topic guide 5.3: equilibrium processes - · pdf filesurroundings. 2 it is dynamic ......

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Unit 5: Chemistry for Applied Biologists

Many biochemical processes, for example, osmosis, partitioning between lipids and extracellular fluid and the redox reactions of respiration, are reversible and ideas about chemical equilibrium are important in understanding how these systems respond to change.

Among the most important of these reversible processes are those involving acid–base processes that affect the pH of a biological environment. Simple mathematical techniques enable useful calculations to be carried out to help understand the factors that affect the pH of these environments.

On successful completion of this topic you will: • understand the features of equilibrium processes (LO3)

To achieve a Pass in this unit you need to show that you can: • explain the features of equilibrium processes (3.1) • interpret the values of calculated equilibrium constants (3.2) • explain the effects of changes in conditions on the position of chemical

equilibrium (3.3) • explain the relationship between acid dissociation constant and pH (3.4).

Equilibrium processes5.3

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5.3: Equilibrium processes

1 Features of equilibrium processesLipophilicity is one of the simplest concepts to grasp of all those needed to understand drug action mechanisms.

The lipophilicity of a substance is measured by allowing molecules of a substance (M) to partition between two solvents – usually water and octan-1-ol.

M

Octan-1-ol

Water

M

M

MM M

MM

MMM

The process occurring in the system shown in Figure 5.3.1 can be represented as:

M(aq) ⇌ M (octan-1-ol)

This illustrates several key features of a chemical equilibrium.

1 It occurs in a closed system – no matter can be exchanged with the surroundings.

2 It is dynamic – at the interface between the two solvents both the forward and backward processes are occurring.

3 The concentrations of the reactants and products remain constant because the rates of the forward and backward reactions are equal at equilibrium.

4 At equilibrium, the concentrations of products and reactants may be very different. In the case of the partitioning of M illustrated in Figure 5.3.1, there is a higher concentration of products than reactants – the equilibrium position lies over to the right.

5 The equilibrium can be approached from either the product or reactant side of an equation. So if an aqueous solution of M was shaken together with an octan-1-ol solvent, the equilibrium position reached would be identical to that produced when a solution of M in octan-1-ol was shaken with water.

Key termLipophilicity: The tendency of a drug to dissolve in a lipid.

Figure 5.3.1: The partitioning of a solute between two solvents is a simple

example of a dynamic equilibrium.

ActivityResearch or revise the process of osmosis. How well do the features listed above apply to the process of osmosis?

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2 Equilibrium constantsPartition coefficientsWhen a solute, for example, paracetamol, is allowed to partition between two solvents, equilibrium will be reached:

paracetamol (aq) ⇌ paracetamol (octan-1-ol)

At equilibrium, the ratio of concentrations in two solvents remains constant (at a given temperature). This ratio is known as the partition coefficient (K

ow) (the

subscripts ‘o’ and ‘w’ refer to octan-1-ol and water).

Kow

= [paracetamol(octan-1-ol)]

[paracetamol(aq)] = 3.0

I work for a company that supplies analytical services to the pharmaceutical industry, in particular to research teams developing new drug treatments. Modern drug discovery and development relies heavily on computer modelling to predict the properties and effectiveness of molecules which may be candidates for drug development. One of the important parameters which must be known about a drug molecule is its lipophilicity, and this can be predicted with some degree of reliability by computer models. But before confirming a drug’s suitability for further work, this computer prediction must be verified by experimental data.

We can analyse samples of new compounds and provide an accurate figure for the octan-1-ol/water partition coefficient. The assay uses less than 1 mg of the compound; we dissolve it in DMSO (dimethyl sulfoxide) and then shake this together with octan-1-ol and an aqueous solvent. The relative concentrations in each layer are measured by a combination of high-performance liquid chromatography (HPLC) and ultraviolet spectroscopy. From these concentrations we can calculate an extremely accurate value for the partition coefficient within a rapid turnaround time of less than two weeks, providing a fast and reliable service for our clients.

Alison Stokes, Analytical Chemist

Writing and interpreting equilibrium constantsThe partition coefficient we have just seen is a simple example of an equilibrium constant. The size of the equilibrium constant gives information about the position of equilibrium – in this case it is clear that the concentration of paracetamol in the octan-1-ol layer is greater than the concentration of paracetamol in the aqueous layer, and so the equilibrium position lies to the right.

Writing equilibrium constants

For a more complex equation:

pA + qB ⇌ rC + sD

Kc (the equilibrium constant) =

[C]r[D]s

[A]p[B]q

Kc has a constant value at any given temperature.

Key term[paracetamol(aq)]: This means ‘the concentration of paracetamol in the aqueous layer’. The use of square brackets to represent concentration terms is a commonly-used shorthand device.

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ActivityEthanoic acid (CH

3COOH) and ethanol (C

2H

5OH) react together in the presence of an acid catalyst

to form an ester, ethyl ethanoate (CH3COOC

2H

5) and water (H

2O).

The equation for the reaction is:

CH3COOH(l) + C

2H

5OH(l) ⇌ CH

3COOC

2H

5(l) + H

2O(l)

In one experiment, the concentrations of the four substances were as shown in Table 5.3.1. • Write down an expression for the equilibrium constant, K

c, corresponding to the equation

above. • Use the data in the table to calculate a value for K

c under the conditions of the experiment.

• Comment on the position of equilibrium of this reaction under these conditions.

Substance Concentration / mol dm–3

CH3COOH 0.12

C2H

5OH 0.29

CH3COOC

2H

50.38

H2O 0.38

Table 5.3.1

Many biochemical reactions can be treated as equilibrium processes. For example, in the reduction of pyruvate to lactate (which you also met in Topic guide 5.2), the equation for the reaction can be written as:

NADH + H+ + pyruvate ➝ lactate + NAD+

Kc for this reaction is quoted as 3.6 × 104, so the equilibrium position lies very

much over to the right-hand side.

Interpreting the values of equilibrium constants

The size of the equilibrium constant provides important information about the position of equilibrium. Table 5.3.2 shows how to interpret the values of equilibrium constants.

Value of Kc Interpretation

<<1 (e.g. 10–10) Equilibrium position almost completely over the LHS. Reaction appears not to occur at all.

<1 (e.g. 10–1 to 10–4) Equilibrium position favours LHS.

Close to 1 Equilibrium position favours neither products nor reactants.

>1 (e.g. 101 to 104) Equilibrium position favours RHS.

>>1 (e.g. 1010) Equilibrium position almost completely over the RHS. Reaction appears to go to completion.

Take it furtherWriting K

c expressions for biochemical reactions can present some difficulties. Some possible

approaches to these problems are suggested at http://www.chem.qmul.ac.uk/iubmb/newsletter/1996/news7.html.

Many reactions occurring in living organisms do not strictly reach equilibrium and this can present difficulties in interpreting thermodynamic data (such as ΔG values) and equilibrium constants. One approach to this issue is discussed at http://sandwalk.blogspot.co.uk/2011/10/better-biochemistry-near-equilibrium.html.

Table 5.3.2: Some values of K

c and their interpretation.

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You will probably get the opportunity to carry out some experimental work which will enable you to calculate a value for an equilibrium constant.

Portfolio activity (3.1, 3.2)Carry out experimental work on an equilibrium system. Systems involving the formation of esters from carboxylic acids and alcohols can provide suitable results. Use the results of your experiment to calculate values for the concentration of the substances in the equilibrium system.

Alternatively, use the following data, which relates to the equilibrium for the reaction of ethanol and propanoic acid to form ethyl propanoate.

Equation: C2H

5OH + CH

3CH

2COOH ⇌ CH

3CH

2COOC

2H

5 + H

2O

The equilibrium concentrations of the substances present at equilibrium are given in Table 5.3.3. Write an interpretation of the results of the experiment. In your answer you should:

• describe the main features of this equilibrium process • explain how to write an expression for the equilibrium constant corresponding to the equation

for this process • calculate a value for the equilibrium constant using suitable data • comment on the value of the equilibrium constant.

Substance Concentration / mol dm–3

CH3CH

2COOH 0.14

C2H

5OH 0.34

CH3CH

2COOC

2H

50.06

H2O 0.46

Table 5.3.3

3 Changing conditionsBiological systems need some mechanism for ensuring that certain key conditions remain approximately constant. Although quite complex mechanisms exist for maintaining, for example, constant temperature or constant blood glucose concentration, some other conditions remain constant because of the way in which systems at equilibrium respond to changes in conditions.

When conditions change, the equilibrium is disrupted, but a new equilibrium will often be established. This in turn may affect the concentration of products or reactants and hence the position of equilibrium.

Le Chatelier’s PrincipleLe Chatelier’s Principle states that ‘when a system at equilibrium is subject to a change, the position of equilibrium will shift in such a way as to oppose the change, as far as possible.’

A simple example to illustrate this would be the dissolving of oxygen in the bloodstream:

O2(g) ⇌ O

2(aq)

If the pressure of oxygen is increased then the position of equilibrium will shift to the right, favouring the formation of dissolved oxygen.

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Effects of changing conditions

Figure 5.3.2a shows a summary of the effects of changing concentration and Figure 5.3.2b shows a summary of the effects of changing pressure and temperature.

Fe3+(aq) + SCN–(aq) ⇌ Fe(SCN)2+(aq)

Change concentration New equilibrium established, Kc unchanged

Increase [Fe3+] Decrease [Fe3+]

Equilibrium position shifts to RHS (to

remove Fe3+)

Equilibrium position shifts to LHS (to

replace Fe3+)

Equilibrium position shifts to LHS (side with fewest moles of gas molecules)

Equilibrium position shifts to RHS (side with greatest moles of gas molecules)

Increase pressure

Change pressure

No effect

Decrease pressure

Equilibrium shifts to the RHS (the endothermic direction)

Equilibrium shifts to the LHS (the exothermic direction)

Increase temperature Decrease temperature

New equilibrium established, Kc unchanged

Add, or change, a catalyst

New equilibrium established, value of Kc changes

N2O4(g) ⇌ 2NO2(g) ∆H = +57 kJ mol–1

Change temperature

Take it furtherA fuller discussion of Le Chatelier’s principle and its applications can be found at http://www.chemguide.co.uk/physical/equilibria/change.html.

The same website also has a helpful section explaining how Le Chatelier’s Principle can be applied to industrial processes, such as the Haber Process for the synthesis of ammonia (http://www.chemguide.co.uk/physical/equilibria/haber.html).

Figure 5.3.2a: Summary of the effects of changing concentration.

Figure 5.3.2b: Summary of the effects of changing pressure and temperature.

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ActivityHere is the equation for the hydration of ethene. The reaction is usually carried out at 300 °C, 60 bar pressure and with a phosphoric acid catalyst:

CH2=CH

2 (g) + H

2O (g) ⇌ CH

3CH

2OH (g) ΔH = –45 kJ mol–1

Predict the effect on the equilibrium position if the following changes are made:1 increasing the pressure2 decreasing the temperature3 changing the catalyst.

Portfolio activity (3.3)The contact process is a chemical reaction used as one of the stages in the manufacture of sulfuric acid.

The equation for the reaction occurring in the contact process is:

2SO2(g) + O

2(g) ⇌ 2SO

3(g) ΔHƟ = –196 kJ mol–1

The reaction is usually carried out at 450 °C, 1 bar pressure and with a catalyst of vanadium pentoxide.

Discuss ways in which the position of equilibrium could be shifted to favour the formation of products.

In your answer: • list the changes which could be made • state Le Chatelier’s Principle and explain how this principle can be used to predict the effect of

any change you suggest.

4 pH and acid dissociationMany biochemical molecules contain acidic groups, or contain ionic groups that are the result of the ionisation of an acid group.

Brønsted–Lowry theory Brønsted–Lowry theory states that:

1 an acid can be defined as a substance able to donate H+ ions

2 a base can be defined as a substance able to accept H+ ions.

For example, hydrochloric acid:

HCl(aq) + H2O(l) ➝ H

3O+(aq) + Cl–(aq)

In this process, the acid (HCl) dissociates to form H+ and Cl− and the H+ is then accepted by the water molecule (which therefore acts as a base).

Often this process is simply written as:

HCl(aq) ➝ H+(aq) + Cl–(aq)

This process goes to completion so the HCl is fully dissociated when dissolved in water. This means that it is described as a strong acid.

Key termStrong acid: An acid which is completely ionised (fully dissociated).

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pH

Solutions with a higher concentration of H+ ions are more acidic. The pH value of a solution quantifies this:

pH = −log10

[H+]

So a solution with [H+] = 10−3 mol dm–3 has a pH of 3; solutions with a low pH value are more acidic.

This calculation can also be done in reverse: knowing the pH value of a solution enables you to calculate the [H+] in that solution:

[H+] = 10−pH

Activity1 Calculate the pH of solutions that have the following [H+] values (in mol dm–3):

(a) 1 × 10–8 (b) 3.6 × 10–3 (c) 0.016 (d) 6.4 × 10–11 (e) 1.40

2 Calculate the [H+] values (in mol dm–3) of solutions that have the following pH values: (a) 5 (b) 8.2 (c) 1.3 (d) 13.4 (e) −0.4

Acidic groups in biological molecules

These include carboxyl groups and phosphate groups (which are derived from phosphoric acid) as well as ammonium groups (R–NH

3+):

R–COOH ⇌ R–COO− + H+

R–NH3

+ ⇌ R–NH2 + H+

R–PO4H

2 ⇌ R–PO

4H− + H+

Other important acids in biological systems include carbonic acid (represented by H

2CO

3), which plays a critical role in the buffering of blood pH:

H2CO

3 ⇌ HCO

3− + H+

Molecules containing carboxyl groups and phosphate groups are weak acids – this means that only a small proportion of the groups ionise in aqueous solution.

Acid dissociation constants and pKa

Because the dissociation of a weak acid is an equilibrium process, the strength of the acid can be quantified by the magnitude of the acid dissociation constant.

In general, for any weak acid:

HA(aq) ⇌ H+ (aq) + A−(aq)

The acid dissociation constant, Ka =

[H+][A–][HA]

The larger the value of Ka, the stronger the acid, which means that a greater

proportion of the acid will be found in its ionised form.

Chemists often find it helpful to use pKa values for weak acids, as the K

a values are

often inconveniently small. The pKa of a weak acid is defined as:

pKa = −log

10 K

a

Key termWeak acid: An acid in which only a small proportion of the molecules are ionised.

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Some values of Ka and pK

a for weak acids are shown in Table 5.3.4.

Acid Ka at 298 K / mol dm–3 pKa at 298 K

phosphoric acid, H3PO

47.9 × 10–3 2.1

dihydrogenphosphate, H2PO

4− 6.3 × 10–7 6.2

ethanoic acid, CH3COOH 1.7 × 10–5 4.8

ammonium, NH4

+ 5.6 × 10–10 9.3

carbonic acid, H2CO

34.5 × 10–7 6.4

Notice that phosphoric acid will therefore lose two H+ ions to exist as a doubly charged ion at pH values above 6.2. It is therefore described as a diprotic acid (the other acids in Table 5.3.4 are monoprotic acids):

H3PO

4 ⇌ H

2PO

4− + H+ ⇌ HPO

42− + 2H+

ActivityUse K

a data to calculate the pK

a values of these weak acids:

• Nitrous acid: Ka = 4.7 × 10−4 mol dm−3

• Phenol: Ka = 1.28 × 10−10 mol dm−3

• Butanoic acid: Ka = 1.5 × 10−5 mol dm−3

Amino acids

All amino acids contain two groups that can take part in acid–base reactions, the amino group (–NH

2) and the carboxyl group (–COOH). Some amino acids also

contain acidic or basic groups in the R side group.

The pattern of ionisation of these amino and carboxyl group acids changes according to the pH of the solution.

Consider the carboxyl group:

R–COOH ⇌ R–COO− + H+

If the environment of an amino acid solution becomes more basic (higher pH), then the concentration of H+ ions will be lower. According to Le Chatelier’s Principle, this will cause the position of equilibrium to shift to favour the products. So a greater proportion of the carboxyl groups will be in the ionised form (R–COO−).

A similar argument can be used for the amine group (R–NH2):

R–NH3

+ ⇌ R–NH2 + H+

At higher pH, the equilibrium will favour the products and so a greater proportion of the amine groups will be in the un-ionised form (R–NH

2).

Figure 5.3.3 shows how the ionisation of these groups changes when the pH of the environment is altered.

Table 5.3.4: Ka and pK

a values

of some weak acids.

Key termsDiprotic acid: An acid molecule that donates two protons to water in aqueous solution.

Monoprotic acid: An acid molecule that donates one proton to water in aqueous solution.

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H+ is lost as pH increases

HNH

HR

H

C

CO OH H+ is lost as

pH increases

+ + OH–

+ H+HNH

HR

H

C

CO O–

+HNH

HR

H

C

CO O–

+N

HR C

CO O–

N

HR C

CO O–

+ OH–

+ H+

HH

Increasing pH

2 7 10

Notice that at pH values of around 7, most of these amino acid molecules actually exist in a form known as a zwitterion, in which both the amino and carboxyl groups are ionised. This helps to explain the very high solubility of amino acids in water.

Ionisation and pKa

You were introduced to the term pKa earlier in this section. pK

a values can be

particularly useful when studying biochemical systems, because the pKa of an

acidic group represents the pH at which 50% of the groups are ionised.

You can easily see why this is by using the general form of the equation for Ka:

Ka =

[H+][A–][HA]

If exactly 50% of the acid groups are ionised, then [A−] = [HA].

So the equation then becomes:

Ka = [H+]

or −lg Ka = −lg [H+]

which can be written as pKa = pH.

For most amino acids, the pKa of the carboxyl group is in the range 2.1–2.3; the

pKaof the amino group is in the range 9.1–9.8.

About seven amino acids also have ionisable groups in their side chains; pKa values

can also be calculated for these groups.

Figure 5.3.3: As the pH increases, different groups on the amino

acid become charged.

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ActivityThe amino acid lysine has a side group containing an amino group. The structure of lysine and the pK

a values of the ionisable groups are shown in Figure 5.3.4.

Predict the most likely form of each group (ionised or un-ionised) at the following pH values:

(a) 1.0 (b) 3.0 (c) 8.0 (d) 9.5 (e) 12.0.

H

C

CH2

CH2

CH2

CH2

pKa = 10.53

pKa = 2.18pKa = 8.95

NH2

O

OHCN

H

H

Figure 5.3.4: The structure of lysine and the pK

a values

of the ionisable groups.

Take it furtherA list of the pK

a values for the ionisable groups in different amino acid molecules can be found at:

http://www.cem.msu.edu/~cem252/sp97/ch24/ch24aa.html.

Calculating pH of weak acidsUnlike for strong acids, the calculation of [H+] for a weak acid solution (and hence the pH of the solution) requires knowledge of both the concentration of the acidic molecule and its K

a value. (It is important to be clear about the difference between

acid strength and acid concentration – see the Key terms here and pages 7–8.)

For pure solutions of relatively weak acids (with a pKa of 3 or greater), the [H+] can

be calculated from the simple formula shown below:

[H+] = √Ka × [HA]

So for a 0.1 mol dm–3 solution of ethanoic acid, the [H+] is given by

[H+] = √(1.7 × 10–5) × (0.1) = 1.30 × 10−3

So pH = −log10

(1.30 × 10−3) = 2.9.

Take it furtherConsult a suitable level 3 Chemistry textbook or websites such as http://www.science.uwaterloo.ca/~cchieh/cact/c123/wkacids.html.

Find out the assumptions that are made to arrive at this equation, and also research the more complex mathematical method that can be used when these assumptions are not justified (for example, when the acid has a pK

a of less than about 3.0).

Key termsConcentrated acid: An acid in which the ratio of solute to water solvent is relatively high.

Dilute acid: An acid in which the ratio of solute to water solvent is relatively low.

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ActivityCalculate the pH values of the following solutions:

• 0.20 mol dm−3 methanoic acid (Ka = 1.6 × 10−4)

• 1.0 mol dm−3 chloric acid (Ka = 3.7 × 10−8)

• 2.5 × 10−3 mol dm−3 phenol (Ka = 1.28 × 10–10)

Henderson–Hasselbalch equation

As noted above, as the pH of a solution increases, the proportion of acid molecules in the ionised form (A−) also increases.

Mathematically, this is expressed by the Henderson–Hasselbalch equation:

pH = pKa + log

10 ( [A–]

[HA])So the pH at which 90% of ethanoic acid molecules are ionised is given by:

pH = 4.8 + log10

(9010) = 5.8

Ionisation of waterYou have already seen that water acts as a base in the process that results in the ionisation of an acid:

HCl−(aq) + H2O(l) ➝ H

3O+(aq) + Cl−(aq)

In fact it can also behave as a very weak acid, with a Ka value of

1.8 × 10−16 mol dm−3 at 298 K:

H2O(l) ⇌ H+(aq) + OH−(aq)

A more significant way of expressing this equilibrium numerically is to calculate the ionic product, K

w:

Kw

= [H+][OH−] = 1.00 × 10−14 mol2 dm−6 at 298 K.

This explains why a neutral solution has a pH of 7 – in a neutral solution

[H+] = [OH−] and so [H+] = √1.00 × 10–14 = 10−7 mol dm−3

An alkaline solution is usually taken to mean one in which the [OH−] > [H+], so alkaline solutions will have [OH−] of greater than 10−7 mol dm−3 and [H+] of less than 10−7 mol dm−3.

The Kw

expression allows you to calculate the pH values of solutions of some familiar alkalis.

Worked exampleCalculate the pH of 0.2 mol dm−3 NaOH at 298 K.

[OH−] = 0.2 mol dm−3

From the Kw

expression, [H+] =

Kw

[OH–] = 1.00 × 10–14

0.2 = 5 × 10−14 mol dm−3

pH = −log10

(5 × 10−14) = 13.3

ActivityCalculate the pH of these solutions of alkalis at 298 K (K

w at 298 K =

1 × 10−14 mol2 dm−6): • 2.5 × 10−3 mol dm−3 KOH • 0.1 mol dm−3 Ca(OH)

2 (hint: think

carefully about the [OH−] value in this solution).

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Buffer solutionsBuffer solutions play a critical role in biological systems because many biochemical reactions, for example the action of enzymes, are very sensitive to the pH of their environment. The pH of human blood plasma, for example, needs to be maintained at a constant value of about 7.4.

Buffering systems

These are based on equilibria involving a weak acid in ionised and un-ionised forms.

The cytoplasm of cells relies on a buffering system based on the dihydrogen phosphate ion:

H2PO

4−(aq) ⇌ HPO

42−(aq) + H+(aq)

Buffering action depends on the existence of reservoirs of both the ionised and un-ionised form.

So ethanoic acid by itself, for example, could not act as a buffer solution because it is a weak acid.

CH3COOH ⇌ CH

3COO− + H+ K

a = 1.7 × 10−5 mol dm−3

This means that the concentration of the ethanoate ion (the ionised form of the acid) is very low. In this case, although there will be a ‘reservoir’ of the un-ionised form of the acid, there is no such ‘reservoir’ of the ionised form. However, if ethanoic acid is mixed with a source of ethanoate ions, for example, the salt sodium ethanoate, there will be a reservoir of both the ionised and un-ionised forms.

Explaining the action of buffers

Le Chatelier’s Principle can then be used to explain the buffering action:

If the [H+] increases, due to the addition of an acidic substance, the equilibrium position will shift to the left-hand side as H+ ions are accepted by HPO

42−, restoring

the original [H+] (see Figure 5.3.5).

Equilibrium position shifts to LHS to oppose change

H

2PO

4−(aq) ⇌ HPO

42−(aq) + H+(aq)

Extra acid added, [H+] increases

If the [H+] decreases, due to the addition of an alkali which will remove H+ ions, the equilibrium position will shift to the right as H

2PO

4− ions dissociate, restoring the

original [H+] (see Figure 5.3.6).

Equilibrium position shifts to RHS to oppose change

H

2PO

4−(aq) ⇌ HPO

42−(aq) + H+(aq)

OH– combines with H+ to form water. [H+] decreases

It is clear that the buffering action requires sufficient reservoirs of both H2PO

4− ions

(to dissociate) and HPO4

2− ions (to accept the additional H+ ions).

Key termBuffer solution: A system that resists changes in pH when small amounts of acid or alkali are added.

Figure 5.3.5: Effect of adding an acidic substance on equilibrium position.

Figure 5.3.6: Effect of adding an alkaline substance on equilibrium position.

14



The pH or [H+] of a buffer can be calculated from the Ka value of the weak acid

and the ratio of the concentrations of the ionised and un-ionised forms, using the Henderson–Hasselbalch equation:

pH = pKa + log

10 ( [A–]

[HA])Worked exampleA buffer solution is made by mixing together equal volumes of 0.2 mol dm−3 ethanoic acid and 0.1 mol dm−3 sodium ethanoate.

The Ka of ethanoic acid is 1.7 × 10−5 mol dm−3.

Assuming that the concentration of ethanoate ions from the ethanoic acid solution is negligible:

• the ratio [A–][HA] =

0.10.2 = 0.5

• pKa = −log 1.7 × 10−5 = 4.8

• hence pH = 4.8 + log10

(0.5) = 4.8 – 0.3 = 4.5.

Buffers are most effective at pH values close to the pKa of the acid. This can be

explained using the equation above; to achieve a pH of 2 units greater than the pK

a value using the ethanoic acid / sodium ethanoate system would require the

ratio of [A–][HA]

to be 100:1 (log10

(100) = 2).

To achieve such a ratio, the ‘reservoir’ of HA would be too small to provide effective buffering.

ActivityCalculate the pH of the following buffer solutions:1 A buffer made by mixing equal volumes of 0.1 mol dm−3 methanoic acid and 0.1 mol dm−3

sodium methanoate (pKa of methanoic acid = 3.8).

2 A buffer which contains ammonium ions at a concentration of 0.6 mol dm−3 and ammonia at a concentration of 0.2 mol dm−3 (pK

a of the ammonium ion is 9.3).

3 Calculate the ratio of ionised to un-ionised form needed to produce a pH of 5.2 in a buffer solution made from a mixture of ethanoic acid and sodium ethanoate (pK

a of ethanoic acid

= 4.8).

Portfolio activity (3.4)Research a biologically important molecule which is a weak acid (it is best to choose a monoprotic acid). A suitable example could be pyruvic acid, C

3H

4O

3.

• Write an equation for the dissociation of the molecule. • Write an expression for the acid dissociation constant and find its value at 298 K. • Calculate the pH of the acid at a range of concentrations, commenting on any assumptions you

have made in the calculation. • Explain how you can predict the likely ratio of ionised to un-ionised form at a pH of 3.0.

ActivityResearch the system that is used to buffer the blood at pH 7.4. Use data about the pK

a of the weak acid

present to calculate the required ratio of ionised to un-ionised form if this buffer solution maintains the pH at 7.4.

15



ChecklistAt the end of this section you should be familiar with the following ideas:

at equilibrium, the rates of forward and backward reactions are equal and concentrations of the components of the system remain constant

values of equilibrium constants give information about the position of equilibrium

the equilibrium position of a reaction shifts to oppose changes in conditions

many biological molecules contain acidic groups that donate H+ ions; these groups are weak so only a small fraction of molecules dissociate

values of acid dissociation constants enable calculations of pH to be made for systems involving weak acids

many biological systems are buffer solutions based on weak acids.

Further readingPractice at a suitable level in calculations for situations involving equilibria (including acid–base equilibria) can be found in Chapters 7 and 8 of Calculations in AS/A level Chemistry (Clark, 2000).

Chapters 11 and 12 of Advanced Chemistry (Clugston and Flemming, 2000) cover the material from this topic guide on equilibria.

Equilibria are also covered in the Physical Chemistry section of the Chemguide website (http://www.chemguide.co.uk).

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