Proton transfer between acids and bases is fast in both directions, so the dynamic equilibria
HF(aq) + H2O(l) H3O+(aq) + F−(aq) H2O(l) + NH3(aq) NH4+(aq) + OH−(aq)
give a more complete description of the behaviour of the acid HF and the base NH3 in water than the forward reaction alone. The central feature of Brứnsted acid–base chemistry in aqueous solution is that of rapid attainment of equilibrium in the proton-transfer reaction, and we concen- trate on this aspect.
(a) Conjugate acids and bases
KEY POINTS When a species donates a proton, it becomes the con- jugate base; when a species gains a proton, it becomes the conjugate acid. Conjugate acids and bases are in equilibrium in solution.
The form of the two forward and reverse reactions given above, both of which depend on the transfer of a proton from an acid to a base, is expressed by writing the general Brứnsted equilibrium as
Acid1 + Base2 Acid2 + Base1
The species Base1 is called the conjugate base of Acid1, and Acid2 is the conjugate acid of Base2. The conjugate base of an acid is the species that is left after a proton is lost. The conjugate acid of a base is the species formed when a proton is gained. Thus, F− is the conjugate base of HF and H3O+ is the conjugate acid of H2O. There is no fundamental distinc- tion between an acid and a conjugate acid or a base and a conjugate base: a conjugate acid is just another acid and a conjugate base is just another base.
EXAMPLE 5.1 Identifying acids and bases
Identify the Brứnsted acid and its conjugate base in the following reactions:
(a) HSO4−(aq) + OH−(aq) → H2O(l) + SO42−(aq) (b) PO43−(aq) + H2O(l) → HPO42–(aq) + OH−
Answer We need to identify the species that loses a proton and its conjugate partner. (a) The hydrogensulfate ion, HSO4−, transfers a proton to hydroxide; it is therefore the acid and the SO42− ion produced is its conjugate base. (b) The H2O molecule transfers a proton to the phosphate ion acting as a base; thus H2O is the acid and the OH− ion is its conjugate base.
Self-test 5.1 Identify the acid, base, conjugate acid, and conjugate base in the following reactions:
(a) HNO3(aq) + H2O(l) → H3O+(aq) + NO3−(aq) (b) CO32−(aq) + H2O(l) → HCO3–(aq) + OH− (c) NH3(aq) + H2S(aq) → NH4+(aq) + HS−(aq).
(b) The strengths of Brứnsted acids
KEY POINTS The strength of a Brứnsted acid is measured by its acidity constant, and the strength of a Brứnsted base is measured by its basic- ity constant; the stronger the base, the weaker is its conjugate acid.
Throughout this discussion, we shall need the concept of pH, which we assume to be familiar from introductory chemistry:
pH = −log [H3O+], and hence [H3O+] = 10−pH (5.1) The strength of a Brứnsted acid, such as HF, in aqueous solution is expressed by its acidity constant (or ‘acid ioniza- tion constant’), Ka:
HF(aq) + H2O(l) H3O+(aq) + F−(aq) =[H O ][F ]+ − [HF]
a 3
K
More generally:
HX(aq) + H2O(l) H3O+(aq) + X−(aq) =[H O ][X ]+ − [HX]
a 3
K
(5.2) In this definition, [X−] denotes the numerical value of the molar concentration of the species X− (so, if the molar con- centration of HF molecules is 0.001 mol dm−3, then [HF] = 0.001). A value Ka <<1 implies that [HX] is large with respect to [X–], and so proton retention by the acid is favoured. The experimental value of Ka for hydrogen fluoride in water is 3.5 × 10−4, indicating that under normal conditions, only a very small fraction of HF molecules are deprotonated. The actual fraction deprotonated can be calculated as a function of acid concentration from the numerical value of Ka.
EXAMPLE 5.2 Calculating acidity constants
The pH of 0.145 M CH3COOH(aq) is 2.80. Calculate Ka for ethanoic acid.
Answer To calculate Ka we need to calculate the concentrations of H3O+, CH3CO2−, and CH3COOH in the solution. The concentration A NOTE ON GOOD PRACTICE
In precise work, Ka is expressed in terms of the activity of X, a(X), its effective thermodynamic concentration. The acidity constant is based on the assumption that the solutions are sufficiently dilute for it to be permissible to write a(H2O) = 1.
of H3O+ is obtained from the pH by writing [H3O+] = 10−pH, so in a solution of pH = 2.80, the molar concentration of H3O+ is 1.6 × 10−3 mol dm−3. Each deprotonation event produces one H3O+ ion and one CH3CO2− ion, so the concentration of CH3CO2− is the same as that of the H3O+ ions (provided the autoprotolysis of water can be neglected). The molar concentration of the remaining acid is 0.145 − 0.0016 mol dm−3 = 0.143 mol dm−3. Therefore
K 1.6 10
0.143 1.7 10
a
3 2
( ) 5
= × − = × −
This value corresponds to pKa = 4.77 (units are usually omitted).
Self-test 5.2 For hydrofluoric acid Ka = 3.5 × 10−4. Calculate the pH of 0.10 M HF(aq).
Likewise, the proton transfer equilibrium characteristic of a base, such as NH3, in water can be expressed in terms of a basicity constant, Kb:
NH3(aq) + H2O(l) NH4+(aq) + OH−(aq) =[NH ][OH ]+ − [NH ]
b 4
3
K
More generally:
B(aq) + H2O(l) HB+(aq) + OH−(aq) =[HB ][OH ]+ [B]
b
K –
(5.3) If Kb<<1, then [HB+] << [B] at typical concentrations of B and only a small fraction of B molecules are protonated.
Therefore, the base is a weak proton acceptor and its conju- gate acid is present in low concentration. The experimental value of Kb for ammonia in water is 1.8 × 10−5, indicating that under normal conditions, only a very small fraction of NH3 molecules is protonated. As for the acid calculation, the actual fraction of base protonated can be determined from the value of Kb.
Because water is amphiprotic, a proton transfer equi- librium exists even in the absence of added acids or bases.
The proton transfer from one water molecule to another is called autoprotolysis (or ‘autoionization’). Proton transfer in water is very fast because it involves the interchange of weak hydrogen bonds between neighbouring molecules (Sections 5.8 and 10.6). The extent of autoprotolysis and the composi- tion of the solution at equilibrium is described by the auto- protolysis constant (or ‘autoionization constant’) of water:
2 H2O(l) H3O+(aq) + OH−(aq) Kw = [H3O+][OH−] The experimental value of Kw is 1.00 × 10−14 at 25°C, indi- cating that only a very tiny fraction of water molecules are present as ions in pure water. Indeed, we know that because the pH of pure water is 7.00, and [H3O+] = [OH−], then [H3O+] = 1.0 × 10−7 mol dm−3. Tap and bottled water have a pH slightly below 7 due to dissolved CO2.
An important role for the autoprotolysis constant of a solvent is that it enables us to relate the strength of a base to the strength of its conjugate acid, thereby allowing use of a single constant to express both acidity and basicity strength. Thus, the value of Kb for the ammonia equilibrium in which NH3 acts as a base is related to the value of Ka for the equilibrium
NH4+(aq) + H2O(l) H3O+(aq) + NH3(aq) in which its conjugate acid acts as an acid by
KaKb = Kw (5.4)
The implication of eqn 5.4 is that the larger the value of Kb, the smaller the value of Ka. That is, the stronger the base, the weaker is its conjugate acid. It is conventional to report the strength of a base in terms of the acidity constant Ka of its conjugate acid.
A BRIEF ILLUSTRATION
The Kb of ammonia in water is 1.8 × 10−5. It follows that Ka of the conjugate acid NH4+ is
K K
K
1 10
1.8 10 5.6 10
a w
b
14 5
= = × 10
× −− = × − .
Because, like molar concentrations, acidity constants span many orders of magnitude, it is convenient to report them as their common logarithms (logarithms to the base 10) like pH by using
pK = −log K (5.5)
where K may be any of the constants we have introduced.
At 25°C, for instance, pKw = 14.00. It follows from this defi- nition and the relation in eqn 5.4 that
pKa + pKb = pKw (5.6)
A similar expression relates the strengths of conjugate acids and bases in any solvent, with pKw replaced by the appro- priate autoprotolysis constant of the solvent, pKsol.
(c) Strong and weak acids and bases
KEY POINTS An acid or base is classified as either weak or strong de- pending on the size of its acidity constant.
Table 5.1 lists the acidity constants of some common acids and conjugate acids of some bases in water. A substance is classified as a strong acid if the proton transfer equilibrium lies strongly in favour of donation of a proton to the solvent.
Thus, a substance with pKa < 0 (corresponding to Ka > 1 and usually to Ka >> 1) is a strong acid. Such acids are commonly regarded as being fully deprotonated in solution (but it must never be forgotten that that is only an approximation). For example, hydrochloric acid is regarded as a solution of
H3O+ and Cl− ions, and a negligible concentration of HCl molecules. A substance with pKa > 0 (corresponding to Ka
< 1) is classified as a weak acid; for such species, the pro- ton transfer equilibrium lies in favour of nonionized acid.
Hydrogen fluoride is a weak acid in water, and hydrofluoric acid consists of hydronium ions, fluoride ions, and a high proportion of HF molecules. Carbonic acid (H2CO3), the hydrate of CO2, is another weak acid.
A strong base is a species that is almost fully protonated in water. An example is the oxide ion, O2−, which is immedi- ately converted into OH− ions in water. A weak base is only partially protonated in water, an example being NH3. The conjugate base of any strong acid is a weak base, because it is thermodynamically unfavourable for such a base to accept a proton.
The pH of a dilute solution containing a simple mixture of acid HX and its conjugate base X− is given by
pH p log[X ]
= Ka+ [HX]− (5.7)
(d) Polyprotic acids
KEY POINTS A polyprotic acid loses protons in succession, and suc- cessive deprotonations are progressively less favourable; a distribution diagram summarizes how the fraction of each species present depends on the pH of the solution.
A polyprotic acid is a substance that can donate more than one proton. An example is hydrogen sulfide, H2S, a diprotic acid. For a diprotic acid, there are two successive proton donations and two acidity constants:
H2S(aq) + H2O(l) HS−(aq) + H3O+(aq) HS−(aq) + H2O(l) S2−(aq) + H3O+(aq)
=
=
+ −
+ −
−
[H O ][HS ] [H S]
[H O ][S ] [HS ]
1 3
2
2 3 2
K K
a
a
From Table 5.1, Ka1 = 9.1 × 10−8 (pKa1 = 7.04) and Ka2 ≈ 1.1 × 10−19 (pKa2 = 19). The second acidity constant, Ka2, is TABLE 5.1 Acidity constants for species in aqueous solution at 25ºC
Acid HA A− Ka pKa
Hydriodic HI I− 1011 −11
Perchloric HCIO4 ClO4− 1010 −10
Hydrobromic HBr Br− 109 −9
Hydrochloric HCl Cl− 107 −7
Sulfuric H2SO4 HSO−4 102 −2
Nitric HNO3 NO3− 102 −2
Hydronium ion H3O+ H2O 1 0.0
Chloric HCIO3 ClO3− 10−1 1
Sulfurous H2SO3 HSO3− 1.5 × 10−2 1.81
Hydrogensulfate ion HSO4− SO42− 1.2 × 10−2 1.92
Phosphoric H3PO4 H PO2 4− 7.5 × 10−3 2.12
Hydrofluoric HF F− 3.5 × 10−4 3.45
Formic HCOOH HCO2− 1.8 × 10−4 3.75
Ethanoic CH3COOH CH3CO2− 1.74 × 10−5 4.76
Pyridinium ion HC5H5N+ C5H5N 5.6 × 10−6 5.25
Carbonic H2CO3 HCO3− 4.3 × 10−7 6.37
Hydrogen sulfide H2S HS− 9.1 × 10−8 7.04
Dihydrogenphosphate ion H PO2 4− HPO42− 6.2 × 10−8 7.21
Boric acid* B(OH)3 B(OH)4− 7.2 × 10−10 9.14
Ammonium ion NH4+ NH3 5.6 × 10−10 9.25
Hydrocyanic HCN CN− 4.9 × 10−10 9.31
Hydrogencarbonate ion HCO3− CO32− 4.8 × 10−11 10.32
Hydrogenarsenate ion HAsO2−4 AsO43− 3.0 × 10−12 11.53
Hydrogenphosphate ion HPO2−4 PO43− 2.2 × 10−13 12.67
Hydrogensulfide ion HS– S2− 1.1 × 10−19 19
*The proton transfer equilibrium is B(OH)3(aq) + 2 H2O(l) H3O+(aq) + B(OH)4−(aq).
always smaller than Ka1 (and hence pKa2 is larger than pKa1). The decrease in Ka is consistent with an electrostatic model of the acid in which, in the second deprotonation, a proton must separate from a centre with one more negative charge than in the first deprotonation. Because additional electrostatic work must be done to remove the positively charged proton, the deprotonation is less favourable.
EXAMPLE 5.3 Calculating concentrations of ions in polyprotic acids
Calculate the concentration of carbonate ions in 0.10 M H2CO3(aq). Ka1 is given in Table 5.1; Ka2 = 4.6 × 10−11.
Answer We need to consider the equilibria for the successive deprotonation steps with their acidity constants:
H2CO3(aq) + H2O(l) HCO3−(aq) + H3O+(aq) HCO3−(aq) + H2O(l) CO32−(aq) + H3O+(aq)
=
=
+ −
+ −
−
K K
H O HCO H CO H O CO
HCO
a
a 1
3 3
2 3
2
3 32
3
We suppose that the second deprotonation is so slight that it has no effect on the value of [H3O+] arising from the first deprotonation, in which case we can write [H3O+] = [HCO3−]. These two terms therefore cancel in the expression for Ka2, which results in
Ka2 = [CO32−]
independent of the initial concentration of the acid. It follows that the concentration of carbonate ions in the solution is 4.6 × 10−11 mol dm−3.
Self-test 5.3 Calculate the pH of 0.20 M H2C4H4O6(aq) (tartaric acid),given Ka1 = 1.0 × 10−3 and Ka2 = 4.6 × 10−5.
The clearest representation of the concentrations of the species that are formed in the successive proton transfer equilibria of polyprotic acids is a distribution diagram, a diagram showing the fraction of solute present as a speci- fied species X, f(X), plotted against the pH. Consider, for instance, the triprotic acid H3PO4, which releases three pro- tons in succession to give H2PO4−, HPO42−, and PO43−. The fraction of solute present as intact H3PO4 molecules is
= + − + − + −
(H PO ) [H PO ]
[H PO ] [H PO ] [HPO ] [PO ]
3 4 3 4
3 4 2 4 42
43
f (5.8)
The concentration of each solute at a given pH can be calcu- lated from the pKa values.1 Figure 5.1 shows the fraction of
1 For the calculations involved, see P. Atkins and L. Jones, Chemical principles. W.H. Freeman & Co. (2010).
all four solute species as a function of pH and hence summa- rizes the relative importance of each acid and its conjugate base at each pH. Conversely, the diagram indicates the pH of the solution that contains a particular fraction of the spe- cies. We see, for instance, that if pH < pKa1, corresponding to high hydronium ion concentrations, then the dominant species is the fully protonated H3PO4 molecule. However, if pH > pKa3, corresponding to low hydronium ion concentra- tions, then the dominant species is the fully deprotonated PO43− ion. The intermediate species are dominant when pH values lie between the relevant pKa values.
(e) Factors governing the strengths of Brứnsted acids and bases
KEY POINTS Proton affinity is the negative of the gas phase proton- gain enthalpy. The proton affinities of p-block conjugate bases de- crease to the right along a period and down a group. Proton affinities (and accordingly, the strengths of bases) are influenced by solvation, which stabilizes species carrying a charge.
A quantitative understanding of the relative acidities of X–H protons can be obtained by considering the enthalpy changes accompanying proton transfer. We shall consider gas-phase proton transfer reactions first and then consider the effects of the solvent.
The simplest reaction of a proton is its attachment to a base, A− (which, although denoted here as a negatively charged species, could be a neutral molecule, such as NH3), in the gas phase:
A−(g) + H+(g) → HA(g)
The standard enthalpy of this reaction is the proton-gain enthalpy, ΔpgH○−−. The negative of this quantity is often reported as the proton affinity, Ap (Table 5.2). When ΔpgH−−○ is large and negative, corresponding to an exothermic FIGURE 5.1 Distribution diagram for the various forms of the triprotic acid phosphoric acid in water, as a function of pH.
0 0.2 0.4 0.6 0.8 1.0
Fraction of species f(X)
0 2 4 6 8 10 12 14
pH
pKa1 pKa2 pKa3
2.12 7.21 12.68
H2PO4–
H3PO4 HPO42– PO43–
proton attachment, the proton affinity is high, indicating strongly basic character in the gas phase. If the proton-gain enthalpy is only slightly negative, then the proton affinity is low, indicating a weaker basic (or more acidic) character.
The proton affinities of the conjugate bases of p-block binary acids HA decrease to the right along a period and down a group, indicating an increase in gas-phase acidity.
Thus, HF is a stronger acid than H2O and HI is the strong- est acid of the hydrogen halides. In other words, the order of proton affinities of their conjugate bases is I− < OH− < F−. These trends can be explained by using a thermodynamic cycle such as that shown in Fig. 5.2 in which proton gain can be thought of as the outcome of three steps:
Electron loss from A−: A−(g) → A(g) + e−(g)
−ΔegH○−−(A) = Ae(A) (the reverse of electron gain by A)
Electron gain by H+: H+(g) + e−(g) → H(g)
−ΔiH−−○(H) = −I(H) (the reverse of the ionization of H)
Combination of H and A: H(g) + A(g) → HA(g)−B(H−A) (the reverse of H−A bond dissociation)
The proton-gain enthalpy of the conjugate base A− is the sum of these enthalpy changes:
Overall: H+(g) + A−(g) → HA(g)
ΔpgH○−−(A−) = Ae(A) − I(H) − B(H−A) Therefore, the proton affinity of A is
Ap(A−) = B(H−A) + I(H) − Ae(A) (5.9) The dominant factor in the variation in proton affin- ity across a period is the trend in electron affinity of A, which increases from left to right and hence lowers the proton affinity of A−. Thus, because the proton affinity of A− decreases, the gas-phase acidity of HA increases across a period as the electron affinity of A increases. Because increasing electron affinity correlates with increasing elec- tronegativity (Section 1.7), the gas-phase acidity of HA
also increases as the electronegativity of A increases. The dominant factor when descending a group is the decrease in the H−A bond dissociation enthalpy, which lowers the proton affinity of A− and therefore increases the gas-phase acid strength of HA. The overall result of these effects is a decrease in gas-phase proton affinity of A−, and therefore to increase the gas-phase acidity of HA, from the top left to bottom right of the p-block. On this basis we see that HI is a much stronger acid than CH4.
The correlation we have described is modified when a solvent (typically water) is present. The gas-phase process A−(g) + H+(g) → AH(g) becomes
A−(aq) + H+(aq) → HA(aq)
and the negative of the accompanying proton-gain enthalpy is called the effective proton affinity, Ap′, of A−(aq).
If the species A− denotes H2O itself, the effective proton affinity of H2O is the enthalpy change accompanying the process
H2O(l) + H+(aq) → H3O+(aq)
The energy released as water molecules are attached to a proton in the gas phase, as in the process
n H2O(g) + H+(g) → H+(H2O)n(g)
can be measured by mass spectrometry and used to assess the energy change for the hydration process in solution. It is found that the energy released passes through a maximum value of 1130 kJ mol−1 as n increases, and this value is taken to be the effective proton affinity of H2O in bulk water. The effective proton affinity of the ion OH− in water is simply the negative of the enthalpy of the reaction
OH−(g) + H+(aq) → H2O(l)
which can be measured by conventional means (such as the temperature dependence of its equilibrium constant, Kw).
The value found is 1188 kJ mol−1. TABLE 5.2 Gas phase and solution proton affinities*
Conjugate acid Base Ap/kJ mol−1 A′p/kJ mol−1
HF F− 1553 1150
HCl Cl− 1393 1090
HBr Br− 1353 1079
HI I− 1314 1068
H2O OH− 1643 1188
HCN CN− 1476 1183
H3O+ H2O 723 1130
NH4+ NH3 865 1182
* Ap is the gas phase proton affinity, A′p is the effective proton affinity for the base in water
FIGURE 5.2 Thermodynamic cycle for a proton gain reaction.
H+(g) + A–(g)
H+(g) + A(g) + e–(g)
H(g) + A(g)
HA(g) B(H–A) I(H)
ΔegH⦵(A)
ΔpgH⦵(A–)
The reaction
HA(aq) + H2O(l) → H3O+(aq) + A−(aq)
is exothermic if the effective proton affinity of A−(aq) is lower than that of H2O(l) (less than 1130 kJ mol−1) and—provided entropy changes are negligible and enthalpy changes are a guide to spontaneity—HA will give up protons to the water and be strongly acidic. Likewise, the reaction
A−(aq) + H2O(l) → HA(aq) + OH−(aq)
is exothermic if the effective proton affinity of A–(aq) is higher than that of OH−(aq) (1188 kJ mol−1). Provided enthalpy changes are a guide to spontaneity, A−(aq) will accept protons and will act as a strong base.
A BRIEF ILLUSTRATION
The effective proton affinity of I− in water is 1068 kJ mol−1 compared to 1314 kJ mol−1 in the gas phase, showing that the I− ion is stabilized by hydration. The effective proton affinity is also smaller than the effective proton affinity of water (1130 kJ mol−1), which is consistent with the fact that HI is a strong acid in water.
All the halide ions except F− have effective proton affinities smaller than that of water, which is consistent with all the hydrogen halides except HF being strong acids in water.
The effects of solvation can be rationalized in terms of an electrostatic model in which the solvent is treated as a con- tinuous dielectric medium. The solvation of a gas-phase ion is always strongly exothermic. The magnitude of the enthalpy of solvation ΔsolvH○−− (the enthalpy of hydration in water, ΔhydH−−○) depends on the radius of the ions, the relative permittivity of the solvent, and the possibility of specific bonding (especially hydrogen bonding) between the ions and the solvent.
When considering the gas phase we assume that entropy contributions for the proton transfer process are small and so ΔG○−− ≈ ΔH○−−. In solution, entropy effects cannot be ignored and we must use ΔG○−−. The Gibbs energy of solvation of an ion can be identified as the energy involved in transferring the anion from a vacuum into a solvent of relative permit- tivity εr. The Born equation can be derived using this model:2
8 1 1
solv
A 2 2
0 r
G N z e
ε r ε
∆ ° = −
π −
(5.10)
where z is the charge number of the ion, r its effective radius which includes part of the radii of solvent molecules, NA is Avogadro’s constant, ε0 is the vacuum permittivity, and εr is the relative permittivity (the dielectric constant).
The Gibbs energy of solvation is proportional to z2/r (known also as the electrostatic parameter, ξ) so small, highly
2 For the derivation of the Born equation, see P. Atkins and J. de Paula, Physical chemistry, Oxford University Press and W.H. Freeman & Co.
(2014).
charged ions are stabilized in polar solvents (Fig. 5.3). The Born equation also shows that the larger the relative permit- tivity the more negative the value of ΔsolvG○−−. This stabiliza- tion is particularly important for water, for which εr = 80 (and the term in parentheses is close to 1), compared with nonpolar solvents for which εr may be as low as 2 (and the term in parentheses is close to 0.5).
Because Δsolv G○−− is the change in molar Gibbs energy when an ion is transferred from the gas phase into aque- ous solution, a large, negative value of Δsolv G○−− favours the formation of ions in solution compared with the gas phase (Fig. 5.3). The interaction of the charged ion with the polar solvent molecules stabilizes the conjugate base A− relative to the parent acid HA, and as a result the acidity of HA is enhanced by the polar solvent. On the other hand, the effec- tive proton affinity of a neutral base B is higher than in the gas phase because the conjugate acid HB+ is stabilized by solvation. Because a cationic acid, such as NH4+, is stabilized by solvation, the effective proton affinity of its conjugate base (NH3) is higher than in the gas phase. The acidity of cationic acids is therefore lowered by a polar solvent.
The Born equation ascribes stabilization by solvation to Coulombic interactions. However, hydrogen bonding (Section 5.8) is an important factor in protic solvents such as water and leads to the formation of hydrogen-bonded clusters around some solutes. As a result, water has a greater stabilizing effect on small, highly charged ions than the Born equation predicts. This stabilizing effect is particularly great for F−, OH−, and Cl−, with their high charge densities and for which water acts as a hydrogen-bond donor. Because a water molecule has lone pairs of electrons on O, it can also be a hydrogen-bond acceptor. Acidic ions such as NH4+ are stabilized by hydrogen bonding and consequently have a lower acidity than predicted by the Born equation.
FIGURE 5.3 Correlation between ΔsolvG○−− and the dimensionless electrostatic parameter ξ (= 100z2/(r/pm)) for selected anions.
0 1 2
Gibbs energy of hydration, –ΔhydG⦵/(1000 kJ mol–1) 0.5
1 1.5
2 2.5
3
0 3
4 PO43–
SO42–
F– OH– Cl– CN– I–
Electrostatic parameter,ξ