cm 3 0.10 mol l -1 NaOH added to 50.0 cm 3 of 0.10 mol l -1 HCl
Acids have been described as substances that dissolve in water to form H + (aq) ions, whilst bases are substances that react with acids. However, a better, broader definition was produced independently by Bronsted and Lowry. any substance that is capable of donating hydrogen ions (protons) to another substance acids are proton donors. any substance that is capable of accepting hydrogen ions (protons) from another substance bases are proton acceptors. In general an acid, with the formula HA, will dissociate in water according to This equation shows the hydronium ion, H 3 O + (aq). For simplicity, this ion usually written in the shorthand way ---------- H + (aq) The equation shown above can be written more simply as In the forward reaction, HA loses a proton and so behaves as an acid. In the reverse reaction, A - accepts a proton, i.e. behaves as a base. A - is said to be the conjugate base of the acid HA. For every acid, there is a conjugate base formed by loss of a proton {H + (aq) ion}. In addition, every base has a conjugate acid.
This idea can be summarised in a general equation for any acid/base pair. To highlight this point more clearly consider what happens to ethanoic acid when it is added to water. Remember the Bronsted Lowry definition of an acid: an acid is a proton donor. In this example ethanoic acid donates its proton to H 2 O in the forward reaction, resulting in the formation of the ethanoate ion. However, in the reverse reaction, the ethanoate ion accepts a proton from H 3 O + to revert to ethanoic acid molecules. CH 3 COOH + H 2 O CH 3 COO - + H 3 O + ethanoic acid water ethanoate ion hydronium ion From the Bronsted Lowry definition, a base is a proton acceptor. Thus, the ethanoate ion acts as a base in the reverse reaction, and is called the conjugate base of ethanoic acid. Ethanoic acid and the ethanoate ion are known as an acid base conjugate pair. Interestingly enough, there is a second acid base conjugate pair in the reaction above. Notice how in the reverse reaction, H 3 O + donates a proton (making it an acid), while in the forward reaction, H 2 O accepts a proton (making it a base). H 2 O and H 3 O + are a conjugate acid base pair, with H 3 O + being the conjugate acid and H 2 O being the conjugate base.
This reaction can be written more simply if we use the hydrogen ion, H + (aq), instead of the hydronium ion. CH 3 COOH(aq) CH 3 COO - (aq) + H + (aq) ethanoic acid ethanoate ion conjugate base of ethanoic acid hydrogen ion Ethanoic acid is clearly the acid in this reaction it has lost a proton. The ethanoate ion is clearly the conjugate base of ethanoic acid as its formula is identical to ethanoic acid minus the hydrogen ion. 1. Identify the acid-conjugate base pairs in the following equations. The key to identifying the acid- conjugate base pairs is to spot the movement of the hydrogen ion and apply the Bronsted-Lowry definition. 2. Write an equation showing how the H 2 PO 4 - ion can act as an acid.
1. Write the formula for the species which is the conjugate base and for the species which is the acid in the following reactions. a. HCN + OH - CN - + H 2 O b. NH 2 NH 2 + H 2 O OH - + + NH 2 NH 3 2. Each of the following species can act as an acid. 2-2 (i) HAsO 4 (ii) HNO 3 (iii) HVO 4 a. Write an equation for each of these species showing how they react with water to form a conjugate base and the hydronium ion. b. Write an equation for each of these species showing how they form a conjugate base and a hydrogen ion. 3. Write equations to show how the following species can act as bases. Circle the conjugate acid in each equation (i) ClO - (ii) HN 2- (iii) AsH 3 4. Consider the reaction shown below. 2- SO 3 + + NH 4 - HSO 3 + NH 3 a. Which species is the acid in this reaction? b. Explain your answer to part a. c. Which species is the conjugate acid in this reaction? 5. Write the formula for the conjugate base of the following acids. a. CH 3 CH 2 COOH b. CH 3 NH 3 + c. H 3 O + d. H 2 S
A compound is described as amphoteric if it has the ability to behave as an acid and as a base. In certain conditions these substances will be proton donors while in different conditions they can also be proton acceptors Water has the ability to self-ionise. When this happens one of the water molecules loses a proton (acts like an acid) while the other gains a proton (acts like a base). The amphoteric nature of water is also shown in the following reactions.
Acids are proton donors. You learned at Higher that some acids are classified as WEAK and some are classified as STRONG. The STRENGTH of an acid depends on the DEGREE OF IONISATION - the ability of the unionised acid molecule to produce hydrogen ions The more an acid molecule tends to produce hydrogen ion, H + (aq), the stronger it will be. Strong acids, like hydrochloric acid, nitric acid and sulfuric acid totally dissociate in water all the acid molecules become ions. This is shown by a one way arrow in the equation for the ionisation of the acid. Strong acid HA H + + A - E. g. HCl H + + Cl - HNO 3 H + + - NO 3 The strength of an acid is NOT RELATED to the CONCENTRATION of the acid. The concentration of an acid is related to the volume of water that has been added to the undiluted acid. More water will produce a more dilute acid but it does not alter the STRENGTH of the acid. Weak acid HA H + + A - E. g. CH 3 COOH H + + CH 3 COO - HCN H + + CN - A weak acid is an equilibrium mixture of unionised acid molecules, hydrogen ions and the conjugate base of the acid. Equations for weak acids must always be written with two-way arrows. These arrows signify the only a proportion of the acid molecules IONISE. Any acid, weak or strong, which is capable of donating only one hydrogen ion per molecule is termed a monoprotic or monobasic acid. An acid like sulfuric acid which can donate two hydrogen ions is called diprotic or dibasic.
Water molecules self-ionise according to the following equation H 2 O(l) H + (aq) + OH - (aq) The equilibrium expression for this reaction is termed the ionic product of water and has the form. K w = [H + (aq)] [OH - (aq)] The value of K w at 25 o C is 1 x 10-14 In any aqueous solution this equilibrium reaction is responsible the ph of the solution. ph is measure of the concentration of hydrogen ions [H + (aq)] in an aqueous solution. These quantities are related by the following formulae. and The table shows the values for ph, H + and OH - at 25 o C. A neutral solution is one which contains equal concentrations of hydrogen ions and hydroxide ions. At 25 o C the ph of any neutral solution is 7. At other temperatures this value will alter as the value of K w is temperature dependent.
The table shows how the value of K w changes with temperature and the effect this has on the ph of water. H 2 O(l) H + (aq) + OH - (aq) The equation shows as water molecules ionise, they produce one hydrogen ion for every hydroxide ion. Therefore, the equilibrium expression for the ionic product of water becomes K w = [H + (aq)] 2 This rearranges to give [H + (aq)] = K w At 60 o C, K w = 9.61 x 10-14 [H + (aq)] = 9.61 x 10-14 = 3.1 x 10-7 mol l -1. As ph = -log [H + (aq)] ph = -log 3.1 x 10-7 = 6.51. It is very important to realise that although the ph of water is 6.51 it is still neutral - the concentration of hydrogen ions and hydroxide ions are still equal. At 60 o C a ph greater than 6.51 would indicate an alkaline solution and a ph of less than 6.51 would be acidic. In addition, it is obvious from the table that K w increases with increasing temperature. This indicates that the ionisation of water is an endothermic process. The value of K w is 51.3 x 10-14 at 100 o C. Calculate the ph of water at this temperature. What does the very small value of K w indicate about the water equilibrium?
1. Calculate the ph of 0.030 mol l -1 hydrochloric acid, HCl. ph = -log [H + (aq)] ph = -log [0.03] = 1.5 2. Calculate the concentration of hydrogen ions in solution of ph 8.5. [H + (aq)] = 10 -ph [H + (aq)] = 10-8.5 = 3.0 x 10-9 mol l -1. 3. Calculate the ph of a solution of 0.25 mol l -1 sodium hydroxide, NaOH. For alkalis the ionic product of water must be used in the calculation. Taking the negative log of both sides of the K w expression. K w = [H + (aq)] [OH - (aq)] = 1 x 10-14 K w = ph + poh = 14 poh = -log[oh - ] poh = -log [OH - (aq)] poh = -log [0.25] = 0.60 ph = 14 poh ph = 14 0.60 = 13.4 4. 25.0 cm 3 of a solution of a strong alkali contained 1.50 10 3 mol of hydroxide ions. Calculate the ph of this solution. C = n/v = 1.50 x 10-3 /0.025 = 0.025 mol l -1 poh = -log [OH - (aq)] poh = -log [0.025] = 1.60 ph = 14 poh ph = 14 1.60 = 12.4
1. Calculate the ph of the following acids a. 0.16 mol l -1 HCl b. 0.45 mol l -1 HNO3 2. Calculate the ph of the following alkalis a. 0.32 mol l -1 NaOH b. 1.2 mol l -1 KOH 3. 7.3 g of HCl is dissolved in water and the solution made up to 500 cm 3. Calculate the ph of this solution. 4. 8.0 g of NaOH was dissolved in 100 cm 3 of water and the solution made up to 250 cm 3. Calculate the ph of this solution. 5. If 25.6 cm 3 of 0.0025mol l -1 of NaOH is added to 50.9 cm 3 of 0.005 mol l -1 HCl. Calculate the ph of the resulting solution. NaOH + HCl NaCl + H2O Hint: This is an excess question you must calculate how many moles of hydrogen ions or hydroxide ions would be left at the end of the reaction. Calculate the concentration of these excess ions. Do the ph calculation as in the examples above. 6. 250 cm 3 of 0.1 mol l -1 of potassium hydroxide is added to 125 cm 3 of 0.05mol l -1 of hydrochloric acid. Calculate the ph of the resulting solution. 7. A solution of a strong acid was found to have a ph of 0.5 a. Calculate the concentration of hydrogen ions in this solution b. Calculate the volume of water which must be added to 25.0 cm 3 of this solution to increase its ph from 0.5 to 0.7. Hint: You need to know the hydrogen ion concentration in both solutions. Use dilution formula C1V1=C2V2. 8. At 25 C, Kw has the value 1.00 10 14. Calculate the ph at 25 C of a. 0.150 mol l -1 solution of potassium hydroxide, b. the solution formed when 35.0 cm 3 of this solution of potassium hydroxide is mixed with 40.0 cm 3 of a 0.120 mol l -1 solution of hydrochloric acid. KOH + HCl KCl + H2O
9. A 30.0 cm 3 sample of a 0.480 mol l -1 solution of potassium hydroxide was partially neutralised by the addition of 18.0 cm 3 of a 0.350 mol l -1 solution of sulfuric acid. KOH + H2SO4 K2SO4 + H2O a. Calculate the initial number of moles of potassium hydroxide. b. Calculate the number of moles of sulfuric acid added. c. Calculate the number of moles of potassium hydroxide remaining after mixing. d. Calculate the concentration of hydroxide ions in the solution formed. e. Calculate the ph of the solution formed. 10. Water molecules dissociate into hydrogen and hydroxide ions. a. Write the expressions for the ionic product of water, Kw. b. At 318 K, the value of Kw is 4.02 10 14 and hence the ph of pure water is 6.70 State why pure water is not acidic at 318 K. c. Use the value of Kw given above and calculate the ph of the solution formed when 2.00 cm 3 of 0.500 mol l -1 aqueous sodium hydroxide are added to 998 cm 3 of pure water at 318 K. 11 a. Calculate the change in ph when 0.250 mol l -1 hydrochloric acid is diluted with water to produce 0.150 mol l -1 hydrochloric acid. b. Calculate the volume of water which must be added to 30.0 cm 3 of 0.250 mol l -1 hydrochloric acid in order to reduce its concentration to 0.150 mol l -1. 12. A mass, m, of solid KOH is added to 755 cm 3 of 0.0120 mol l -1 HCl. The ph after this addition is 11.60, measured at 25 C. The volume of the resulting solution is still 755 cm 3. KOH + HCl KCl + H2O a. Calculate the number of moles of KOH needed to neutralise exactly the HCl present in the 755 cm 3 of 0.0120 mol l -1 HCl. b. Calculate the number of moles of NaOH in excess when the ph is 11.60 c. Use these results to calculate the total number of moles of KOH added. d. Deduce the value of m. This is you if you get part d correct
The Acid dissociation constant, K a, is the equilibrium constant for the reaction in which a weak acid is in equilibrium with its conjugate base in aqueous solution. Ethanoic acid dissociates according to the following equation CH3COOH(aq) + H2O(l) CH3COO - (aq) + H3O + (aq) This can also be written as CH3COOH(aq) CH3COO - (aq) + H + (aq) The equilibrium expression for this reaction is Ka = [CH3COO - (aq)] [H + (aq)] [CH3COOH(aq)] + The value of K a for ethanoic acid at 25 o C is 1.7 x 10-5. The size of K a gives an indication of the degree of ionisation the acid ungergoes and hence can be used as a measure of acid strength. The larger the value of Ka,the stronger the acid For most weak acids the K a value is a fairly small value and so to give more readable numbers K a values are often expressed as pk a values. Just as ph = -log[h + (aq)] pka = -logka The larger the value of pka,the weaker the acid Hydrofluoric acid, HF(aq), is a weak acid with a Ka value of 6.3x10-4. a. Write an equation which shows how hydrofluoric acid dissociates in water. HF H + + F - b. Write the equilibrium expression for the dissociation constant Ka. Ka = [H + ][F - ] [HF] c. Calculate the value for pka. pka = -logka pka = -log 6.3x10-4 = 3.2 d. State whether hydrofluoric acid is stronger or weaker than ethanoic acid. Stronger(larger Ka)
The table shows the dissociation constants for several weak acids. 1. Which of the acids is the strongest? 2. Which of the acids is the weakest? 3. Write the formula for the conjugate base of hydrocyanic acid. 4. Write the equilibrium expression for the hydrogencarbonate ion. 5. The conjugate base of boric acid can also act as an acid. Write an equation which shows how the conjugate base of boric acid dissociates in water. 6. What is the relationship between Ka and pka? 7. Hydrochloric acid is a strong acid. a. Calculate the ph of 0.001 mol l -1 hydrochloric acid. b. The ph of of 0.001 mol l -1 propanoic acid is 3.92. Explain why the ph of 0.001 mol l -1 hydrochloric acid is different from the ph of 0.001 mol l -1 propanoic acid. c. Would the ph of 0.001 mol l -1 phenol be higher or lower than 0.001 mol l -1 propanoic acid?
The ph of several acids of known concentration were determined using a ph meter. Complete the table with the results. Acid Concentration/ mol l -1 Ka pka ph ph meters The formula ph = -log[h + (aq)] CANNOT be used to calculate the ph of weak acids. In a monoprotic strong acid there will be 0.1 mo l -1 of hydrogen ions in a 0.1 mol l -1 solution because the acid totally ionises in solution. In a weak acid only a small proportion of the molecules ionise and so in a 0.1 mol l -1 solution of a monoprotic weak acid there will be much less than 0.1 mol l -1 of hydrogen ions. When calculating the ph of a weak acid the Ka or pka value must be included because, together with the concentration of the acid, this gives a measure of the concentration of hydrogen ions in solution. The formula used to calculate the ph of a weak acid is Use this formula to calculate the theoretical ph of the acid solutions used in the experiment. Do your answers match the experimental values?
Calculate the ph of a 0.36 mol l -1 solution of ethanoic acid. Use From data booklet pka = 4.76 ph = ½ (4.76) - ½ log (0.36) ph = 2.38 ½ (-0.44) ph = 2.38 + 0.22 = 2.60 Note that a strong acid of concentration 0.36 mol l -1 would have a ph of 0.44. 1. The ph of a 0.15 M solution of a weak acid, HA, is 2.82. a.. Write an expression for the acid dissociation constant, Ka, of HA, and determine the value of Ka for this acid. b. The dissociation of HA into its ions in aqueous solution is an endothermic process. How would its ph change if the temperature were increased? Explain your answer. 2. Phenol is a weak acid. The dissociation of phenol in aqueous solution is represented by the following equation: C 6 H 5 OH(aq) + H 2 O(l) H 3 O + (aq) + C 6 H 5 O (aq) a. Write an expression for the acid dissociation constant, Ka, for phenol. b. The value of the acid dissociation constant, Ka, for phenol is 1 10 10 mol l -1. Calculate the pka value of phenol. c. Calculate the ph of 0.0025 mol l -1 aqueous solution of phenol. 3. The acid dissociation constant, Ka, for propanoic acid has the value of 1.35 10 5. a. Calculate the ph of a 0.117 mol l -1 aqueous solution of propanoic acid. b. Write the formula for the conjugate base of propanoic acid. c. Use the data booklet to find an acid which could described to be weaker than propanoic acid.
4. An acid, HA, has a pka = 4.20. a. Calculate the value of the dissociation constant, Ka, for this acid. b. Calculate the ph of a 0.40 mol l -1 aqueous solution of this acid. 5. The hydrogen halides all react with water to form acids. Hydrogen fluoride forms a weak acid while the others all form strong acids. a. Write equations to show the reactions that occur when hydrogen fluoride and hydrogen chloride are dissolved in water. b. Calculate the ph of an aqueous solution of hydrofluoric acid of concentration 0.050 mol l -1, given that Ka = 5.6 10 4. c. When hydrogen fluoride is dissolved in pure nitric acid, a reaction takes place that can be represented by the equation: HNO3 + HF + H2NO3 + F d. State, with a reason, which reactant acts as a Bronsted-Lowry acid in this reaction and give the formula of its conjugate base. 6. Fizzy drinks contain carbon dioxide dissolved in water which dissociates, as shown, to produce carbonic acid. a. What is the Bronsted-Lowry definition of an acid? b. Write the expression for the dissociation constant, Ka, for carbonic acid. c. Calculate the ph of a 0.10 mol l -1 solution of carbonic acid. 7. When an ant bites, it injects methanoic acid (HCOOH) a. Methanoic acid is a weak acid (i) (ii) What is the conjugate base of methanoic acid? Write the expression for the dissociation constant, Ka, for methanoic acid. b. In a typical bite, an ant injects 3.6 x 10-3 g of methanoic acid. Assuming that the methanoic acid dissolves in 1 cm 3 of water in the body, calculate the ph of this methanoic acid solution.
What are they? Buffer solutions have a fixed ph. A buffer solution resists changes to its ph when small volumes of acid or alkali are added to it. The ph of a buffer solution will not change when it is diluted with water. Different buffer solutions will have different ph values What are the ingredients of a buffer solution? Buffer solutions are formed by mixing (in the proper concentration ratio) two different chemicals. Acidic Buffer These are made by mixing a weak acid and its salt. E.G. ethanoic acid/sodium ethanoate or benzoic acid/potassium benzoate. Basic Buffer These are made by mixing a weak base and its salt. E.G. ammonia/ammonium chloride or ammonia/ammonium sulphate. Buffer solutions can be thought of as either a weak acid and it conjugate base or a weak base and its conjugate acid. Buffer solutions cannot be made with strong acids or alkalis.
What are buffers used for? Their resistance to changes in ph gives buffers many uses in biochemcal systems and chemical reactions. The blood in our bodies is buffered at a ph value of 7.36-7.42 due to bicarbonate - carbonic acid buffer. A mere change of 0.2 ph units can cause death. Certain enzymes get activated only at certain definite ph values. Industrially, buffer solutions are used in fermentation processes and in setting the correct conditions for dyes used in colouring fabrics. They are also used in chemical analysis (e.g. EDTA titration) and in the calibration of ph meters. How do buffers work? If a buffer is to stabilise ph, it must have a component which is able to react with any extra hydrogen ions added to the buffer and a component which is able to react with any extra hydroxide ions which are added to the buffer. The two components ( weak acid and the salt) dissociate as follows 1 HA(aq) H + (aq) + A - (aq) Weak acid 2 NaA(s) Na + (aq) + A - (aq) Salt In the buffer there will be large concentration of undissociated HA(aq) molecules from reaction 1. There will also be a large concentration of the conjugate base A - (aq) from reaction 2. Add acid,h + (aq), to the buffer Any additional hydrogen ions added to the buffer will react with the large concentration of A - (aq) present due to the complete ionisation of the salt. This will cause reaction 1 to shift to left removing the extra hydrogen ions and maintaining the buffer ph. Add alkali, OH - (aq), to the buffer Any additional hydroxide ions added to the buffer will react with the hydrogen ions present. This will cause reaction 1to shift to the right maintaining the ph of the buffer.
The ph of a buffer solution. The ph of a buffer is determined by two factors; 1. The equilibrium constant K a of the weak acid and 2. The ratio of the concentration of the weak acid [Acid] to the concentration of the conjugate base [Base] in the solution. Remember - the salt and the conjugate base are the same thing. These facts are formalised in the Henderson-Hasselbalch equation: In practice to obtain a buffer of a certain ph, a weak acid with a pk a value around the desired ph would be selected. The concentrations of both the acid and the base can then be manipulated to fine tune the buffer solution to the desired ph. We can deduce several facts from this equation: 1. The equation allows calculation of ph of an acid buffer from its composition and acid dissociation constant, or calculation of composition from the other two values. Values for K a and pk a are available in the data booklet. 2. If the [ACID] = [BASE] then ph = pka (since the log of 1 is 0) 3. If the buffer mixture is diluted with water, the ratio [ACID]/[BASE] will not change as both components will be diluted equally. This means that diluting the buffer will NOT alter its ph value.
Calculate the ph of a buffer solution made with 0.10 mol l -1 ethanoic acid (pk a = 4.8) and 0.30 mol l -1 sodium ethanoate. ph = 4.8 - log [0.1]/[0.3] ph = 4.8 - log 0.333 ph = 4.8 - (-0.477) ph = 5.28 Calculate the concentration of ethanoic acid and sodium ethanoate required to make a buffer solution with ph 4.9 (pk a = 4.8) 4.9 = 4.8 - log[acid]/[base] log[acid]/[base] = - 0.1 [Acid]/[Base] = 10-0.1 [Acid]/[Base] = 0.79 Therefore the [Acid] should be 0.79 mol l -1 and the [Base] should be 1 mol l -1. This buffer solution could be made by mixing any volume of 0.79 mol l -1 ethanoic acid with any volume of 1 mol l -1 sodium ethanoate. Alternatively any volume of 0.79 mol l -1 ethanoic acid could be measured out in a beaker and the appropriate mass of sodium ethanoate could be mixed with the acid. For example, if 500 cm 3 of the buffer was required, 0.5 mol of sodium ethanoate would be added to 500 cm 3 of 0.79 mol l -1 ethanoic acid.
1. Calculate the ph of a buffer solution containing 0.1 mol l -1 ethanoic acid and 0.50 mol l -1 sodium ethanoate. 2. Calculate the concentration of methanoic acid and sodium methanoate required to make a buffer solution with a ph of 4.0. 3. 25 cm 3 of 0.10 mol l -1 of a weak acid and 25 cm 3 of 0.40 mol l -1 of the sodium salt of the weak acid are mixed together. The resulting buffer solution has a ph 5.35. Calculate the dissociation constant, Ka, of the weak acid. 4. Calculate the ph of a buffer solution which contains 7.20g of sodium benzoate, C6H5COONa, dissolved in one litre of 0.02 mol l -1 benzoic acid. 5. Buffer solutions are important in Chemistry. a. State what is meant by the term buffer solution. b. Identify a reagent which could be added to a solution of ammonia in order to form a buffer solution. c. An acidic buffer solution is obtained when sodium ethanoate is dissolved in aqueous ethanoic acid. (i) (ii) Explain how this buffer can resist a decrease in ph when a small amount of acid is added to it. Explain why this ph of this buffer is unaffected when the buffer is diluted with water. (iii) Calculate the ph of the buffer solution formed at when 0.125 mol of sodium ethanoate is dissolved in 250 cm 3 of a 1.00 mol l -1 solution of ethanoic acid. 6. A mixture of the acid HA and the sodium salt of this acid, NaA, can be used to prepare a buffer solution. a. Write an ionic equation for the reaction which occurs when a small volume of a dilute acid is added to this buffer solution. b. The concentration of HA in a buffer solution is 0.250 mol l -1. The dissociation constant, Ka, for this acid is 1.45 10 4. Calculate the concentration of the conjugate base, A, in this buffer solution when the ph is 3.59.
Acid/base indicators (or simply indicators) are themselves weak acids which change colour depending on the ph of the solution. HIn can be used as a general formula for an indicator and its dissociation can be represented by the equation: In a good indicator, the undissociated acid, HIn, will have a distinctly different colour from its conjugate base, In -. For example; the indicator litmus, HIn is red and In - is blue. As this is an equilibrium reaction, adding ACID (low ph) will shift the reaction to the left and the indicator colour will be that of HIn. Adding ALKALI (high ph) will have the opposite effect and the colour will be that of In - The acid dissociation constant for an indicator HIn is given the symbol K In and is represented by: When the concentrations of In - and HIn are equal the indicator will change from one colour to the other. At this point ph = pk In which means the indicator changes colour at the ph which is equivalent to its pk In. The ph of the solution is determined by the pk In of the indicator and the ratio of [In - ] to [HIn]. Since these are different colours, the ratio of [In - ] to [HIn] determines the overall colour of the solution. For a given indicator, the overall colour is dependent on the ph of the solution.
There are a huge range of ph indicators. The table below shows some data for three common indicators. Indicator pkin Colour at low ph Colour at high ph ph range Methyl Orange 3.7 Red Yellow 3.1 4.4 Bromothymol Blue 7.0 Yellow Blue 6.0-7.6 Phenolphthalein 9.3 Colourless Pink 8.3 10.0 The ph range of indicators Indicators don't change colour sharply at one particular ph (given by their pkind). Instead, they change over a narrow range of ph. Assume the indicator equilibrium is firmly to one side, but now you add something to start to shift it. As the equilibrium shifts, you will start to get more and more of the second colour formed, and at some point the eye will start to detect it. For example, suppose you had methyl orange in an alkaline solution so that the dominant colour was yellow. Now start to add acid so that the indicator equilibrium begins to shift. At some point there will be enough of the red form of the methyl orange present that the solution will begin to take on an orange tint. As you go on adding more acid, the red will eventually become so dominant that you can no longer see any yellow. There is a gradual smooth change from one colour to the other, taking place over a range of ph. As a rough "rule of thumb", the visible change takes place about 1 ph unit either side of the pkind value.
Some more indicators and their ph ranges are shown above. Universal indicator is a mixture of different indicators each with its own pka value and colour change. A universal indicator is typically composed of water, propan-1-ol, phenolphthalein, sodium hydroxide, methyl red, bromothymol blue, and thymol blue.
In general when an acid reacts with an alkali a salt and water are produced. acid + alkali salt + water E.g. hydrochloric acid + potassium hydroxide potassium chloride + water HCl + KOH KCl + H2O ethanoic acid + sodium hydroxide sodium ethanoate + water CH3COOH + NaOH CH3COONa + H2O When enough acid is added to completely react with an alkali the EQUIVALENCE POINT has been reached. Suppose 25 cm 3 of a 0.10 mol l -1 solution of a monoprotic acid {HA} is titrated with 0.01 mol l -1 sodium hydroxide {NaOH}. HA + NaOH KCl + H 2 O The equivalence point would occur when 25 cm 3 of NaOH had been added. This equivalence point would be the same for any monoprotic acid regardless of its strength. Burette containing However the ph at the equivalence point is dependent on the strength of the acid and base used in the reaction. HA(aq) The ph at the equivalence point depends on the nature of the salt formed during the reaction. This is summarised in the table below. Acid Base equivalence point ph strong strong neutral (= 7) strong weak acidic (< 7) weak strong alkaline (> 7) If the acid used was CH3COOH and the base was NaOH, the salt formed would be CH3COONa. When dissolved in water this salt will form free CH3COO - ions. These ions will react with the hydrogen ions present in the water, leaving excess hydroxide ions and hence the solution is alkaline. If the base used was NH3 and the acid was HCl the salt formed would be NH4Cl. When dissolved in water this salt forms free NH4 + ions. These can react with the hydroxide ions in the water leaving excess hydrogen ions and hence the solution is acidic.
If the ph is monitored when an acid is added to a base, an acid base titration curve can be plotted. Strong acid v Strong base The graph shows the ph curve for adding a strong acid to a strong base. For example; adding 0.1 mol l -1 hydrochloric acid to 25 cm 3 of 0.1 mol l -1 sodium hydroxide would produce this curve the equivalence point would be when 25 cm 3 of hydrochloric acid had been added. Superimposed on the graph are the ph ranges for methyl orange and phenolphthalein indicators 3. 25 cm 3 1. The equivalence point occurs half way down the steepest part of the curve. 2. As this titration involved a strong acid and a strong base the equivalence point occurs at ph = 7. The ph changes from around 12.5 to 1.5 when a very small volume of acid is added. Therefore at the equivalence point there is a large change in the concentration of hydrogen ions. This is ideal if an indicator is to be used to determine the equivalence point. HIn H + + In - A large change in [H + ] will cause the equilibrium to move greatly one way or the other and hence a rapid change in the colour of the indicator will occur. 4. 5. Using PHENOLPHTHALEIN, the indicator would start off PINK and you would titrate until it just becomes COLOURLESS (at ph 8.3) because that is as close as you can get to the equivalence point. On the other hand, using METHYL ORANGE, the indicator would start off YELLOW and you would titrate until there is the very first trace of ORANGE (a mixture of yellow and red make orange) in the solution this would happen at ph 4.4. If the solution becomes red, you are getting further from the equivalence point.
6. Ideally the indicator colour change should happen as close to the equivalence point as possible. In this case the indicator should have a pkin of 7 as the equivalence point is at ph 7 You can see that neither indicator changes colour at the equivalence point. However, the graph is so steep at that point that there will be virtually no difference in the volume of acid added whichever indicator you choose. For this reason both indicators would be suitable in this titration. Weak acid v Strong base To be effective the indicator must change colour on the steep part of the curve as it is here that the hydrogen ion concentration is changing rapidly. The indicator equilibrium will swing from one side to the other resulting in a rapid colour change. The PHENOLPHTHALEIN changes colour (pink to colourless) exactly where you want it to the ideal indicator. The equivalence point is at ph 9.2 because the titration is between a strong base and a weak acid. This time, the METHYL ORANGE is hopeless! As the ph never gets low enough for the indicator to change colour. Strong acid v Weak base This time it is obvious that PHENOLPHTHALEIN would be completely useless as it would change from pink to colourless well before the equivalence point. METHYL ORANGE starts to change from yellow to orange very close to the equivalence point. The pkin for methyl orange is 4.4. Ideally the indicator for this titration should have a pkin of 4.8. Methyl orange is suitable here as it changes colour on the steep section of the curve.
Weak acid v Weak base This curve is for a case where the acid and base are both equally weak (their dissociation constants are the same) - for example, ethanoic acid and ammonia solution this gives an equivalence ph of seven In other cases, the equivalence point will be at some other ph. You can see that neither indicator is any use. Phenolphthalein will have finished changing well before the equivalence point, and methyl orange falls off the graph altogether. Notice that the ph change at the equivalence point here is much more gradual than in any of the other titrations. This means that the concentration of hydrogen ions will also be changing gradually. This will lead to a very slow colour change in the indicator, often over several millilitres of added acid even if the indicator had a pkin exactly on the equivalance point. Therefore there is no indicator which can be used to accurately detect the equivalence point in the titration of a weak acid with a weak base. Summary Table Titration Equivalence point ph Suitable indicator pkin Strong acid/strong Base 7 Several 3-11 Weak acid/strong Base More than 7 Phenolphthalein 8-10 Strong acid/weak Base Less than 7 Methyl Orange 3-6 Weak acid/weak Base Depends on Ka of both reactants None None There will be a range of suitable indicators for acid/base titrations the important idea to remember is that the pkin of the indicator should match, as closely as possible, the equivalence point ph of the titration.
All the titration curves we have looked at so far have been for MONOPROTIC acids (acids which yield only one hydrogen ion) like HCl, HNO 3 and CH 3 COOH. Ethanedioic acid was also known as oxalic acid. It is a diprotic acid, it can donate 2 protons (hydrogen ions) to a base. The reaction of sodium hydroxide with oxalic takes place in two stages because one of the hydrogen ions is easier to remove than the other. The two successive reactions are: then The ph curve for the titration of 25 cm 3 of 0.1 mol l -1 oxalic acid with 0.1 mol l -1 sodium hydroxide is shown below. This titration may need a mixture of two separate indicators to detect both equivalence points one with a pk In of 2.5, the other with a pk In of 8.5.
1. The graph below shows how the ph changes when 0.12 mol l -1 NaOH is added to 25.0 cm 3 of a solution of a weak monoprotic acid, HA. The equation for this reaction is NaOH + HA NaA + H2O 14 13 12 11 10 9 ph 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Volume of 0.12 M NaOH/cm 3 volume of 0.12 mol l -1 NaOH/cm 3 a. What volume of alkali was required to reach the equivalence point of the reaction? b. Use the graph to determine the ph at the equivalence point of the reaction. c. Use your answer to part a to calculate the initial concentration of the weak acid HA. d. Use the graph and your answer to part c to calculate the value of the dissociation constant, Ka, of the weak acid HA. 2. The ph indicator phenolphthalein is a weak acid which can be represented by the formula HIn. It dissociates in solution and has a pka value 9.3 HIn H + + In- colourless red a. State the colour of the indicator at the following ph values (i) 6.3 (ii) 7.9 (iii) 9.8
b. Explain why phenolphthalein is unsuitable for use in a titration between a strong acid and a weak base. c. Explain why phenolphthalein is unsuitable for use in a titration between a weak acid and a weak base. d. Calculate the value of the dissociation constant Ka for phenolphthalein. 3. A sample of 0.005 mol l -1 was propanoic acid was titrated with potassium hydroxide. a. Sketch the ph graph that would be obtained for this titration. Your graph should clearly show the ph at the equivalence point. b. From the list below select the best indicator for this titration and justify your choice. Indicator ph range bromophenol blue 3.0-4.6 methyl red 4.2-6.3 bromothymol blue 6.0-7.6 thymol blue 8.0-9.6 c. Calculate the ph of the propanoic acid before any potassium hydroxide was added. 4. The ph indicator thymol blue has different structures at different ph values. red yellow blue a. The formula of the unionised molecule responsible for the red colour of the indicator can be shown as HIn. Use this information to write an equation for the dissociation which takes place when the indicator changes colour from yellow to blue.
b. Explain why this indicator would be suitable for both the titration of a strong acid with a weak base and the titration of a weak acid with a strong base. 5. Titration curves labelled A,B, C and D for combinations of different acids and bases are shown below. All solutions have a concentration of 0.1 mol l -1. ph 14 12 10 8 6 4 2 0 0 10 20 30 40 50 ph 14 12 10 8 6 4 2 0 0 10 20 30 40 50 Volume/cm 3 Volume/cm 3 A B ph 14 12 10 8 6 4 2 0 0 10 20 30 40 50 ph 14 12 10 8 6 4 2 0 0 10 20 30 40 50 Volume/cm 3 Volume/cm 3 C D a. Select from A,B,C or D the curve produced by the addition of (i) (ii) (ii) aqueous ammonia to 25 cm 3 of hydrochloric acid benzoic acid to 25 cm 3 of potassium hydroxide sodium hydroxide to 25 cm 3 of nitric acid. b. A table of acid base indicators and the ph ranges over which they change colour is shown. Indicator ph range thymol blue 1.2 2.8 bromophenol blue 3.0 4.6 methyl red 4.2 6.3 cresolphthalein 8.2 9.8 thymolphthalein 9.3 10.5 Select from the table an indicator which could be used in the titration which produces curve A but not in the titration which produces curve B.
6. The ph curve below was obtained when a 0.05 mol l -1 solution of sodium hydroxide was added to 25 cm 3 of a diprotic acid H2A. The equation for the reaction is 2NaOH + H2A 2H2O + Na2A 10 20 30 40 a. Why are there two steeper parts on the graph? b. What volume of sodium hydroxide was required to completely neutralise the acid? c. Calculate the initial concentration of the diprotic acid. 7. The table gives some information about three different ph indicators. Indicator K indicator Colour change (low ph to high ph) Methyl yellow 5.1 x 10 4 Red to yellow Bromothymol blue 1.0 x 10 7 Yellow to blue Thymol blue 1.3 x 10 9 Yellow to blue a. At what ph would the colour change be expected to take place in each of the indicators. b. Which indicator would be most suitable in a titration of ammonia solution and hydrochloric acid?
A list of learning outcomes for the topic is shown below. When the topic is complete you should review each learning outcome. Your teacher will collect your completed notes, mark them, and then decide if any revision work is necessary. Need Help Revise Understand State the Bronsted - Lowry definition of an acid and a base. Define the terms strong and weak acid or base. Write equations to represent the dissociation of weak acids. Identify the conjugate base of an acid and the conjugate acid of a base. State that water is amphoteric and that an amphoteric substance can act as an acid or a base. Write an equation which represents the ionic product of water. State that the ionic product of water, K w, is an equilibrium constant and that its value is temperature dependent. State that water and any solution which contains [H + ] =[OH - ] is neutral but the ph of a neutral solution is only 7 when the temperature is 25 o C. State that ph = -log[h + ] and use this expression along with the ionic product of water to calculate the ph of any strong acid or strong base solution. State that weak acids have a dissociation constant, K a, and that this can be converted to a more readable pk a value using pk a = -logk a. State that as the value of K a increases the strength of the acid increases. Use the expression to calculate the ph of a weak acid. State that a buffer solution is one which resists ph changes when small volumes of acid or base are added to it and to state some uses of buffer solutions. State that a buffer solution is prepared from a weak acid and its salt or a weak base and its salt. Explain how a buffer solution is able to resist ph changes. Explain why the ph of a buffer solution is unaffected by dilution with water.
Be able to use the equation ph of a buffer of the composition of a buffer. to calculate either the State that ph indicators are weak acids. State that the pk a value of a particular indicator is equal to the ph at which the indicator will change colour. State the approximate equivalence point ph in a titration based on knowledge of the acid and base used in the titration. Be able to select the most suitable indicator for an acid/base titration from a selection of indicators based on the relative strengths of the acid and base. State that ph indicators are of little use in the titration of a weak acid with a weak base as at the equivalence point the ph does not change rapidly enough. I have discussed the learning outcomes with my teacher. Teacher comments. Date. Pupil Signature. Teacher Signature.
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