Acids and Bases. When an acid loses a proton, the resulting species is its conjugate base. For example, NH 3 + H +

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1 Acids and Bases Definitions An acid is a proton donor, e.g. HCl. For example, consider the reaction between HCl and H 2 O. HCl + H 2 O H 3 O + + Cl - Acid In this reaction, HCl donates a proton to H 2 O. A base is a proton acceptor, e.g. NH 3 For example, consider the reaction between NH 3 and H 2 O. NH 3 + H 2 O NH OH - Base In this reaction, NH 3 accepts a proton from H 2 O. Conjugate acid-base pairs When an acid loses a proton, the resulting species is its conjugate base. For example, CH 3 CO 2 H - CH 3 CO 2 Acid Conjugate base + H + NH 4 + Acid NH 3 + H + Conjugate base Its called the conjugate base because if you imagine the reaction going backwards, it would be the species acting as a base (whether the reaction is actually an equilibrium or not!) When a base gains a proton, the resulting species is its conjugate acid. For example, NH 3 + H + + NH 4 Base Conjugate acid OH - + H + H 2 O Base Conjugate acid In an exam, you will most likely be asked to label both pairs of conjugate acids and bases in a given equation. CH 3 CO 2 H + H 2 O - CH 3 CO 2 + H 3 O + Acid1 Base2 Con base1 Con acid2 Page 1 of 28

2 Acids and bases The terms conjugate acid and conjugate base are useful for describing certain situations, for example A solution of the conjugate base of a weak acid will be alkaline A solution of the conjugate acid of a weak base will be acidic A buffer can be made by mixing a solution of a weak acid with a solution of its conjugate base Label the conjugate acid-base pairs in the following reactions, using appropriate symbols CH 3 CO 2 H + HCl CH 3 CO 2 H Cl - HNO 3 + H 2 SO 4 H 2 NO HSO 4 - HCl + HF Cl - + H 2 F + HCO H 2 O CO H 3 O + NH 3 + H 3 O + NH H 2 O Page 2 of 28

3 Acids and Bases Strong and weak The terms strong and weak when applied to an acid or base indicate whether all or some of the molecules dissociate in aqueous solution. A strong acid fully dissociates in aqueous solution. For example, HCl + H 2 O H 3 O + + Cl - The important point here is that if, for example, you make up a 0.2 mol dm -3 solution of hydrochloric acid, the concentration of H 3 O + will also be 0.2 mol dm -3. We are interested in knowing the concentration of H 3 O + so we can work out ph of acidic solutions. A strong base also fully dissociates in aqueous solution. For example, NaOH Na + + OH - Again, this means we know the concentration of OH - ions if we know the concentration of the solution of NaOH, they will be the same. We are interested in knowing the concentration of OH - because it is related to the concentration of H +, so if we know it we can work out the ph of alkaline solutions. A weak acid does not fully dissociate in aqueous solution, even at high dilution. For example, CH 3 CO 2 H + H 2 O CH 3 CO H 3 O + In this case, we cannot work out the concentration of H 3 O + ions directly form the concentration of CH 3 CO 2 H, we also need to know the equilibrium constant for the acid dissociation. (To give an indication of what is going on in this situation, in a 0.1 mol dm -3 solution of ethanoic acid for example, it turns out that only 1.34% of the acid molecules are dissociated. The rest can dissociate, for example when it is neutralised by a base, also the reaction is in dynamic equilibrium, so isn t always the same 1.34% of molecules that are dissociated at any one time). A weak base also does not fully dissociate, even at high dilution. For example, NH 3 + H 2 O NH OH - As above, we cannot work out the concentration of OH - (and therefore H 3 O + ) without knowing the equilibrium constant for the above reaction. The point about even at high dilution is that if a weak acid or base is diluted, they do dissociate a little more, but they never manage to dissociate fully. Page 3 of 28

4 Acids and bases The p function The topic acids and bases involves lots of small numbers, which have to be written in exponential form, such as 1 x To make life easier (!) chemists often talk about the p function of a value rather, than the value itself. The p function of any value is the log 10 of it. What is the p value of 1.00 x 10-3? Answer = 3.00 Most commonly, you will be using the p value of the concentration of H 3 O + ions, which is known as ph. However, you will need to be comfortable using the p value of other things, such as poh, pk w, pk a, pk b and pk in If a solution has a concentration of H 3 O + ions of mol dm -3, calculate the ph of this solution. If the value of K w at 398K is 1.20 x 10-14, calculate pk w at this temperature. Answer = Answer = 13.9 You also need to be able to work back from a p value to the original value itself. What you do is change the sign of the p value work out the inverse log 10 of the p value What is the value of K w at 298 K if pk w is 14.0? What is the value of [H 3 O + ] if the ph of a solution is 1.00? Answer = 1.00 x mol 2 dm -6 Answer = mol dm -3 Page 4 of 28

5 Acids and Bases The auto-ionisation of water Water reacts with itself, reaching dynamic equilibrium with aqueous protons and hydroxide ions. H 2 O + H 2 O H 3 O + + OH - The equilibrium constant for this reaction, K w is 1.0 x mol 2 dm -6 at 298K NB it is a different value at different temperatures! The equilibrium expression for this reaction is K w = [H 3 O + ][OH - ] The ph of pure water In pure water, there is no additional source of protons or hydroxide ions. From the top equation, you can see that for every aqueous proton, there will also be an aqueous hydroxide ion, so their concentrations will be equal. In pure water, the concentrations of [H 3 O + ] and [OH - ] are the same Therefore, K w = [H 3 O + ] 2 and K w = [OH - ] 2 The concentration of [H 3 O + ] and [OH - ] is the square root of the K w value Calculate the ph of pure water at 298K, where K w is 1.00 x mol 2 dm -6 Calculate the ph of pure water at 398K, where K w is 1.1 x mol 2 dm -6 Answer = 7.00 What would be the poh? Answer =6.98 Answer = also ph and poh are equal in pure water to each other whatever the temperature. Using 14-pH only works at 298K. Page 5 of 28

6 Acids and bases Auto-ionisation of water, endothermic or exothermic? Based on the calculation on the previous page, you can deduce whether the autoionisation of water is endothermic or exothermic. When heated, the equilibrium constant increases, which means the position of equilibrium has shifted to the right-hand side Increased heat favours the endothermic pathway (Le Chatelier) Therefore the forward reaction is endothermic NB you could refer to the decrease in ph as being evidence that the position of equilibrium has shifted to the right-hand side. Some important points In pure water, the [H 3 O + ] = [OH - ] What makes water neutral is that the concentrations of aqueous protons and hydroxide ions are the same, which makes the ph and poh also the same For pure water at 298K, ph and poh are both 7, which is neutral for this temperature. At different temperatures, for pure water, ph and poh will be equal to each other, but not equal to 7 Acidic and alkaline solutions An acidic solution has a greater concentration of aqueous protons than hydroxide ions. An alkaline solution has a greater concentration of aqueous hydroxide ions. In any aqueous solution, whether neutral, acidic or alkaline, the K w expression can relate the concentration of protons to hydroxide ions. To work out the ph of an acidic solution, once you have worked out the [H 3 O + ] you simply work out the p value and that is the ph (see examples later). To work out the ph of an alkaline solution, you need to work out the [OH - ] ions and then relate that to the [H 3 O + ] using the K w expression (see examples later). At 298K, [H 3 O + ] = 1 x or [OH - ] or ph + poh = 14. How you calculate the [H 3 O + ] or[oh - ] depends on whether you or dealing with a strong or weak acid or base. Page 6 of 28

7 Acids and Bases Strong acids As we have seen, strong acids dissociate fully in aqueous solution, so it is very easy to work out the concentration of aqueous protons from the concentration of the acid. Calculate the ph of mol dm -3 HNO 3. Answer = Calculate the ph of mol dm -3 HCl. Calculate the ph of mol dm -3 H 2 SO 4, assuming both protons fully dissociate. Answer = 2.00 Answer = 1.00 In reality, the dissociation of sulphuric acid is more complicated. The first proton fully dissociates, whereas the resulting HSO 4 - ion acts as a weak acid. H 2 SO 4 + H 2 O HSO H 3 O + HSO H 2 O SO H 3 O + The second dissociation is shifted backwards by the H 3 O + ions in the first one, so does not dissociate as much as it would in the absence of the first reaction. This is an important point, it has come up in examinations. Another consequence of the fact that HSO 4 - is a weak acid is that a solution of sodium hydrogen sulphate will be acidic, due to the second reaction above. This fact enable HSO 4 - ions to be distinguished from SO 4 2- ions. Both form a white precipitate with BaCl 2 solution but in the case of a hydrogensulphate ion being present the initial solution will be acidic to litmus paper. Page 7 of 28

8 Acids and bases Strong bases Strong bases also fully dissociate, so the concentration of OH - ions can be worked out easily from the concentration of the base. Assume that the temperature is 298K. Calculate the ph of mol dm -3 NaOH. Calculate the ph of mol dm -3 KOH Answer = 13.3 Calculate the ph of mol dm -3 Ba(OH) 2. Answer = 12.0 Answer = 13.0 NB At 298K [H 3 O + ] = 1 x / [OH - ] ph + poh = 14. Page 8 of 28

9 Acids and Bases Weak acids Weak acids do not fully dissociate, in fact, the vast majority of a weak acid s molecules are not dissociated in aqueous solution. for ethanoic acid CH 3 CO 2 H + H 2 O CH 3 CO H 3 O + the acid dissociation constant is K a = [CH 3 CO - 2 ][H 3 O + ] units mol dm -3 [CH 3 CO 2 H] When calculating the ph of a solution of a weak acid, we have to make the following assumptions the concentrations of [CH 3 CO 2 - ] and [H 3 O + ] are the same. This is true because for each aqueous proton we also get an ethanoate ion, and also the concentration of aqueous protons due to auto-ionisation of water is negligible (1 x 10-7 mol dm -3 ). there is no significant difference between the [CH 3 CO 2 H] at equilibrium and its concentration when the solution was made up. This is true because very few of the ethanoic acid molecules are dissociated at any one time. For calculations, the equation becomes K a = [H 3 O + ] 2 [CH 3 CO 2 H] start If you are asked for a definition of Ka, give the top expression. When doing a weak acid calculation, use the equation above. Calculate the ph of mol dm -3 ethanoic acid, given that K a for ethanoic acid is 1.80 x 10-5 mol dm -3 Calculate the K a of propanoic acid, given that the ph of a mol dm -3 solution is 3.40 Answer = 2.87 Answer = 1.58 x 10-6 mol dm -3 Page 9 of 28

10 Inductive effects in weak acids Acids and bases The relative strength of a weak acid is determined by the stability of the ion that is created when it loses its proton (its conjugate base). From the two examples on the previous page, how can you tell that propanoic acid is a weaker acid than ethanoic acid? Now Hint: compare the K a values and look at the K a expression. Write equations to show the dissociation of ethanoic acid and propanoic acid. indicate the electron releasing effect in propanoic acid. this effect increases the strength of the O-H bond and destabilises the propanoate ion. which in turn makes protons less likely to be released by propanoic acid, so it is less acidic. Electronegative atoms are electron withdrawing, such as Cl. How would you expect the K a for chloroethanoic acid and the ph of a mol dm -3 solution of it to differ to the case for ethanoic acid? K a would be bigger and the ph would be lower. Page 10 of 28

11 Acids and Bases Buffer solutions A buffer solution is a solution that resists changes in ph when small amounts of acid, base or water are added to it. A buffer typically consists of a weak acid (e.g. CH 3 CO 2 H) whose dissociation is suppressed by a comparable concentration of its conjugate base (CH 3 CO 2 - ). Addition of acid can be countered by the following reaction: CH 3 CO H + CH 3 CO 2 H So, the added acid is replaced by undissociated ethanoic acid molecules. Addition of base can be countered by the following reaction: CH 3 CO 2 H + OH - CH 3 CO H 2 O So, the added base is neutralised by undissociated ethanoic acid molecules. Calculations on buffers Doing past paper exam questions will show you that this kind of buffer can be made simply by mixing solutions of the weak acid, e.g. ethanoic acid and its conjugate base in the form a salt, e.g. sodium ethanoate. this kind of buffer is also formed when a weak acid is titrated with a base, as there will be a point in the titration when a mixture of the acid and the salt is formed. Obviously it makes your life easier if you spot which of the above situations you are in! In either case, the K a expression for the weak acid is still true, but it is no longer possible to say that the concentration of the aqueous protons and ethanoate ions are the same. K a = [CH 3 CO - 2 ][H 3 O + ] [CH 3 CO 2 H] which for a buffer is usually thought of as K a = [Salt][H 3 O + ] [Acid] Some people prefer to use the equation in its log form, which is ph = pk a + log { [salt]/[acid] } See the next page for calculation examples. Page 11 of 28

12 Acids and bases Buffers formed by mixing solutions of acid and salt molecules Calculate the ph of a buffer formed by mixing 6.00 g of ethanoic acid and 8.20g of sodium ethanoate in distilled water, made up to 500 cm 3, given that K a for ethanoic acid is 1.80 x 10-5 mol dm -3. calculate the moles of each substance present, by using their RMMs calculate the concentration of ethanoic acid and sodium ethanoate present using the K a expression for buffers, calculate the [H + ] and the ph Answer = 4.74 Note from the above example, that when the concentration of salt and acid is the same, the ph of the buffer is simply the same as the pk a of the weak acid you are making it from. Also, a buffer works best at this special point, so by knowing the pka values for a selection of weak acids you could choose the best one to make a buffer with a certain ph. Sometimes you are given information about the relative amounts of salt and acid species. Calculate the ph of a buffer made from ethanoic acid and sodium ethanoate where the ratio of salt : acid concentrations is 2:3, given that Ka for ethanoic acid is 1.80 x 10-5 mol dm -3. Answer = 4.56 Page 12 of 28

13 Acids and Bases Buffers formed during titrations, by mixing acid and alkali solutions. Calculate the ph of a buffer made when 12.5 cm 3 of sodium hydroxide is added to 25.0 cm 3 of ethanoic acid, given that K a for ethanoic acid is 1.80 x 10-5 mol dm -3, and both solutions are mol dm -3. write the equation for the reaction calculate the moles of ethanoic acid and sodium hydroxide present in each solution before mixing calculate the moles of ethanoic acid and sodium ethanoate present in the solution after they are mixed and the reaction has taken place moles of sodium ethanoate formed moles of ethanoic acid remaining un-neutralised calculate their concentrations work out the ph of the solution using the K a expression for buffers Answer = 4.74 NB these questions are much faster if you spot that a 1:1 ratio of salt and acid would be formed, therefore pk a = ph. However, if it is not a 1:1 buffer, these methods will find you the right answer. This situation occurs when a weak acid is titrated with the alkali in the burette. Page 13 of 28

14 Acids and bases Other buffering systems Amino acids can act as buffers, as they have an acidic and a basic group present in the same molecule. See Acids and Bases 2 for a full discussion. One of the reactions that buffers blood is a mixture of hydrogen carbonate and carbonate ions. The equations for this system are dissociation equation HCO H 2 O CO H 3 O + K a expression K a = [CO 2-3 ][H 3 O + ] [HCO - 3 ] when acid is added CO H + HCO 3 - when base is added HCO OH - CO H 2 O Another buffer is formed during titration of NaHSO 4. The HSO 4 - ion is a weak acid and its conjugate base SO 4 2- formed the other part of the buffering system. A mixture of HCN (a weak acid) and CN - (its conjugate base) is often used in organic chemistry for addition of HCN to carbonyl compounds. Page 14 of 28

15 Acids and Bases Titration curves In an acid-base titration, an acid and a base are reacted together. One goes in the burette, and the other goes in the conical flask. In acids and bases questions, we are concerned with plotting the ph variation during the titration versus the volume of solution added from the burette. To understand titration curves, it is very important to be able to visualise the contents of the conical flask, so you can understand how to work out or estimate the ph. The shapes of titration curves are categorised by the sorts of acids and bases that are reacting. The combinations are Strong acid with strong base Weak acid with strong base Strong acid with weak base Weak acid with weak base Hydrochloric acid and sodium hydroxide Ethanoic acid and sodium hydroxide Hydrochloric acid with ammonia Ethanoic acid with ammonia Each of the curves (except the last one) has a rapid change in ph (steep section) about the point at which moles of acid = the moles of base. The point at which moles of acid = moles of base is known as the equivalence point. The ph at the equivalence point is only seven for a strong acid with strong base titration, otherwise it will vary depending on what salt is formed in the conical flask when the acid and base have neutralised each other. At the equivalence point the mixture contains the salt that is produced when the acid and base neutralise each other. An indicator is a substance that changes colour over a particular ph range of about two units, for example phenolphthalein is colourless at ph 8 and below, but bright pink above ph 10. Different indicators have a different ph range about which they change colour. The indicator is chosen so that it changes colour about a ph which matches the point during the titration when the ph is changing rapidly. This way, the end point (when the indicator changes colour) will indicate the equivalence point (when the moles of acid = the moles of base). During titrations where one or more of the components is weak, a buffer will form. The buffer will be most effective when the moles of salt and acid molecules are equal. If the reaction equation is 1:1 between acid and base, and the concentration of acid and base solutions are the same, the volume at the equivalence point will be equal to the volume of solution that starts off in the conical flask. In this topic you know the concentration of both the acid and alkali solutions. In reality, the whole point of a titration is that usually one of the concentrations is unknown. By finding out what volume of a solution with a known concentration reacts with another volume of a solution of unknown concentration, you can work out the unknown concentration, as long as you know the equation! Page 15 of 28

16 Strong acid versus strong base Acids and bases A 25.0 cm 3 sample of mol dm -3 hydrochloric acid is titrated against 50.0 cm 3 of mol dm -3 sodium hydroxide solution. Sketch a curve of ph versus volume of sodium hydroxide added. This is a strong acid strong base curve, which makes things a little more simple than they might otherwise be, for two reasons. there will no buffer to worry about the ph at the equivalence point will be 7 (as NaCl does not affect the ph of water) Work out the ph in the conical flask before any sodium hydroxide solution has been added. The concentration of H + ions is also 0.1 mol dm -3, because hydrochloric acid is a strong acid. So the ph is log[0.1] = 1. Work out the ph in the conical flask after 50 cm 3 of the sodium hydroxide solution has been added. At this point in the reaction, the hydrochloric acid has been swamped with excess sodium hydroxide, so it is acceptable for a sketch to simply use the ph of the sodium hydroxide solution in the burette. The concentration of OH- ions is 0.1 mol dm -3 because sodium hydroxide solution is a strong base. The poh is therefore log[0.1] = 1, so providing the solution is at room temperature the ph will be 14-1= 13. Sketch the curve below. ph Finish ph 14--log[0.1] 13 Start ph Equivalence point Volume = 25 cm 3 ph = 7 (strong acid-strong base) = -log[0.1] = 1 Page 16 of 28 Volume of NaOH added / cm

17 Acids and Bases Weak acid versus strong base A 25.0 cm 3 sample of mol dm -3 ethanoic acid is titrated against 50.0 cm 3 of mol dm -3 sodium hydroxide solution. Sketch a curve of ph versus volume of sodium hydroxide added. K a for ethanoic acid is 1.80 x 10-5 mol dm -3. As this titration involves a weak acid, a buffer will form when the ratio of salt:acid is 1:1. To understand the curve, it is important to have a good idea about what is in the contents of the conical flask at various points during the titration. NB the reaction occurring is CH 3 CO 2 H + NaOH CH 3 CO 2 Na + H 2 O At the start: there is only aqueous ethanoic acid in the conical flask. We can calculate its ph using the method for a weak acid. Show this Answer = 2.87 After 12.5 cm 3 of sodium hydroxide solution has been added: there is a mixture of sodium ethanoate formed by the above reaction (salt) and ethanoic acid that has not yet been neutralised because not enough sodium hydroxide has been added. This mixture of salt and acid is a buffer, so we can work out the ph at this point using the buffer equation. In fact, at this point the amounts of sodium ethanoate and ethanoic acid in the flask are exactly equal, so the ph turns out to be the same as the pk a value for the acid. Show this Answer = 4.74 After 25 cm 3 of sodium hydroxide solution has been added: the equivalence point has been reached and the salt formed in the conical flask is sodium ethanoate. Sodium ethanoate is the salt of a weak acid made with a strong base. The ethanoate ion is a weak base, so the solution of sodium ethanoate formed at the equivalence point has the following reaction occurring in it. CH 3 CO H 2 O CH 3 CO 2 H + OH - The creation of OH - ions in the solution makes the ph at the equivalence point above seven. There is a way of calculating exactly what the ph would be, but for a sketch it is good enough simply to use a ph of 9 for the equivalence point in a weak acid-strong base titration. After 50 cm 3 of sodium hydroxide solution has been added: the ethanoic acid has been swamped by an excess of sodium hydroxide, so it is acceptable for a sketch to use the ph of the sodium hydroxide solution in the burette, i.e. 14--log[0.1] 13. Page 17 of 28

18 Acids and bases Weak-acid vs. strong-base curve weak acid in conical flask. Sketch the curve for mol dm -3 ethanoic acid vs sodium hydroxide (described above). ph Finish ph Buffer ph Equivalence point ph Start ph Volume of NaOH added / cm Page 18 of 28

19 Weak acid versus strong base continued Acids and Bases A 25.0 cm 3 sample of mol dm -3 sodium hydroxide is titrated against 50.0 cm 3 of mol dm -3 ethanoic acid solution. Sketch a curve of ph versus volume of ethanoic acid added. K a for ethanoic acid is 1.80 x 10-5 mol dm -3. This example is very similar to the previous one, only this time the acid is being added to the base. NB the reaction occurring is still CH 3 CO 2 H + NaOH CH 3 CO 2 Na + H 2 O At the start: there is only sodium hydroxide solution in the conical flask, so the ph can be worked out using the method for a strong base. Show this Answer = 13.0 As the reaction proceeds and ethanoic acid is added to the conical flask, it is converted into sodium ethanoate. The mixture of un-neutralised sodium hydroxide and sodium ethanoate formed is not a buffer. After 25 cm 3 of ethanoic acid solution has been added: the equivalence point has been reached and the salt formed in the conical flask is sodium ethanoate. As in the previous example, the ph of the aqueous sodium ethanoate is above seven, due to the basic nature of the ethanoate ion. CH 3 CO H 2 O CH 3 CO 2 H + OH - After 50 cm 3 of ethanoic acid solution has been added: there is a mixture of excess ethanoic acid and sodium ethanoate formed during the titration, and this mixture is a buffer. In fact, at this point the amounts of sodium ethanoate and ethanoic acid in the flask are exactly equal, so the ph turns out to be the same as the pk a value for the acid. Show this NB the point at which the buffer forms and the ph = pk a for the acid will either occur at the half-equivalence point (previous example) or at twice the equivalence point (this example). NB to help you work out which side of the graph the buffer will be on, if a weak acid is involved, then the buffer formed will have an acidic ph. So, if you are starting with a base in the conical flask for example, then the buffer will have to occur on the other side of the graph, because the ph values on the left hand side will be alkaline rather than acidic. Page 19 of 28

20 Acids and bases Weak-acid strong-base curve strong base in conical flask. Sketch the curve for mol dm -3 sodium hydroxide vs. ethanoic acid (described above). ph Start ph Equivalence point ph Buffer / End ph Volume of CH 3 CO 2 H added / cm Page 20 of 28

21 Titrations involving weak bases Acids and Bases As weak bases are no longer on the Edexcel syllabus you could choose to ignore this, but they could be asked about by analogy with weak acids in synoptic questions. Weak bases do not fully dissociate, even at high dilution. Using ammonia, the dissociation equation is NH 3 + H 2 O NH OH - The base dissociation equation is K b = [NH + 4 ][OH - ] [NH 3 ] How this equation is used depends on the situation: If you are working out the ph of a weak base, then the concentration of [NH 4 + ] = [OH - ], and the starting concentration of the ammonia can be used in place of the actual equilibrium concentration (see the weak acids section for a full explanation of these approximations). Therefore, the equation to use is K b = [OH - ] 2 [NH 3 ] Calculate the ph of a mol dm -3 solution of ammonia, given that K b for ammonia is 2.00 x 10-5 mol dm -3. Answer = 11.1 If you are working out a buffer formed from a mixture of salt (e.g. NH 4 Cl) and weak base (e.g. NH 3 ), then the equation must stay in its original form. Calculate the ph of a buffer solution in which the concentrations of ammonia and ammonium chloride are mol dm -3 and mol dm -3 respectively, given that K b for ammonia is 2.00 x 10-5 mol dm -3. Answer = 9.30 NB when the concentration of the salt and base molecules is the same, the buffer is most effective and the ph is equal to 14-pK b. Give the equations that occur when a basic buffer has acid and base added to it. Page 21 of 28

22 Acids and bases Weak base versus strong acid A 25.0 cm 3 sample of mol dm -3 ammonia solution is titrated against 50.0 cm 3 of mol dm -3 hydrochloric acid solution. Sketch a curve of ph versus volume of hydrochloric acid added. K b for ammonia solution is 2.00 x 10-5 mol dm -3. NB the reaction occurring is NH 3 + HCl NH 4 Cl At the start: There is only aqueous ammonia solution in the flask, so we can work out its ph using the method for a weak base. Show this Answer = 11.1 After 12.5 cm 3 of hydrochloric acid has been added: there is a mixture of ammonium chloride formed and un-reacted ammonia solution. This is a buffer, so we can work out the ph using the buffer equation. In fact, there is an exactly equal amount of ammonia molecules as ammonium ions, so the ph is equal to 14-pK b Show this After 25 cm 3 of hydrochloric acid has been added: the equivalence point has been reached and the salt in the conical flask is ammonium chloride. As ammonia is a weak base, the ammonium ion is a weak acid and the solution in the conical flask has the following reaction occurring inside it. NH H 2 O NH 3 + H 3 O + The creation of H 3 O + ions in the solution makes the ph at the equivalence point below seven. There is a way of calculating exactly what the ph would be, but for a sketch it is good enough simply to use a ph of 5 for the equivalence point in a strong acid-weak base titration. After 50 cm 3 of hydrochloric acid has been added: the ammonia solution has been swamped with hydrochloric acid so for a sketch the ph is taken to be the same as the ph of the hydrochloric acid in the burette, i.e. log[0.1] = 1. Page 22 of 28

23 Acids and Bases Weak-base vs. strong-acid curve Weak base in conical flask. Sketch the curve for mol dm -3 previous page). ammonia solution vs. hydrochloric acid (described on ph Start ph Buffer ph Equivalence point ph Finish ph Volume of HCl added / cm Page 23 of 28

24 Weak base versus strong acid continued Acids and bases A 25.0 cm 3 sample of mol dm -3 hydrochloric acid is titrated against 50.0 cm 3 of mol dm -3 ammonia solution. Sketch a curve of ph versus volume of ammonia solution added. K b for ammonia solution is 2.00 x 10-5 mol dm -3. This is very similar to the previous example but the opposite way round. NB the reaction occurring is still NH 3 + HCl NH 4 Cl At the start: There is only hydrochloric acid in the flask so we can use the method for a strong acid to work out the ph. Show this Answer = 1.00 As ammonia solution is added to the conical flask, mixture of ammonium chloride formed and un-reacted hydrochloric acid is created, and this mixture is not a buffer! When 25 cm 3 of ammonia solution has been added: the equivalence point has been reached and the conical flask contains the salt ammonium chloride. As this is the salt of a weak base with a strong acid, it makes the solution acidic, due to the following reaction: NH H 2 O NH 3 + H 3 O + When 50 cm 3 of ammonia solution has been added: there is a mixture of additional ammonia in the conical flask and ammonium chloride formed at the equivalence point. This is a buffer, so we can find out its ph using the buffer equation. In fact, the amount of ammonia and ammonium chloride is exactly the same at this point, so the ph is equal to the 14-pK b. Show this Page 24 of 28

25 Acids and Bases Weak-base vs. strong-acid curve strong acid in conical flask. Sketch the curve for mol dm -3 page). hydrochloric acid vs. ammonia (described on previous ph Finish and buffer ph Equivalence point ph Start ph Volume of NH 3 added / cm Page 25 of 28

26 Acids and bases Indicators Indicators are weak acids or bases where the protonated and deprotonated forms of the molecule are different colours. For example, imagine is a weak acid where the acid molecule (HA) is red, whereas the conjugate base (A - ) is blue, and the acid dissociation constant is 1.0 x 10-7 mol dm -3. The dissociation equation is HA A - + H + red blue To understand how such an acid can act as an indicator, imagine how the equilibrium would respond if it were occurring in an acidic solution or an alkaline solution. In an acidic solution: The increased concentration of H+ ions shifts the position of equilibrium to the left hand side, so the solution appears red. In an alkaline solution: The increased concentration of OH - ions react with the H + ions the equilibrium position shifts to the right hand side, so the solution appears blue. To find out the ph about which an indicator changes colour we can use the acid dissociation equation for the indicator. The acid dissociation equation is K a = [A - ][H + ] [HA] The indicator changes colour when the concentration of [A - ] and [HA] are equal, because at this point an excess of either colour molecule would be obtained if a drop of acid or base were added to the solution. At this point, you can see from the equation that the values of [A - ] and [HA] cancel, leaving K a = [H + ], so the ph about which an indicator changes colour is given by ph = pk a. For the indicator we are using as an example, the K a is 1 x 10-7 mol dm -3 so the ph about which it changes colour is pk a = -log[1 x 10-7 ] = 7. In order for the solution to be identifiably one colour or the other, there must be a 10:1 excess of whichever species. So, the ph at which the solution is blue can be found by making [A - ] = 10 and [HA] = 1. Show this the ph at which the solution is red can be found by making [A - ] = 1 and [HA] = 10. Show this Consequently an indicator changes colour over a ph range of about two units. Page 26 of 28

27 Acids and Bases Indicators continued An indicator is chosen so that it changes colour at a ph about which the ph of the solution in the conical flask is changing rapidly. As you have seen from titration curves, the steep section (rapid change of ph) occurs at the equivalence point, so an indicator s end-point must also occur at the same volume, thereby indicating the volume at which the acid or base has in the conical flask has been neutralised. For a given concentration of solutions involved in the titration, the strong acidstrong base curve has the longest steep section, so the largest number of indicators can be used. As long as the solutions of acid and base are not very dilute, any indicator will work. The weak acid-strong base curve has a shorter steep section that is above seven, so an indicator must be chosen that changes colour at a ph above seven, such as phenolphthalein. When adding alkali from the burette the colour change is colourless to pink. The weak base strong acid curve also has a shorter steep section that is below seven, so an indicator must be chosen that changes colour at a ph below seven, such as methyl orange. When adding acid from the burette the colour change is orange to red. Page 27 of 28

28 Acids and bases The remaining factor to be considered is that the acid and salt molecules of an indicator are of significantly different colours and that the colours are intense, to make it easier to detect the end point. NB A titration involving a weak acid and a weak base does not have a steep section, so any indicator would change colour gradually and be no use in indicating the equivalent point of the reaction. For this reason, we never titrate weak acids with weak bases in volumetric analysis. Page 28 of 28

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