NO3+(l) Note HNO3(l) now behaves like a base. Task 4.7b Write equations for equilibria reactions between: hydrogen ions and hydroxide ions, ammonium ions and water molecules, methanoic acid and water, water and carbonate ions. 4.7c conjugate acid-base pairs 4.7c demonstrate understanding of the BrnstedLowry theory of acid-base behaviour, and use it to identify conjugate acid-base pairs Any acid which loses a proton forms a species which can gain its proton back. This new species is a base, a so called conjugate base. CH3COOH(aq) + H2O(l) ( CH3COO-(aq) + H3O+(aq) acid 1 base 2 conjugate base 1 conjugate acid 2 HCl(aq) + H20 (l) <--------> H3O+ (aq) + Cl-(aq) acid 1 base 2 acid 2 base 1 conjugate acid conjugate base H3O+ called oxonium ion. HCl and CH3COOH, can only donate one proton each so are called monoprotic acids and each only have a single conjugate base so are monobasic acids. Water, which accepts the proton, is a base. In the reverse reaction, oxonium ion is an acid because it donates a proton to acetate ion and acetate ion is a base because it accepts a proton. An acid which donates more than one proton is polyprotic. An acid which has two conjugate bases is dibasic e.g. H2SO4 Water can also act as an acid by donating a proton to a weak base. H2O(aq) + NH3(aq) <-----> NH4+(aq) + OH-(aq) acid 1 base 1 conjugate acid 2 conjugate base 2 Task 4.7c Identify the acids and bases in the following reactions: H2SO4(l) + HNO3(l) <-----> HSO4-(l) + H2NO3+(l) equationH2SO4 + H2O  INCLUDEPICTURE "http://dl.clackamas.edu/ch105-04/rtarrow.gif" \* MERGEFORMATINET HSO4- + H3O+HCl + OH-  INCLUDEPICTURE "http://dl.clackamas.edu/ch105-04/rtarrow.gif" \* MERGEFORMATINET Cl- + H2OHCl + NH3  INCLUDEPICTURE "http://dl.clackamas.edu/ch105-04/rtarrow.gif" \* MERGEFORMATINET Cl- + NH4+NH3 + H2O  INCLUDEPICTURE "http://dl.clackamas.edu/ch105-04/rtarrow.gif" \* MERGEFORMATINET NH4+ + OH- What are the conjugate bases of these acids? original acidconjugate baseHNO3H2OH3O+H2SO4HBrHCO3-What are the conjugate acids of these bases? original baseconjugate acidOH-H2OHCO3-SO42-ClO4-  Label the acids, conjugate acids, bases and conjugate bases in the following equationH2SO4 + H2O  INCLUDEPICTURE "http://dl.clackamas.edu/ch105-04/rtarrow.gif" \* MERGEFORMATINET HSO4- + H3O+HCl + OH-  INCLUDEPICTURE "http://dl.clackamas.edu/ch105-04/rtarrow.gif" \* MERGEFORMATINET Cl- + H2OHCl + NH3  INCLUDEPICTURE "http://dl.clackamas.edu/ch105-04/rtarrow.gif" \* MERGEFORMATINET Cl- + NH4+NH3 + H2O  INCLUDEPICTURE "http://dl.clackamas.edu/ch105-04/rtarrow.gif" \* MERGEFORMATINET NH4+ + OH- 4.7d pH, Ka and Kw, pKa and pKw 4.7d define the terms pH, Ka and Kw, pKa and pKw, and be able to carry out calculations relating the pH of strong acids and bases to their concentrations in mol dm-3 pH = -log[H+(aq)] where [H+] = the hydrogen ion concentration at equilibrium What is the pH of a solution where [H+] is 0.0250M i.e. 2.50*10-2moldm-3? pH = -log[H+] pH = -log(2.50*10-2 ) pH = -(-1.60) = 1.60 What is [H+] of a solution of HCl of pH 2.30? pH = -log[H+] 2.30 = -log[H+] log[H+] = -2.30 [H+] = antilog -2.30 [H+] = 10 -2.30 = 5.01*10-3 moldm-3 In a pure sample of water, some of the molecules will spontaneously transfer a proton from one water molecule to another. The equilibrium reaction is written as: 2 H20 (1) <----> H3O+(aq) + OH-(aq) Kw= [H3O+(aq)][ OH-(aq)] = 1.0 X 10-14 mol2dm-6 Kw is known as the ionisation constant or ionic product of water. It is a measure of how many water molecules ionise spontaneously. Measurements have shown that this equilibrium is far to the left, with only a small concentration of the ions. For the general weak acid HA we can write the following equilibrium: HA(aq) + H2O(l) <-----> H3O+(aq) + A-(aq) The equilibrium constant (acid dissociation constant) Ka is Ka= [ H3O+][ A-] / [HA] units moldm-3 In each case the concentration of water is incorporated into Kc so Kw =Kc[H2O] and Ka=Kc[H2O] Ka and Kw are normally very small inconvenient numbers. E.g. Ka for ethanoic acid is about 10-5 mol dm-3 and Kw is about 10-14 mol2dm-6. pKa and pKw give more convenient numbers in the same way that pH values are easier than hydrogen ion concentrations. pKa = -logKa pKw = - logKw Task 4.7d.1 Calculate the pH of a solution with [H+] = 5 x 10-5 M. 2. Calculate the pH of a solution with [H+] = 1 M. 3. Calculate the pH of a 0.01 M solution of HCl. 4. Calculate the pH of a 0.05 M solution of NaOH.. 5. Calculate the pH of a 7.5 x 10-6 M solution of Mg(OH)2. 6. Find [H+] of a solution with pH = 3. 7. Find [OH-] of a solution with pH = 8. 8. A 1.0 L solution of HCl has a pH = 1. How many litres of distilled water must be added to change the pH to 2? 9. 6 g of LiOH is added to water to make 500 ml of solution. What is the pH? Answers: 1. 4.4, 2. 0, 3. 2, 4. 1.30, 5. 9.176, 6. 10-3, 7. 10-6. 8. 9dm3, 9.13.7 What is the pH of a 0.2M solution of sodium hydroxide given that Kw=1.0 * 10-14 mol2dm-6? 4.7e The equilibrium law, Ka and KwUnit 4.7e demonstrate an understanding that weak acids and bases are only slightly dissociated in aqueous solution, and apply the equilibrium law to deduce the expressions for the equilibrium constants Ka and KwUnit 4 A weak acid forms few hydrogen ions. In the equilibrium reaction which is its dissociation the position of the equilibrium lies far to the left and the concentration of hydrogen ion is low. HA(aq) + H2O(l) ( H3O+(aq) + A-(aq) The equilibrium constant Kc can be calculated from: Kc=[ H3O+(aq)][ A-(aq)] / [HA(aq)][ H2O(l)] Kc[ H2O(l)]=[ H3O+(aq)][ A-(aq)] / [HA(aq)] The concentration of a pure liquid like water [ H2O(l)] is constant so Kc[ H2O(l)] is also a constant called the acid dissociation constant Ka. Ka = [ H3O+(aq)][ A-(aq)] / [HA(aq)] Pure water can be viewed as a weak acid. H2O(l)] + H2O(l)] ( H3O+(aq) + OH-(aq) The hydrogen ion concentration is very low as little water dissociates. Kc=[ H3O+(aq)][ OH-(aq)] / [H2O(l)][ H2O(l)] [H2O(l)] is a constant so Kc[H2O(l)][ H2O(l)]=[ H3O+(aq)][ OH-(aq)] Kc[H2O(l)][ H2O(l)] is also therefore a constant called the ionic product of water Kw. Kw = [ H3O+(aq)][ OH-(aq)] A weak base is also poorly reacted with water. NH3(aq) + H2O(aq) ( NH4+(aq) + OH-(aq) Many molecules of ammonia remain unreacted so the hydroxide concentration is low. The position of this equilibrium is also far to the left. Task 4.7e.1 Write equilibrium equations to show the dissociation of methanoic acid, hydrogen carbonate ion and the equilibrium between carbonate ions and water to show weakly basic properties. Write expressions for Ka or Kc in each case. Task 4.7e.2 Complete the table for the weak acid HX and H2X acid[HX] or [H2X][H]equ[X]equKaHX10.010.01HX0.0030.0020.00003H2X20.0020.003H2X1.50.0000050.000002 4.7f Analysing experimental results: 4.7f analyse the results obtained from the following experiments: i measuring the pH of a variety of substances, eg equimolar solutions of strong and weak acids, strong and weak bases and salts ii comparing the pH of a strong acid and a weak acid after dilution 10, 100 and 1000 times Results solutionpH1M HClCalc= exp =10cm3 1M HCl in 100cm3 solutionCalc= exp =1cm3 1M HCl in 100cm3 solutionCalc= exp =0.1cm3 1M HCl in 100cm3 solutionCalc= exp =1M NaOHCalc= exp =1M CH3COOH Ka=1.74*10-5moldm-31M NH31M NaCl1M NH4Cl1M Na2CO3Equal volumes 1M HCl + 1M NaOHEqual volumes 1M CH3COOH + 1M NaOHEqual volumes 1M HCl + 1M NH3 Results solutionpH1M HClCalc= 0 Exp= 0.24 10cm3 1M HCl in 100cm3 solutionCalc= 1 exp= 1.101cm3 1M HCl in 100cm3 solutionCalc= 2 Exp=2.150.1cm3 1M HCl in 100cm3 solutionCalc= 3 Exp=2.951M NaOHCalc= Exp=13.501M CH3COOH Ka=1.74*10-5moldm-3 Exp=3.111M NH3 exp=12.111M NaCl Exp=6.691M NH4Cl Exp=5.231M Na2CO3 Exp=10.10Equal volumes 1M HCl + 1M NaOHEqual volumes 1M CH3COOH + 1M NaOHEqual volumes 1M HCl + 1M NH3Task 4.7f Calculate the expected pH of each HCl and NaOH solution above 4.7g Calculating Ka from pH for a weak acid HA. 4.7g analyse and evaluate the results obtained from experiments to determine Ka for a weak acid by measuring the pH of a solution containing a known mass of acid, and discuss the assumptions made in this calculation What is Ka for a 0.100M solution of a weak acid HA of pH 5? pH =-log[H+(aq)] 5=-log[H+(aq)] log[H+(aq)] = -5 [H+(aq)] = 0.00001moldm-3 HA(aq) ( H+(aq) + A-(aq) Initial concs 0.100M 0.00 0.00 Equ concs 0.1-0.00001M 0.00001M 0.00001M Assume 0.100-0.00001 = 0.100 so Ka=[H+(aq)] * [A-(aq) / [HA(aq)] Ka = 0.00001*0.00001/0.100 = 1*10-9moldm-3 Task 4.7g Calculate Ka for the following weak acids: 0.0400 moldm-3 HX if its pH is 4.50 0.0100M carbonic acid solution H2CO3 if its pH is 4.17 5.0*10-3moldm-3 benzoic acid C6H5COOH if its pH is 3.28 Answers: (a) 2.49*10-8 moldm-3 (b) 4.57*10-7 moldm-3 (c) 5.51*10-11 moldm-3 4.7h Calculating the pH of a solution of a weak acid using Ka 4.7h calculate the pH of a solution of a weak acid based on data for concentration and Ka, and discuss the assumptions made in this calculation What is the pH of 0.100moldm-3 ethanoic acid if Ka =1.74*10-5moldm-3? CH3COOH(aq) ( CH3COO-(aq) + H+(aq) Initial conc 0.100 0 0 Equilib conc 0.100 - [H+] [H+] [H+] For weak acid 0.100 - [H+] H" 0.100 Ka= [CH3COO-(aq)][H+(aq)]/[ CH3COOH(aq)] Ka= [H+(aq)]2/[ CH3COOH(aq)] 1.74*10-5moldm-3 = [H+(aq)]2/ 0.100moldm-3 [H+(aq)] = 1.32*10-3moldm-3 pH = -log[H+] = -log 1.32*10-3 pH = 2.33 Task 4.7h Calculate the pH for (a) 0.200moldm-3 HF if Ka=5.62*10-4moldm-3 1.97 (b) 0.0200moldm-3 chloric(I) acid HOCl if Ka=3.72*10-8moldm-3 4.56 (c) 0.0500moldm-3 nitrous acid HNO2 if Ka =4.57*10-4 moldm-3 (d) 1.00moldm-3 iodoethanoic acid, CH2ICOOH if Ka =6.76*10-4 moldm-3 (e) 0.100moldm-3 propanoic acid C2H5COOH of pKa = 4.87 4.7i Titration curves 4.7i measure the pH change during titrations and draw titration curves using different combinations of strong and weak monobasic acids and bases weak acid weak base weak acid strong base  INCLUDEPICTURE "http://www.drbateman.net/asa2sums/sum4.4/sum4.41.jpg" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.drbateman.net/asa2sums/sum4.4/sum4.42.jpg" \* MERGEFORMATINET  strong acid weak base strong acid strong base  INCLUDEPICTURE "http://www.drbateman.net/asa2sums/sum4.4/sum4.43.jpg" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.drbateman.net/asa2sums/sum4.4/sum4.44.jpg" \* MERGEFORMATINET  Task 4.7i Plot pH against vol of alkali for the following data to determine whether the acid and alkali involved in the titration are strong or weak. The table shows pH when 0.100moldm-3 monobasic alkali is added to 25.0cm3 of 0.100moldm-3 monobasic acid. Volume of alkali/cm3pH0.02.883.03.896.04.269.04.5112.04.7315.04.9418.05.1721.05.4824.06.1424.56.45.24.97.1625.08.7325.111.3025.512.0027.012.59HW 4.7i Sketch curves for titrations with the named acids and alkalis below: HCl and NaOH, HNO3 and KOH, HCl and NH4OH, CH3COOH and KOH. 4.7j Selecting indicators 4.7j use data about indicators, together with titration curves, to select a suitable indicator and the use of titrations in analysis indicatorAcid colourpH rangeAlkali colourMethyl orangered2.9-4.0yellowCongo redviolet3.0-5.0redBromothymol blueyellow6.0-7.6bluephenolphthaleincolourless8.2-10.0redAlizarin yellowyellow10.1-13.0redMethyl orange changes colour at a low pH. This is suitable for a titration between a strong acid (e.g. HCl) and a weak base (e.g. ammonia) because the end point of the reaction is at low pH. Methyl orange is not suitable for a weak base strong acid titration as it will change colour long before the end point is reached. Use site below to see how pH curves determine indicator to use.  HYPERLINK "http://www.avogadro.co.uk/chemeqm/acidbase/titration/phcurves.htm" http://www.avogadro.co.uk/chemeqm/acidbase/titration/phcurves.htm Task 4.7j Sketch the 4 acid base titration curves and label the pH ranges over which suitable named indicators change colour. Indicators are generally weak acids, HIn, in an equilibrium with water. The colour depends on the position of the equilibrium which in turn depends on the total [H30+], that is on the pH of the solution to which the indicator is added. HIn(aq) + H2O(l) -----> H30+(aq) + In-(aq) colour 1 colour 2 The pH at which the colour changes depends on the acid dissociation constant Ka, for the indicator known as the indicator constant KIn. KIn = [H3O+][In-] / [HIn] pKIn = -logKIn The colour of the indicator changes from one colour to another at its end point. At the end point [HIn] = [In-] Therefore, at the end point for an indicator (this is when half the indicator is in acid form and half is in the form of its conjugate base.): KIn = [H3O+] therefore -log KIn = -log[H3O+] and pKIn = pH HW4.7.j Match suitable indicators to titrations with selected acids and alkalis 4.7k Buffer solutions 4.7 k explain the action of buffer solutions and carry out calculations on the pH of buffer solutions, eg making buffer solutions and comparing the effect of adding acid or alkali on the pH of the buffer in cars... the cooling system that uses water also has buffer in it to prevent it from going too acidic or basic. in washing... to a certain extent the soaps that you use buffer the water so that the detergents and soap work to clean your dishes or clothes. in swimming pools or hot tubs... the water is carefully buffered to 6.8-7.6 A buffer is a solution that absorbs small amounts of added acid or base while changing little in pH. A common buffer mixture contains a weak base and its salt or a weak acid and its salt e.g. a mixture of ethanoic acid and sodium ethanoate is a good buffer. Two equilibria exist in this buffer. CH3COOH(aq) <-----> CH3COO-(aq) + H+(aq) If a small amount of H+(aq) is added it reacts with CH3COO- which shifts the position of equilibrium to the left thus keeping [H+], and hence pH, almost constant. If a small amount of OH-(aq) is added it reacts with H+ which shifts the position of equilibrium to the right producing more H+ thus keeping the pH almost constant. [CH3COO-(aq)] is large and kept constant because the equilibrium below lies far to the right. CH3COONa(aq) -----> CH3COO-(aq) + Na+(aq) Buffered solutions do change in pH upon addition of H+ or OH- ions. However, the change is much less than that would occur if no buffer were present. Solutions which are very strongly acidic or very strongly basic are rarely deliberately buffered by addition of a chemical buffer because it already takes a relatively large amount of strong acid or strong base to change their pH significantly. Calculate the pH of a buffer made by dissolving 0.200 mol of ethanoic acid and 0.200mol of sodium ethanoate to give 1 dm3 of buffer solution given Ka ethanoic acid = 1.7*10-5moldm-3 initial conc 0.200 0.200 0 CH3COOH(aq) <----> CH3COO-(aq) + H+(aq) equilib conc 0.200 - x 0.200 + x x assume 0.200 - x and 0.200 + x = 0.200 Ka =[H+][CH3COO-]/[CH3COOH] [H+] = Ka [CH3COOH]/[CH3COO-] -log[H+] = -logKa -log[CH3COOH]/[CH3COO-] pH = pKa -log[CH3COOH] /[ CH3COO-]= -logKa - log(0.200/0.200) pH = -log 1.7*10-5- log(0.200/0.200) = 4.77 Task 4.7k What is the pH of a buffer solution made by adding 2.05g sodium ethanoate to 1000cm3 of 0.09M ethanoic acid (Ka ethanoic acid = 1.7*10-5moldm-3) pH=4.21 4.7l Buffers and Titration curves use titration curves to show the buffer action and to determine Ka from the pH at the point where half the acid is neutralised The buffer range is the part of the titration curve where the pH change is gradual. A mixture of weak acid and salt is present here. Acid <-> conjugate base + H+ Ka=[H+][base]/[acid] When half of the acid is neutralised [base] = [acid] so Log Ka = log[H+] -Log Ka = -log[H+] -Log Ka = pH pKa = pH Task 4.7l.1 When a solution of ethanoic acid is half neutralised the pH of the solution is 4.77. What is Ka for ethanoic acid? 1.70*10-5 moldm-3 Task 4.7l.2 When a solution of benzoic acid is half neutralised the pH of the solution is 4.21. What is Ka for benzoic acid? 6.17 *10-5 moldm-3  4.7m Buffers in biological systems 4.7m explain the importance of buffer solutions in biological environments, eg buffers in cells and in blood (H2CO3/HCO3-) and in foods to prevent deterioration due to pH change (caused by bacterial or fungal activity). N Normal blood pH is about 7.4. Carbon dioxide from cells can react and drop the pH interfering with blood chemistry. Carbonic acid forms if acidity rises but an excess of hydrogen carbonate ions form if acidity falls. The blood contains moderate concentrations of sodium hydrogen carbonate to provide sufficient base for buffer action. bloNormal blood pH is aH2CO3(aq)  INCLUDEPICTURE "http://scifun.chem.wisc.edu/CHEMWEEK/arrowdbl.gif" \* MERGEFORMATINET H+(aq) + HCO3-(aq) bout The phosphate buffer system operates in the internal fluid of all cells. This buffer system consists of dihydrogen phosphate ions (H2PO4-) as hydrogen-ion donor (acid) and hydrogen phosphate ions (HPO42-) as hydrogen-ion acceptor (base). These two ions are in equilibrium with each other as indicated by the chemical equation below. 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