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 Equilibria Part 6

 "Acid-base titrations, indicators, buffers and non-aqueous media"

GCE-AS-A2-IB Advanced Level Theoretical-Physical Chemistry revision notes

GCSE Notes on reversible reactions-equilibrium * Advanced Part 1. Equilibrium, Le Chatelier's Principle-rules * Part 2. Kc and Kp equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * Part 4. Partition, solubility product and ion-exchange * Part 5. pH, weak-strong acid-base theory and calculations * Part 6 sub-index: 6.1 Salt hydrolysis * 6.2 Acid-base indicators and titrations * 6.3 Buffers - definition, formulation and action * 6.4 Buffer calculations * 6.5 Case studies of buffer function * 6.6 Acids-bases in non-aqueous media * Part 7. Redox equilibria, half-cell electrode potentials, electrolysis and electrochemical series * Part 8 Phase equilibria-vapour pressure, boiling point and intermolecular forces * The K and ΔS-ΔG connection with EØcell will be dealt with via new thermodynamics pages later, but an example of a ΔS-ΔG calculation is given at the end of the advanced kinetics pages and ΔG for cells is mentioned in Equilibria Part 7. M = old fashioned shorthand for mol dm-3 * EMAIL query?comment


6.1 Salt Hydrolysis

  • Despite being taught at lower academic levels that salts e.g. sodium chloride, dissolve in water to form neutral solutions of pH 7.

  • In reality, and looking at a wider variety of 'salts', the picture is much more complicated and a 'salt' solution may be acid, neutral or alkaline depending on the nature of the interaction of the salt ions with water.

  • The reasons are quite clear when you consider the possible Bronsted-Lowry interactions that can take place between the ions of the salt and water.

  • 6.1.1 Examples of acidic salt solutions: pH <7

    • 6.1.1a: Hexa-aqa metal cations often show acidic behaviour particularly with their salts with strong acids.

      • e.g. the hydrated aluminium ion from aqueous aluminium chloride/sulphate/nitrate.

      • [Al(H2O)6]3+(aq) + H2O(l) (c) doc b [Al(H2O)5(OH)]2+(aq) + H3O+(aq) 

        • or more simply: [Al(H2O)6]3+(aq) (c) doc b [Al(H2O)5(OH)]2+(aq) + H+(aq) 

      • The hydrated metal ion acts as an acid (proton donor) and water acts as the base (proton acceptor) and so aqueous hydrogen/oxonium ions are formed.

      • The greater the charge on the central metal ion, the stronger the hexa-aqua ion acid. e.g.

      • [Al(H2O)6]3+(aq) > [Mg(H2O)6]2+(aq) > [Na(H2O)6]+(aq) (the cations of Gps 1-3 on Period 3)

      • or [Fe(H2O)6]3+(aq) > [Fe(H2O)6]2+(aq) (in the 3d-block transition metal example)

        • From left to right, the trend is due to a decreasing charge density effect of the central metal ion on the O-H bond of a co-ordinated water molecule. The charge density decreases as the positive charge of the central metal ion decreases and its ionic radius increases.

        • The sodium ion shows virtually no acidic behaviour.

        • Further discussion of this situation will be on the Transition Metals Appendix page section 1 (currently under production).

      • However, the anion of the salt must not be neglected for a full explanation. The anions derived from the very strong  hydrochloric/sulphuric/nitric acids are all very weak bases and so have little tendency to interact with water in an acid-base reaction. Its a general, and logical rule, that the conjugate base of a very strong acid is very weak.

    • 6.1.1b: Salts of weak bases and strong acids give acidic solutions.

      • e.g. ammonium chloride. The chloride ion is such a weak base that there is no acid-base reaction with water, but the ammonium ion is an effective proton donor. As a general rule, the conjugate acid of a weak base is quite strong. The result here is that ammonium salt solutions have a pH of 3-4.

      • NH4+(aq) + H2O(l) (c) doc b NH3(aq) + H3O+(aq)

      • In zinc-carbon batteries an acidic ammonium chloride paste dissolves the zinc in the cell reaction, though an oxidising agent must be added (MnO2) to oxidise the hydrogen formed into water, or batteries might regularly explode!

      • If you place a piece of magnesium ribbon or a zinc granule in ammonium chloride or ammonium sulphate solution you will see fizzing as hydrogen gas is formed.

        • 2H3O+(aq) + M(s) ==> M2+(aq) + H2O(l) + H2(g)

        • M = zinc or magnesium

  • 6.1.2 Examples of nearly neutral salt solutions: pH approx. 7

    • 6.1.2a: Salts of strong acids and strong bases e.g. sodium chloride

      • Here the hexa-aqua sodium ion shows no acidic behaviour and the chloride ion no base behaviour, so little or no interaction with water to produce either H+(aq) or OH-(aq) to change the pH.

    • 6.1.2b: Salts of weak acids and weak bases: e.g. ammonium ethanoate

      • Here the ammonium ion can act as an acid to form H+(aq) with water, but the ethanoate ion acts as a base to give  OH-(aq) with water, so they effectively neutralise each other.

  • 6.1.3 Examples of alkaline salt solutions: pH>7

    • 6.1.3a: Salts of a weak acid and a strong base e.g. sodium ethanoate

      • The hydrated sodium ion shows no acidic character but the ethanoate ion is a strong conjugate base of a the weak ethanoic acid (pKa = 4.76,  Ka = 1.74 x 10-5 mol dm-3), so an acid-base hydrolysis reaction occurs to generate hydroxide ions to raise the pH to about pH 9.

        • CH3COO-(aq) + H2O(l) (c) doc b CH3COOH(aq) + OH-(aq)

    • 6.1.3b: Potassium cyanide: is the salt of the very strong base potassium hydroxide and the very weak hydrocyanic acid (pKa = 9.31, Ka = 4.9 x 10-10 mol dm-3). The hydrated potassium ion shows no acidic behaviour, but the cyanide ion is a strong conjugate base of the very weak hydrocyanic acid (HCN) which interacts with water to generate hydroxide ions. Hydrocyanic acid (pKa = 9.4) is weaker than ethanoic acid (pKa = 4.76) , so the equilibrium is more on the right, more OH-, and so the pH is more alkaline, i.e. over 9.

      • CN-(aq) + H2O(l) (c) doc b HCN(aq) + OH-(aq)

    • 6.1.3c: Sodium carbonate is the 'salt' of the strong base sodium hydroxide and the very weak 'carbonic acid'

      • Again the hydrated sodium ion shows no acidic character but the carbonate ion is a strong conjugate base of a the weak 'carbonic' acid, so an acid-base hydrolysis reaction occurs to generate hydroxide ions to raise the pH.

      • CO32-(aq) + H2O(l) (c) doc b HCO3-(aq) + OH-(aq)

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6.2 The theory of acid-base indicators and pKind

  • Equilibria part 5 "pH and weak-strong acids and bases in aqueous solution" should have been studied prior to tackling Part 6.

  • 6.2.1 Acid-base titration indicators are quite often weak acids in which the unionised acid (lets call it HIn) and its 'de-protonated' form, or conjugate base, the anion (In-), have different colours.

    • One form can be colourless e.g. phenolphthalein in acid-neutral solutions.

    • The equilibrium can be simply expressed as ....

    • HIn-(aq, colour 1) (c) doc b H+(aq) + In-(aq, colour 2)

      • Applying Le Chatelier's equilibrium principle:

      • Addition of acid favours the formation of more HIn (colour 1)

        • HIn(aq) (c) doc b H+(aq) + In-(aq)

          • because an increase on the right of [H+] causes a shift to left increasing [HIn] to minimise 'enforced' rise in [H+].

      • Addition of alkali favours the formation of more I- (colour 2):

        • HIn(aq) (c) doc b H+(aq) + In-(aq)

          • The increase in [OH-] causes a shift to right because the reaction

          •  H+(aq) + OH-(aq) ==> H2O(l)

          • reduces the [H+] on the right so more HIn ionises to try to increase the [H+] i.e. minimising the change in [H+].

  • 6.2.2 The colour that is observed will depend on the ratio [HIn]/[In-], but at pH extremes i.e. very acid, colour 1 will dominate, or in very alkaline solution, colour 2 will dominate. Therefore the maximum colour 'shade' change from one to the other will occur when [HIn] = [In-], or [colour 1] = [colour 2].

    • The pH when [HIn] = [In-] can be calculated from the dissociation constant, Kind (ka), for the weak acid indicator.

    • Kind = [H+(aq)] [In-(aq)]/[HIn(aq)], but when  [HIn] = [In-] the equilibrium expression simplifies to ...

    • Kind = [H+(aq)], so at this point the pH = -log(Kind)

    • and is referred to as the pKind value.

  • 6.2.3 pH titration curves and choice of indicator

    • The greatest change in indicator colour (per volume of reagent added), will occur at the equivalence point in the titration.

    • Therefore you need to choose an indicator with a pKind close to the pH at the equivalence point (theory above).

    • In fact acid-base titration indicators are usually effective over a range of several pH units but it is essential for accurate titrations that the colour change is sharp at the equivalence point with a small addition of acidic or alkaline titration solution.

    • Universal indicator is NOT suitable for quantitative analysis and the indicator choices tabulated below are explained via the sets of pH graphs shown further down using Graphs A to D. The effective pH range of the indicator is where there is sufficient colour change to give a good sharp end-point and can be above or below, but close to the pKind value of the indicator. Some effective pH ranges for selected indicators are given below.

  • Indicator colour change, from acid to alkali

    pKind pH range example of titration use
    Methyl orange, (red ==> yellow) 3.7 3.1-4.4 weak base - strong acid titration e.g. ammonia titrated with hydrochloric acid
    Bromophenol blue, (yellow ==> blue) 4.0 2.8-4.6 weak base - strong acid titration
    Methyl red, (red ==> yellow) 5.1 4.2-6.3 weak base - strong acid titration
    Bromothymol blue, (yellow ==> blue) 7.0 6.0-7.6 strong acid - strong base titration e.g. hydrochloric acid <=> sodium hydroxide titration
    Phenol red, (yellow ==> red) 7.9 6.8-8.4 strong acid - strong base titration e.g. hydrochloric acid <=> sodium hydroxide titration
    Thymol blue (base form), (yellow ==> blue) 8.9 8.0-9.6 weak/strong acid - strong base titration
    Phenolphthalein, (colourless ==> pinky-red) 9.3 8.3-10.0 weak acid - strong base titration e.g. ethanoic acid titrated with sodium hydroxide
  • 6.2.4 The change in pH through various titrations is illustrated and explained to extend the idea of choosing the right indicator.

    • Below are two graphs of sets of curves showing how the pH changes when weak/strong alkalis are added to weak/strong acids (set A) and vice versa (set B).

    • The curves are a bit simplified and approximate, but show how the pH changes in titrations.

    • By putting two graph sections from (1) to (4) together you can construct an approximate pH curve.

    • This is done below each graph for the four acid-base permutations and it is assumed the acids and bases are monobasic/monoprotic.

      • Note:

        1. The end-point = equivalence point or stoichiometric point.

        2. If a buffer calibrated  pH meter is used rather than an indicator, the end-point is obtained from the graph at the mid-point of the steepest inflexion of the titration curve.

        3. The first two graphs (A and B) assume 20cm3 of the acid/alkali is titrated with an alkali/acid of the same concentration e.g. 0.1 or 1.0 mol dm-3.

        4. For monoprotic/monobasic acids-base titrations there is only one point of steepest inflexion on the pH curve.

        5. However, apart from the strong acid-strong base curves, there are one or two other, but much less steep, points of inflexion due to the formation of a buffer mixture (see determination of Ka of weak acid via titration curve).

        6. The formation of this buffer mixture makes the end-point less sharp because it resists pH change.

        • pHcurve1 Graph A

    • 6.2.5 pH curves - Graph A: The pH change when adding soluble base (alkali) to acid

      • 6.2.5a: Curve A1 (1) + (3): Adding a weak base to a strong acid, end point at (i1), approx. pH 3-5.

        • pH change at end-point reasonable sharp e.g. adding ammonia to hydrochloric acid, though this titration is usually done the other way round, for more details see B4 below.

      • 6.2.5b: Curve A2 (1) + (4): Adding a strong base to a strong acid, end point (i2), approx. pH 7.

        • pH change at end-point very sharp e.g. titrating hydrochloric acid with sodium hydroxide.

        • Suitable indicators: bromothymol blue (pKind 7.0, range 6.0-7.6), phenol red (pKind 7.9, range 6.8-8.4), phenolphthalein (pKind 9.3, range 8.3-10.0, ok for any strong acid - strong base titration because the pH change is so sharp at the end-point i.e. the point of inflexion is very sharp for 1-2 drops of alkali over pH 3-10)

          • This titration has the sharpest pH change at the end-point, hence the sharpest indicator colour change of all the titrations due to lack of buffering effects.

      • 6.2.5c: Curve A3 (2) + (3): Adding a weak base to a weak acid, end point (i2), approx. pH 7.

        • pH change at end-point not very sharp, not practical for any titration e.g. adding ammonia to ethanoic acid.

        • Suitable indicators: None.

          • This titration gives the lowest rate of change of pH approaching the end-point, hence the poorest end-point to detect with indicator. This is due to strong buffering effect of the mixture of a weak base, weak acid and their salt. (for more details see buffer examples 6.3.1 and 6.3.2)

      • 6.2.5d: Curve A4 (2) + (4): Adding a strong base to weak acid, end point (i3), approx. pH 9.

        • pH change at end-point reasonable sharp e.g. you can titrate weak organic acids like ethanoic acid with sodium hydroxide.

        • Suitable indicators:  phenolphthalein (pKind 9.3, range 8.3-10.0), thymol blue (base, pKind 8.9, range 8.0-9.6)

          • The lower rate of change of pH approaching the end-point compared to curve A2 (above) is due to the weak buffering effect of the mixture of a weak acid and the salt of a weak acid-strong base. (for more details see buffer example 6.3.1)

          • Using a pH titration curve to determine the Ka of a weak acid

          • pHcurve1Ka Graph E

          • When a weak acid (HA) is titrated with a strong base (e.g. NaOH) a buffer mixture of A- and HA exists from soon after the titration starts to near the end-point.

          • Therefore, half-way to the equivalence point e.g. on addition of 10 cm3 of alkali of a 20 cm3 titration (Graph E, curve (2) above), it means in terms of concentrations

          • [NaA(aq)]salt = [A-(aq)] = [HA(aq)]unreacted acid

          • Now, the equilibrium expression for a mono basic/protic weak acid is ...

          • Ka =

            [H+(aq)] [A-(aq)]

            ------------

                [HA(aq)]

          • so, at the half-way point, when [A-] = [HA], Ka = [H+(aq)],

          • or at half-way point: pH = pKa and Ka = 10-pKa.

          • In the 'ficticious' case of the weak acid above the pH is 4.2 at this point,

          • therefore: [H+(aq)] = Ka = 6.3 x 10-5 mol dm-3.

      • pHcurve2 Graph B

    • 6.2.6 pH curves - Graph B: The pH change when adding acid to a soluble base (alkali)

    • 6.2.6a: Curve B1 (1) + (3): Adding a weak acid to a strong base, end point (i1), approx. pH 9.

      • Reasonably sharp pH change at end-point, but this titration is usually done the other way round, for more details see A4.

    • 6.2.6b: Curve B2 (1) + (4): Adding a strong acid to strong base, end point (i2), approx. pH 7.

      • pH change at end-point very sharp e.g. titrating sodium hydroxide with hydrochloric acid. For more details see A2 above.

    • 6.2.6c: Curve B3 (2) + (3): Adding a weak acid to weak base, end point (i2), approx. pH 7.

      • The pH change at the end-point NOT very sharp and so not suitable for titrations e.g. adding ethanoic acid to ammonia solution (see A3 above for more details).

    • 6.2.6d: Curve B4 (2) + (4): Adding a strong acid to weak base, end point (i3), approx. pH 3-5

      • e.g. titrating ammonia with hydrochloric acid.

      • Suitable indicators: methyl orange (pKind 3.7, range 3.1-4.0)

        • The lower rate of change of pH approaching the end-point compared to curve B2 (above) is due to buffering effect of the mixture of a weak base and the salt of a weak base-strong acid. (for more details see buffer example 6.3.2)

  • 6.2.7 More complicated pH titration curves

    • 6.2.7a The titration of a weak dibasic acid e.g. 25 cm3 of 0.1 mol dm-3 ethanedioc acid (oxalic acid) titrated with 0.1 mol dm-3 sodium hydroxide

      • pHcurve3 Graph C

      • There are two inflexion points on the pH curve corresponding to the half and full neutralisation of the dibasic/diprotic acid..

        • HOOC-COOH(aq) + NaOH(aq) ==> HCOO-COO-Na+(aq) + H2O(l)

        • ionically: HOOC-COOH(aq) + OH-(aq) ==> HCOO-COO-(aq) + H2O(l)

        • HCOO-COO-Na+(aq) + NaOH(aq) ==> Na+-OOC-COO-Na+(aq) + H2O(l)

        • ionically: HCOO-COO-(aq) + OH-(aq) ==> -OOC-COO-(aq) + H2O(l)

      • To detect the 2nd end-point, and hence the acid quantitatively, phenolphthalein indicator (pKind 9.3, range 8.3-10.0) is used, since it is essentially a weak acid-strong base titration.

        • I'm not sure if the 1st 'end-point' can be detected e.g. using methyl orange (pKind 3.7, range 3.1-4.0) and I doubt if it is of any quantitative use?

      • The lower rate of change of pH approaching the end-point compared to a strong base-strong acid titration is due to the weak buffering effect of the mixture of a weak acid and the salt of a weak acid-strong base. (for more details see  Case study ?)

      • Other acids like propanedioic acid (malonic acid) and butanedioic acid (succinic acid) behave, and be titrated, in the same way.

      • In the case of the tribasic/triprotic phosphoric(V) acid, H3PO4, you would get three points of inflexion on the titration curve of added sodium hydroxide versus pH corresponding to the formation of

        • (i) NaH2PO4, (ii) Na2HPO4 and (iii) Na3PO4

        • as each hydrogen of the acid is successively replaced.

    • 6.2.7b The titration of 25cm3 of 0.1 mol dm-3 sodium carbonate titrated with 0.1 mol dm-3 hydrochloric acid.

      • pHcurve4 Graph D

      • There are two inflexion points on the pH curve.

      • Endpoint (1) corresponds to the 1st stage of neutralisation, the formation of the hydrogencarbonate ion.

        • Na2CO3(aq) + HCl(aq) ==> NaCl(aq) + NaHCO3(aq)

        • ionically: CO32-(aq) + H+(aq) ==> HCO3-(aq)

        • This end-point at around pH 8-9 can be detected with phenolphthalein (pKind 9.3, range 8.3-10.0) or thymol blue - base form (pKind 8.9, range 8.0-9.6)

        • If the conical flask is rapidly swirled on adding the acid, you don't see any gas bubbles of carbon dioxide.

      • End-point (2) corresponds to the 2nd stage of neutralisation, the formation of water and carbon dioxide.

        • NaHCO3(aq) + HCl(aq) ==> NaCl(aq) + H2O(l) + CO2(g)

        • ionically: HCO3-(aq) + H+(aq) ==> H2O(l) + CO2(g)

        • This end-point around pH 3-4 can be detected with methyl orange indicator (pKind 3.7, range 3.1-4.0) or bromophenol blue (pKind 4.0, 2.8-4.6).

      • Overall the reaction is ...

        • Na2CO3(aq) + 2HCl(aq) ==> 2NaCl(aq) + H2O(l) + CO2(g)

        • ionically: CO32-(aq) + 2H+(aq) ==> H2O(l) + CO2(g)

      • Note:

        • A mixture of sodium carbonate and sodium hydrogencarbonate can be analysed using two separate titrations.

        • Titration (i) using phenolphthalein indicator measures the sodium carbonate,

        • and titration (ii) measures the sodium carbonate plus the sodium hydrogencarbonate. So both quantities can be calculated from the titration results.

  • 6.2.8 Back titrations

    • Examples include ...

      • Where a known excess of acid is added to a base/alkali and the unreacted acid is 'backtitrated' with a standard alkali solution,

      • Where a known excess of alkali is added to an acid and the unreacted alkali is 'backtitrated' with a standard acid solution.

  • 6.2.9 Examples of acid-base titration questions with all the answers and working.

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6.3 Buffers - definition, formulation and action

  • 6.3.1 A buffer is a solution that minimises pH change on the addition of small amounts of acid or alkali.

    • Never say "it prevents change in pH".

  • Buffers and their chemical reactions must obey Le Chatelier's Equilibrium Concentration Principle, and act in a way to remove H+ and OH- ions. BUT, they cannot theoretically prevent the pH being lowered/raised by the addition of acid/alkali, however small the change. Any buffer will eventually be 'used up' if large quantities of acid or alkali are added to the solution.

  • 6.3.2 Typical buffers and their action.

    • The buffering chemistry is quite simple in principle and the ideas behind the examples described below can be applied to the design and action of most buffers.

  • Buffering action example 6.3.2a

    • A mixture of a weak acid and the salt of the weak acid with a strong base.

    • Organic acids like methanoic, ethanoic, propanoic, citric, benzenedicarboxylic etc. are frequently used in buffer mixtures i.e. those with the carboxylic acid functional group -COOH

    • The salts are usually those of the strong base-alkalis sodium and potassium hydroxide.

    • e.g. ethanoic acid CH3COOH and sodium ethanoate CH3COO-Na+ gives buffers in the range pH 3.7-5.6

    • CH3COOH and CH3COO- constitute a conjugate acid-base pair.

    • In solution most of the weak acid is NOT ionised and the relatively high concentration of the CH3COO- ion actually inhibits ionisation.

      • It is the relatively high concentration of the weak ethanoic acid that 'removes' any added hydroxide ions:

        • CH3COOH(aq) + OH-(aq) (c) doc b CH3COO-(aq) + H2O(l)

        • Note the use of the 'styled' reversible sign to show a bias to RHS of the equilibrium

    • The salt is fully ionised in solution to give a relatively high concentration of the ethanoate ions.

      • It is the ethanoate ion which removes most of any added hydrogen ions.

        • CH3COO-(aq) + H+(aq) (c) doc b CH3COOH(aq)

        • The conjugate base of a weak acid is relatively strong!

  • Buffering action example 6.3.2b

    • A mixture of a weak base and the salt of the weak base with a strong acid.

    • e.g. ammonia NH3 and ammonium chloride NH4+Cl-

    • NH4+ and NH3 constitute a conjugate acid-base pair.

    • In solution most of the ammonia is NOT ionised (and even suppressed by the ammonium ions from the salt).

      • It is the weak base that 'removes' most of any added hydrogen ions.

        • NH3(aq) + H+(aq) (c) doc b NH4+(aq)

    • The salt is fully ionised in solution giving relatively high concentrations of the ammonium ion.

      • It is the ammonium ion that removes most of any added hydroxide ions.

        • NH4+(aq) + OH-(aq) (c) doc b NH3(aq) + H2O(l)

    • ?

  • 6.3.3 Preparing buffer solutions.

    • Quite often several solutions of salts, weak acids/bases are prepared and then mixed in different ratios to provide buffers of a wide pH range.

    • Sometimes a single salt will do to give a single accurate pH value for calibrating a pH meter. (see Case study 6.5.1)

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6.4 Buffer calculations

  • 6.4.1 Calculations involving a buffer made from a weak acid and its salt with a strong base.

    • Consider the mixture is made from a monobasic acid HA and an alkali metal salt M+A- (e.g. A = CH3COO, M = K or Na):

    • it is reasonable to assume for simple approximate calculations that ..

      1. [A-(aq)] = [salt(aq)] since salt fully ionised and M+ is a spectator ion, and

      2. [HA(aq)]equilib., = [HA(aq)]initial since little of the weak acid is ionised.

    • Therefore the weak acid Ka expression:

      • Ka =

        [H+(aq)] [A-(aq)]
        ------------ mol dm-3
            [HA(aq)]
      • becomes

      • Ka =

        [H+(aq)] [salt(aq)]
        -------------
           [acid(aq)]
      • therefore: [H+(aq)] = Ka [acid(aq)] / [salt(aq)] mol dm-3

        • Note: For a given conjugate pair (HA and A-), the pH of the buffer is determined by the acid/salt ratio, though the more concentrated the buffer, the greater its capacity to neutralise larger amounts of added/formed in a reaction medium.

      • and: pHbuffer = -log(Ka x [acid(aq)] / [salt(aq)])

  • 6.4.2 Buffer calculations

    • Buffer calculation example 6.4.2a

      • A buffer solution was prepared which had a concentration of 0.20 mol dm-3 in ethanoic acid and 0.10 mol dm-3 in sodium ethanoate. If the Ka for ethanoic acid is 1.74 x 10-5 mol dm-3, calculate the theoretical hydrogen ion concentration and pH of the buffer solution.

      • Ka = [H+(aq)] [salt(aq)]/[acid(aq)]

      • 1.74 x 10-5 = [H+(aq)] x 0.10 / 0.20

      • [H+(aq)] = 1.74 x 10-5 x 0.20/0.10 = 3.48 x 10-5 mol dm-3

      • pH = -log(3.48 x 10-5) = 4.46

    • Buffer calculation example 6.4.2b

      • In what ratio should a 0.30 mol dm-3 of ethanoic acid be mixed with a 0.30 mol dm-3 solution of sodium ethanoate to give a buffer solution of pH 5.6?

        • Ka for ethanoic acid is 1.74 x 10-5 mol dm-3

      • [H+(aq)] = 10-pH = 10-5.6 = 2.51 x 10-6 mol dm-3

      • Ka = [H+(aq)] [salt(aq)]/[acid(aq)]

      • [salt]/[acid] = Ka/[H+(aq)] = 1.74 x 10-5/2.51 x 10-6  = 6.93

      • Therefore volume ratio is 6.93 : 1 for salt : acid, e.g. 6.93 cm3 of 0.30M sodium ethanoate is mixed with 1.0 cm3 of 0.30 M ethanoic acid to give a buffer solution of pH 5.6.

      • Note that the pH is determined by the ratio of concentrations, but the buffering capacity of the solution can be increased by increasing the concentrations of both components in the same molar concentration ratio.

    • Buffer calculation example 6.4.2c

      • What is the pH of a buffer solution made from dissolving 2.0g of benzoic acid and 5.0g of sodium benzoate in 250 cm3 of water?

        • Ka benzoic acid = 6.3 x 10-5 mol dm-3, Ar's: H = 1, C = 12, O = 16, Na = 23

        • Molecular masses: Mr(C6H5COOH) = 122, Mr(C6H5COO-Na+) = 144, 250 cm3 = 0.25dm3

        • moles acid C6H5COOH = 2.0/122 = 0.0164 mol, molarity = 0.0656 mol dm-3

        • moles salt C6H5COO-Na+ = 5.0/144 = 0.0347 mol, molarity = 0.139 mol dm-3

        • [H+(aq)] = Ka [acid(aq)]/[salt(aq)]

        • [H+(aq)] = 6.3 x 10-5 x 0.0656 / 0.139 = 2.97 x 10-5 mol dm-3

        • pH = -log[H+(aq)] = -log(2.97 x 10-5) = 4.53

    • Buffer calculation example 6.4.2d

      • Calculate the pH of a buffer made by mixing 100 cm3 of a 0.40 M sodium propanoate and 50 cm3 of 0.2 M propanoic acid solution.

        • Ka propanoic acid = 1.3 x 10-5 mol dm-3, total volume of buffer = 150 cm3

        • molarities in the mixture:

          • [salt] = 0.40 x 100/150 = 0.267 mol dm-3

          • [acid] = 0.20 x 50/150 = 0.0667 mol dm-3

        • [H+(aq)] = Ka [acid(aq)]/[salt(aq)]

        • [H+(aq)] = 1.3 x 10-5 x 0.0667 / 0.267 = 3.24 x 10-6 mol dm-3

        • pH = -log[H+(aq)] = -log(3.24 x 10-6) = 5.48

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6.5 Case studies of the function and uses of buffers in aqueous media

  • Case study 6.5.1 Other common buffer solutions and their use in the laboratory.

    • (c) doc bPotassium hydrogen benzene-1,2-dicarboxylate is an 'all in one' buffer solution of pH 4.0

      • (i) H+(aq) + -OOC-C6H4-COOH(aq) (c) doc b HOOC-C6H4-COOH(aq)

      • (ii) -OOC-C6H4-COOH(aq) + OH-(aq) (c) doc b -OOC-C6H4-COO-(aq) + H2O(l)

      • (i) removes hydrogen ions and (ii) removes hydroxide ions.

      • Buffers can be made by mixing the salt with the original benzene-1,2-dicarboxylic acid to give buffers in the range 2.2-3.8

    • A mixture of salts of a polybasic/polyprotic acid e.g. the salts KH2PO4, Na2HPO4 and Na3PO4 from phosphoric(V) acid (a tribasic/triprotic acid) can give buffer solutions in the range pH 6-12 e.g.

      • from Na2HPO4: HPO42-(aq) + H+(aq) (c) doc b H2PO4-(aq)  (removes hydrogen ions)

      • from Na3PO4: PO43-(aq) + H+(aq) (c) doc b HPO42-(aq) + H2O(l) (removes hydrogen ions)

      • from KH2PO4: H2PO4-(aq) + OH-(aq) (c) doc b HPO42-(aq) + H2O(l) (removes hydroxide ions)

      • from Na2HPO4: HPO42-(aq) + OH-(aq) (c) doc b PO43-(aq) + H2O(l) (removes hydroxide ions)

      • The HPO42- ion is amphoteric, acting both as a proton donor and acceptor and phosphate(V) ions are important in the buffering of intracellular fluids in living organisms (see Case study 6.5.2 below).

    • Buffer solutions are used to accurately calibrate pH meters.

  • Case study 6.5.2 The importance of buffering in biological systems

    • enzyme2The importance of a stable pH cannot be over emphasised for the efficient function of enzyme-proteins which control most of the chemistry of life. Most enzymes can only operate at their maximum catalytic capacity over a narrow range of pH and three examples are shown. Examples of conjugate acid-base pairs that fulfil this important buffering operation are described below.
    • 6.5.2a Optimum enzyme function.

    • 6.5.2b Inside cells hydrogenphosphate(V) ions act as the major intracellular buffer system, with contributions from organic phosphates such as glucose-6-phosphate and ATP.

      • HPO42-(aq) + H+(aq) (c) doc b H2PO4-(aq) or H2PO4-(aq) + OH-(aq) (c) doc b HPO42-(aq)

    • 6.5.2c The major extracellular buffer is the 'carbonic acid'-'bicarbonate' or hydrogencarbonate system which enables e.g. blood, to function as an extraordinary effective buffer operating at about pH 7.

      • (i) CO2(g, lungs) + aq (c) doc b (ii) CO2(aq*) + H2O(l) (c) doc b (iii) H2CO3(aq) (c) doc b (iv) HCO3-(aq) + H+(aq)

      • (v) H2CO3(aq) + OH-(aq) (c) doc b (vi) HCO3-(aq) + H2O(l)  *(aq) = intra/extracellular fluids

      • (iv) to (iii) removes hydrogen ions and (v) to (vi) removes hydroxide ions.

      • The effectiveness of the system depends on the reservoir of dissolved carbon dioxide in the blood plasma and the gas in the lungs.

        • If hydroxide ions are removed via reaction (v) to (iv), the depleted H2CO3 is readily replaced via the reaction sequence (i) to (ii) and (ii) to (iii).

        • If hydrogen ions are removed via (iv) to (iii) the reverse sequence of 1. can restore the system to the original pH.

      • The ability of mammals to maintain a fairly constant [HCO3-]/[H2CO3] ratio in blood plasma is reflected in the rate of CO2 production in the cell oxidation reactions of respiration and the rate of CO2 loss by expiration.

      • The blood plasma of man is about 7.4 and any deviation below 7.0, or above 7.8, as can happen in disease, can cause irreparable damage.

      • Intracellular and extracellular systems are very pH sensitive and small changes in pH can produce ill-effects in living organisms, hence, e.g. the bodies irritation by all except the very weakest of acids and alkalis in contact with the skin.

  • Case study 6.5.3 Shampoos

    • to complete

  • Case study 6.5.4: *

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6.6 Acid and bases in non-aqueous media

  • Liquid ammonia can self-ionise just like water (need low temperatures and high pressure!).

    • 2NH3 (c) doc b NH4+ + NH2-

    • One ammonia molecule acts as the acid and another acts as the acid.

    • The ammonium ion, NH4+, is the conjugate acid.

    • The amide ion, NH2-, is the conjugate base.

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