Chapter 5. Simple mixtures
At equilibrium the chemical potential of a species is the same in every
phase.
How to express the chemical potential of a substance in terms of its
mole fraction in a mixture in Raoult’s and Henry’s laws.
The effect of a solute on certain thermodynamic properties of a solution
(boiling point; melting point, osmosis ).
To express the chemical potential of a substance in a real mixture in
terms of a property is known as the activity.
Chemistry deals with mixtures, including mixtures of substances that
can react together.
 at first stage, a binary mixtures (a mixture of two components) that do
not react together is considered.
The thermodynamic description of mixtures
5.1 Partial molar quantities:
(a) partial molar volume:
At 25oC, 1 mol H2O was added to a huge volume of water, the volume
increases by 18 cm3  18 cm3 mol-1 is the molar volume of pure water.
When 1 mol H2O was added to a huge volume of ethanol, the volume
increases by only 14 cm3  14 cm3 mol-1 is the molar volume in pure ethanol.
 In general, the partial molar volume of a substance A in a mixture is the
change in volume per mole of A added to a large volume of the mixture.
The partial molar volumes of the components of a mixture vary with
composition because the environment of each type of molecule changes as
the compositions from pure A to pure B.  modification of the mutual forces.
VJ = (V/nJ)p,T,n’
the subscript n’ signifies the amount of all
other substances present are constant.
Slope of a plot of V vs. nJ.  VJ
VJ value is dependent on the composition.
When the composition of the mixture is changed by addition of dnA and dnB

dV = (V/nA)p,T,nBdnA + (V/nB)p,T,nA dnB= VA dnA + VB dnB
When the composition is held constant, as the amount of A and B are
increased, the final volume of a mixture can be calculated by integration.
V =  VA dnA +  VB dnB = VA  dnA + VB  dnB
= VAnA + VBnB
Because V is a state function the final result is valid however the solution
is in fact prepared.
Illustration 5.1
A polynomial fit to measurements of the total volume of a water/ethanol
mixture at 25oC that contains 1.00 kg of water.
V = 1002.93 + 54.6664 x – .36394 x2 + 0.028256 x3
VE = (V/nE) = (v/x)
dv/dx = 54.66664 – 2(0.36394) x
+ 0.084768 x2
Molar volumes are always positive
but partial molar volume quantities
need not be.
Exp. The limiting partial molar volume
of MgSO4 in water is – 1.4 cm3 mol-1.
 The salts break up the open structure
of water as ions become hydrated.
(b) Partial molar Gibbs energies
The chemical potential energy is defined as the partial molar Gibbs energy.
J = (G/nJ)p,T,n’
 The chemical potential is the slope
of the Gibbs energy against the
amount of the component J
The total Gibbs energy of a mixture is
G = nA A + nB B.
The fundamental equation of chemical
thermodynamics relates the change
in Gibbs energy to changes in pressure,
temperature, and composition:
dG = Vdp – SdT + AdnA + BdnB + .
At constant pressure and temperature
dG = AdnA + BdnB + .
dG = dwadd, max
 dwadd,max = AdnA + BdnB + .
The additional work (non-expansion, electrical work) can arise from the changing
composition of a system (or electrochemical cell).
(c) The wider significance of the chemical potential:
G = U + pV – TS ; U = – pV + TS + G
 dU = – pdV – v dp + SdT + TdS + Vdp – SdT + AdnA + BdnB + .
= – pdV + TdS + AdnA + BdnB + .
At constant S and V;
dU = AdnA + BdnB + .
 J = (U/nJ)S,V,n’
J = (H/nJ)S,p,n’ ; J = (U/nJ)V,T,n’
(d) The Gibbs-Duhem equation:
At constant pressure and temperature, the total Gibbs energy of a binary mixture is
G = nA A + nB B
dG = AdnA + BdnB + nAdA + nBdB
 dG = AdnA + BdnB
 nAdA + nBdB = 0
The Gibbs–Duhem equation is J nJdμJ = 0.
In a system in which the compositions have mutual interaction or are mutually
exchangeable, the chemical potential of one component can not change independently
of the chemical potentials of the other components.
In a binary mixture,
dB = – (nA/nB) dA
 The same line of reasoning applies to all partial molar quantities.
Exp. Where the Vwater increases, the Vethanol decreases.
In practice, the Gibbs-Duhem equation is used to determine the partial molar
properties of one component in a binary mixture from the second component.
Example 5.1 Using the Gibbs-Duhem equation:
The experimental values of the partial molar volume of K2SO4(aq) at 298 K are
found to fit the expression: vB = 32.280 + 18.216 x1/2
x is the numerical value of the morality of K2SO4 (x =b/b).
Use the Gibbs-Duhem equation to derive an equation for molar volume of
water in the solution. vwater = 18.079 cm3 mol-1
nA dvA + nB dvB = 0  dvA = – (nA/nB) dvB
vA = vA* –
 (nA/nB) dvB
;
vA*: vA/cm3 mol-1 is the numerical value of the molar volume of pure A.
b/b
dvB/dx = 9.108 x–1/2  vA = vA* – 9.108 o (nA/nB) x–1/2 dx
nB/nA = nB/(1 kg)/MA = nBMA/1kg = bMA = x bMA
b/b

1/2
o x
A
 vA = vA* – 9.108 M b

dx = vA*– 2/3 {9.108 MAb(b/b)3/2}
Example 5.1 Using the Gibbs-Duhem equation:
 vA /(cm3 mol-1) = 18.079 – 0.1094 (b/b)3/2
5.2 The thermodynamics of mixing:
At constant temperature and pressure,
systems tend toward lower Gibbs energy.
A spontaneous mixing process is that of
two gases introduced into the same container.
The mixing is spontaneous, so it must
correspond to a decrease in G.
(a) The Gibbs energy of mixing of perfect gases:
 = Gm   =  + RT ln p/p
: standard chemical potential, the  of the
pure gas at 1.0 bar.
For simpler notation, the p/p is replaced by p
p: the numerical value of p in bars.
 =  + RT ln p
The Gibbs energy of the total system:
Gi = nAA + nBB = nA (A + RT ln p) + nB (B + RT ln p)
After mixing the partial pressure of the gases are pA and pB with pA + pB = p.
Gf = nA (A + RT ln pA) + nB (B + RT ln pB)
Gmix = nA RT ln (pA/p) + nB RT ln (pB/p)
Use the relation between partial pressure and mole fraction; pJ/p = J
 Gmix = nRT (A ln A + B ln B)
Because J < 1,
Gmix < 0
Gmix is negative for all compositions.
Exp. 5.2 Calculate a Gibbs energy of mixing
Gi = (3.0 mol) ((H2) + RT ln 3p)
+ (1.0 mol) ((N2) + RT ln p)
Gf = (3.0 mol) ((H2) + RT ln 3/2 p)
+ (1.0 mol) ((N2) + RT ln ½ p)
Gmix = (3.0 mol) RT ln (3/2 p/3p)
+ (1.0 mol) RT ln (½ p/p)
= – 6.9 kJ
Gmix = the mixing itself and the changes in pressure of the two gases.
When 3.0 mol H2 mixes with 1.0 mol N2 at same pressure, with the volumes of the
vessels adjusted accordingly, Gmix = – 5.6 kJ
(b) Other thermodynamic mixing functions:
(G/T)p,n = – S
 The entropy of mixing of two perfect gases
is given by
ΔmixS = (mixG/T)p,nA,nB
= –nR(xA ln xA + xB ln xB) > 0
For perfect gases, the enthalpy of mixing
ΔmixH = 0.
The enthalpy of mixing is zero because there
are no interactions between molecules forming
in the gas mixture.
5.3 The chemical potentials of liquids:
We should denoted quantities to pure substances by a superscript *.
A* is the chemical potential of the liquid A*(l).
A* = A0 + RT ln pA* (relative pressure pA* = pA/p0)
If another substance (solute) is
also present in the liquid,
the chemical potential is changed to A.
A = A0 + RT ln pA (pA = pA/p0)
 A0 = A* – RT ln pA*
 A = A* – RT ln pA* + RT ln pA
= A*+ RT ln pA/pA*
In a series of experiments on mixtures of closely related liquids, the partial vapor
pressure of each component to its vapor pressure of pure liquid, pA/pA* is
approximately equal to the mole fraction of A in the liquid mixture.
Raoult’s law: pA = ApA*
Mixtures that obey the law throughout the composition range from pure A to pure B
are called ideal solution.
 A = A* + RT ln A
Molecular interpretation 5.1
The presence of a second component reduces
the rate at which A molecules leave the surface
of the liquid but does not inhibit the return rate.
rate of vaporization = k A
rate of condensation = k’pA
At equilibrium, k A = k’pA
pA = (k/k’) A
For pure liquid, A = 1  pA* = (k/k’)
Some solutions depart significantly from Raoult’s law.
The law is therefore a good approximation for the properties of the solvent if the
solution is dilute.
(b) Ideal-dilute solution:
William Henry: for real solutions at low concentrations, although the vapor pressure
is proportional to its mole fraction, the
constant of proportionality is not the
vapor pressure of the pure substance.
pB = BKB
Mixtures for which the solute obeys
Henry’s law and the solvent obeys
Raoult’s law, are called ideal-dilute
solutions.
In a dilute solution:
The solvent molecules are in an environment very much like the one they have in
the pure liquid.
In contrast, the solute molecules are surrounded by the solvent molecules, which
is entirely different from their environment when pure.
Unless the solvent and solute molecules
happen to very similar.
 Raoult’s law.
Example. Investigate the validity of Raoult’s and Henry’s laws:
Rault’s law: pJ = JpJ*
Henry’s law: pJ = J KJ
KJ is an empirical constant chosen
so that the plot of the vapor pressure
of J against its molar fraction is tangent
to the experimental curve at J = 0.
K(propane) = 23.3 kPa
K(chloroform) = 22.0 kPa
In practice, Henry’s law is expressed in terms of mobility, b, of the solute.
pB = bB KB
 respiration: oxygen, diving, mountaineering, anesthetics.
I5.1 Gas solubility and breathing:
We inhale about 500 cm3 of air with each breath we take.
As the diaphragm (橫隔膜) is depressed and chest expands  a decrease in
pressure of 100 Pa related to atmospheric pressure.
The total volume in the lungs is about 6 dm3 (L).
Some air remains in the lungs at all time to prevent the collapse of the alveoli (肺泡).
Alveolar gas is in fact a mixture of newly inhaled air and air about to be exhaled.
pO2 in arterial blood is about 40 Torr  pO2 in freshly inhaled air is about 104 Torr.
Arterial blood remains in the capillary passing through the wall of the alveolus for
about 0.75 s, but such is the steepness of the pressure gradient that it becomes fully
saturated with oxygen in about 0.25 s.
If the lungs collect fluid, then the respiratory membrane thickens, diffusion is greatly
slow, and body tissue begin to suffer from oxygen starvation.
CO2 moves in the opposite direction across the respiratory tissue, but the pCO2
gradient is much less, corresponding to about 5 Torr in blood and 40 Torr in air at
equilibrium.
Because the CO2 is much more soluble in the alveolar fluid than oxygen is, equal
amount of oxygen and carbon dioxide are exchanged in each breath.
A hyperbaric oxygen chamber, high pO2, is used to treat certain types of disease,
such as CO poisoning, disease caused by anaerobic bacteria.
In scuba diving, one unfortunate consequence of breathing air at high pressure is
that N2 is much more soluble in fatty tissue than in water, so it tends to dissolve in
central nervous system, bone marrow, and fat reserves.
 nitrogen narcosis (氮醉).
The Properties of Solutions:
5.4 Liquid mixtures:
(a) Ideal solution
The Gibbs energy of mixing of two liquids to form an ideal solution is calculated
as in exactly the same way as for two gases.
Before mixing:
Gi = nAA* + nBB*
After mixing:
Gf = nA {A* + RTln A} + nB {B* + RTln B}
 mixG = nRT {A RTln A + B RTln B} ; n = nA + nB
mixS = nR {A RTln A + B RTln B}
 mixH = mixG + TmixS = 0
The mixV = 0  (G/p)T = V  ( mixG/p)T = mixV = 0
mixG is independent of pressure.
In a perfect gas there are no forces acting between molecules 
In an ideal solutions, there are interactions, but the average energy of A–B
interactions in the mixture is the same as the average energy of A–A and B–B
interactions in the pure liquids.
Real solution:
No only may there be enthalpy and volume changes when liquids mix, but there
may also be an additional contribution to the entropy arising from the way in which
the molecules of one type might cluster together instead of mingling freely with the
others.
Large and positive mixH or if mixS << 0, then the mixG might be positive for mixing.
 separation is spontaneous and the liquid may be immiscible.
When they are miscible only over a certain range compositions, the liquid might
be partially miscible.
(b) Excess functions and regular solutions:
An excess function (XE) is the difference
between the observed thermodynamic
function of mixing and the function for
an ideal solution.
SE = mixS – mixSideal
A regular solution: as one in which two
kinds of molecules are distributed randomly but
have different energies of interactions with each
other.

HE  0 but SE = 0.
Suppose the HE depends on compositions:

HE = nRTAB
HE = nRTAB
 is a dimensionless parameter that is a
measure of the energy AB interactions
related to that of the AA and BB interactions.
If  < 0, mixing is exothermic and
solute-solvent interactions are
more favourable than solvent-solvent
and solute-solute interactions.
If  > 0, mixing is endothermic.
For a regular solution:
mixG = nRT {AlnA + BlnB + AB}
mixG = nRT {AlnA + BlnB + AB}
The important feature is that  > 2
the graph shows two minima separated
by a maximum.
At  >2, the system will separate
spontaneously into two phases with
compositions corresponding to the
two minima.
5.5 Colligative properties: (眾數性質)
A colligative property is a property that depends only on the number of solute
particles present, not their identity.
 the elevation of boiling point, the depression of freezing point, osmotic pressure.
Throughout the following:
(1). The solute is not volatile.
(2). The solute does not dissolve in the solid solvent.
(a) The common feature of colligative properties:
For an ideal-dilute solution, the reduction of the chemical potential of the solvent is
from A* for pure solvent to A* + RT lnA when a solute is present.
There is no direct influence of the solute on the chemical potential of the solvent
vapor and the solid solvent because the solute appears in neither the vapor nor the
solid.
A reduction in chemical potential of solvent
 boiling point is raised
and freezing point is lowered.
Molecular interpretation 5.2
The lowering of the chemical potential is not the energy of interaction of the solute
and solvent particles, because it occurs even in an ideal solution.
If it is not an enthalpy effect,
it must be an entropy effect.
S = kB ln
The vapor pressure reflects the tendency
of the solution towards greater entropy.
When a solute is present, there is an
additional contribution to the entropy of
the liquid, even in an ideal solution.
 S(solution  gas) < S(solvent  gas)
 lower vapor pressure
and higher boiling point.
(b) The elevation of boiling point:
The heterogeneous equilibrium between
the solvent vapor and the solvent in solution
A*(g) = A*(l) + RT ln A
The presence of a solute at a mole fraction
B causes an increase in normal boiling
point from T* to T* + T,
T = KB ; K = RT*2/ vapH
Justification 5.1
ln A = (A*(g) – A*(l))/RT = vapG/RT
 Gibbs-Helmholtz equation: ((G/T)/T)p = – H/T2
 d lnA /dT = 1/R d (vapG/T)/dT = – vapH/RT2
lnA
0 d lnA =
– (1/R) T* vapH/T2 dT
T
 ln A = ln(1 – B ) = vapH/R (1/T – 1/T*)
When B << 1, ln(1 – B)  – B
B = vapH/R (1/T* – 1/T)
1/T* – 1/T = (T – T*)/TT*  T/T*2
For practical applications, we note that the molar fraction of B is proportional to its
molality, b, in the solution.
 T = Kb b
(c) The depression of freezing point:
The heterogeneous equilibrium between
pure solid solvent and solution with
solute present at a mole fraction B:
A*(s) = A*(l) + RT ln A


T = K’B ;
K’ = RT*2/ fusH
T = Kf b
Cryoscopy (Greek "freeze-viewing") :
A technique for determining the molecular weight of a solute by dissolving a known
quantity of it in a solvent and recording the amount by which the freezing point of
the solvent drops.
(d) Solubility:
Solubility is not strictly a colligative
property (identity of the solute).
Saturation is a state of equilibrium, with the
undissolved solute in equilibrium with
the dissolved solute.
In solution B* = B*(l) + RT ln B
B*(s) = B*(l) + RT ln B
RT ln B = {B*(s) – B*(l)}/RT
= – fusG/RT
0 d lnB =
– (1/R) Tf fusH/T2 dT
 ln B = fusH/R (1/Tf – 1/T)
The solubility of B decreases exponentially as the temperature is lowered from
its melting point.
High melting point, large enthalpies
 low solubility.
This predict fails to predict that solutes
will have different solubilities in different
solvents, for no solvent properties appear
in the expression.
(e) Osmosis:
Osmosis is the spontaneous passage of a pure solvent into a solution separated
from it by a semipermeable membrane, a membrane permeable to the solvent but
not to the solute.
The osmotic pressure,  , is the pressure that must be applied to the solution to
stop the influx of solvent.
 transport of fluid through cell membrane,
dialysis (透析), osmometry (the determination
of molar mass of macromolecules).
Equilibrium is reached when the hydrostatic
pressure of the column of solution match the
osmotic pressure.
The chemical potential of the solvent is lowered by the solute, but it is restored to
its “pure” value by the application of pressure.
 The van ’t Hoff equation for the osmotic pressure is Π = [B]RT
[B] = the molar concentration of the solute.
Justification 5.3
On the pure solvent side: A*(p)
One the solution side:
A(A, p + Π) = *A(p + Π) + RT lnA
*A(p + Π) = *A(p) + p Vm dp
 A(A, p + Π) =
*A(p) + p Vm dp + RT lnA
 – RT lnA = p Vm dp  Π = [B]RT
Because the effect is so readily measurable and large, osometry is commonly
used to measure molar mass of macromolecules (exp. Protein…).
As these huge molecules dissolve to produce solutions that are far from ideal,
 Π = [J]RT {1+ B[J] + …………}
B: osmotic viral coefficient.
Exp. 5.4 Osmotic pressure of PVC:
c (g dm–3) vs. h/cm
[J] = c/M; osmotic pressure = gh
 h/c = RT/ gM (1 + Bc/M + …….)
= RT/ gM + (RTB/ gM) c
 To find M, h/c vs. c , and expect
a straight line with intercept RT/ gM at c = 0.
 1.2 x 102 kg mol-1 = 120 kDa (1 Da = 1g mol-1)
I5.2 Osmosis in physiology and Biochemistry:
Cell membranes are semipermeable and allow water, small molecules, and
hydrated ions to pass, while blocking the passage of biopolymers.
The influx of water also keeps the cell swollen, whereas dehydration causes the
cell to shrink.
To maintain the integrity of the cells, solution injected into the blood must be
isotonic with the blood.  the same osmotic pressure.
Dialysis: a common technique for the removal of impurities from solution of
biological macromolecules and for the study of binding of small molecules to
macromolecules.
In a purification experiments, a solution of macromolecules and impurities (ions
and small proteins) is place in a bag made of semipermeable membrane, and the
bag is immersed in a solvent.
In a binding experiment, a solution contains a macromolecules ([M]) and smaller
molecules A ( the total concentration of A = [A]in) in a bag.
The total concentration A is the sum of [A]in = [A]free and [A]bound.
At equilibrium, the chemical potential of free A in macromolecule solution
is equal to that in the solution on the other side.
 [A]free = [A]out
(the activity coefficient of A is the same in both solutions)
The [A]bound = [A]in – [A]out
The average number of A molecules bound to M molecules, v, is then the ratio
v = [A]bound/[M] = ([A]in – [A]out)/M
If there are N identical and independent binding sites on each macromolecule,
each macromolecule behaves like N smaller macromolecules, M’, with same value
of K for each site.
The bound and unbound A molecule to
each binding site in the macromolecule
are in equilibrium, M’ + A  M’A
K = [M’A]/[M’]free[A]free
[M’A]/[M’]free = (v/N) / (1 – v/N)
 K = v/N/{(1 – v/N) [A]out}
 v/[A]out = KN – Kv (Scatchard equation)
If a straight line is not obtained we can
conclude that the binding sites are not
equivalent or independent.
Activities
Account the deviations from ideal behavior.
It is important to be aware of the different definitions of standard states and activities.
5.6 The solvent activity
The general form of the chemical potential of a real or ideal solvent is:
A = A* + RT ln (pA/pA*)
For the ideal solution:
A = A* + RT ln A
The solvent activity of the real solution is related to its chemical potential by
A = A* + RT ln aA
aA is the activity of A, a kind of ‘effective’ mole fraction.
 The activity is defined as aA = pA/pA*.
Illustration 5.3 The vapour pressure of 0.500 M KNO3 at 100oC is 99.95 kPa, so
the activity of the solution at this temperature is
aA = 99.95 kPa/ 101.325 kPa = 0.9864
Because all solvents obey the Raoult’s law at A  1
aA  A as A  1
A convenient way of expressing this convergence is to introduce the activity
coefficient, aA = A A
A  1
as A  1
A = A* + RT ln A + RT ln A
5.7 The solute activity:
For solutes, they approach ideal-dilute (Henry’s law) behavior as B  0.
(a) Ideal-dilute solutions: (Henry’s law: pB = KBB)
The chemical potential of solute, B:
B = B* + RT ln (pB/pB*) = B* + RT ln (KB/pB*) + RT ln B
 a new standard chemical potential B0 = B* + RT ln (KB/pB*)
 B = B0 + RT ln B
For ideal solution: KB = pB* ;  B0 =  B*
(b) Real solution:
We now permit deviations from ideal-dilute solution:
B = B0 + RT ln aB ;
aB = pB/KB
It is sensible to introduce an activity coefficient through
aB = BB
Because the solute obeys Henry’s law: [solute]  0
aB  B and B  1 as B  0.
Deviations of the solute form ideality disappear as zero concentration is
approached.
Example 5.5 Measuring activity
c (chloroform); pc/kPa ; pA (acetone)/kPa
When chloroform is solvent: aC = pC/pC*
If it is solute: aC = pc/Kc ; C = aC/C
Raoult’s law
Henry’s law
(c) Activities in terms of molarities:
In chemistry, compositions are often expressed as molalities, b, in place of mole
fractions.
B = B0 + RT ln bB
The chemical potential of the solute has its standard value of o when molality of
B is equal to bo (1.0 mol kg-1).
as bB  0 , B  – 
 The solute becomes increasingly stabilized.
In practical consequence, it is very difficult to remove the last trace of a solute
from a solution.
aB = B (bB/b0) where B  1 as bB  0
For the chemical potential of a real solution at any molality:
 = 0 + RT ln a
(d) The biological standard state:
The conventional standard state of hydrogen ions (pH = 0; [H+] = 1.0 M) is not
appropriate to normal biological conditions.
In biochemistry, the biological standard state at pH = 7.0 is common to be adapted.
The corresponding standard thermodynamic functions as G; H; ; S.
The biological standard state (pH = 7) is related to the thermodynamic standard
state by H+ = H+0 + RTln aH+ = H+0 – 7RT ln 10.
At 298 K, H+0 – H+ 40 kJ mol-1.
5.8 The activities of regular solutions
The original of deviations from Raoult’s law and the its relation to activity
coefficient.
The Gibbs energy of mixing to form a nonideal solution is
mixG = nRT {A RTln aA + B RTln aB}
If each activity is replaced by 
 mixG = nRT {A RTln A + B RTln B + A ln A + B ln B}
 mixG = nRT {AlnA + BlnB + AB}
 mixG = nRT {AlnA + BlnB + AB(A + B)}
mixG = nRT {AlnA + BlnB + AB2 + BA2}
 ln A = B2 ; ln B = A2 (Margules equations)
A  1 as B  0 and B  1 as A  0
The activity of A:
aA = A A = A exp(B2) = A exp{(1 – A)2}
 pA = pA* [A exp{(1 – A)2}]
 > 0 (endothermic mixing, unfavourable
solute-solvent interaction), give vapour
pressure higher than ideal ( = 0 ).
All the curves approach linearity and coincide
with Raoult’s law as A  1.
When A << 1 ; pA = pA* (exp) A

pA = KAA (Henry’s law)
K = pA* exp
5.9 The activities of ions in solution
Interactions between ions are so strong that the approximation of replacing
activities by molalities is valid only in very dilute solutions (< 10-3 m in total ion
concentration) and in precise work activities themselves must be used.
(a) Mean activity coefficients
The chemical potential of a univalent cation M+ is denoted as + and that of a
univalent anion X– is denoted as –, the Gibbs energy of the ions in the electrically
neutral solution is the sum of these partial molar quantities.
Gm = +ideal + –ideal
For a real solution,
Gm = + + – = +ideal + RT ln + + –ideal + RT ln –
= Gmideal + RT ln 
There is no experimental way of separating the product +– into contributions from
cations and anions.
 For a 1,1-electrolyte, we introduce the mean activity coefficient as geometric
mean of the individual coefficients:
 = (+–)½
The individual chemical potentials of the ions as
+ = +ideal + RT ln  ; – = –ideal + RT ln 
The nonideality is shared equally.
This approach to the case of a compound MpXq ( p cations + q anions):
The molar Gibbs energy of ions is the sum of their partial molar Gibbs energies:
Gm = p + + q – = Gmideal + p RTln + + q RT ln–
 = (+p–q)1/s ; s = p + q
 i = iideal + RT ln   G = p + + q –
Both types of ion now share equal responsibility for the nonideality.
(b) The Debye-Hückel limiting law:
The long range and strength of the Columbic interaction between ions means
that it is likely to be primarily responsible for the departures form ideality in ionic
solutions and to dominate all other contributions to nonideality.
The Debye–Hückel theory (1923):
Oppositely charged ions attract one another.
As a result, anions more likely to be found near
cations in solution, and vice versa.
The time-average, spherical haze around the
central ion, in which counter ions outnumber ions
of the same charge as the central ions, has
a net charge equal in magnitude but opposite
in sign to that on central ion, and is called its ionic atmosphere.
The energy, and chemical potential of any given central ion is lowered as a result
of its electrostatic interaction with its ionic atmosphere.
The stabilization of ions by their interaction with their ionic atmospheres is part of
the explanation why chemists commonly use dilute solutions, in which stabilization is
less important, to achieve precipitation of ions from electrolyte solutions.
The Debye–Hückel limiting law is
log  = – |z+z-| A I1/2 ; where I is the ionic strength,
I = ½ i zi2(bi/b0). (dimensionless)
at 25oC, A = 0.509 for an aqueous solution.
For a solution consisting of two types of ion at molalities b+ and b–
I = ½ (b+z+2 + b–z–2)/b0
The mean activity coefficient of 5.0 x 10-3 mol kg-1 KCl(aq) at 25oC.
I = ½ (b+ + b–)/b0 = b/b0 (b+ = b– = b) ;
log γ± = – 0.509 x (5.0 x 10-3)1/2 = – 0.036   = 0.92.
The ionic strength emphasizes the charges of the ions because the charge
numbers occur as their square.

Ionic solutions of moderate molalities may have activity coefficients that different
from the values given by the eq. 5.69, yet all solutions are expected to conform as
b  0.
The agreement at very low molalities
(less than 1 mmol kg -1 , depending on
charge type) is impressive, and convincing
evidence in support of the model.
(c) The extended Debye-Hückel law
The extended Debye–Hückel law is
ln ± = – |z+z–|AI1/2/(1 + BI1/2) + CI.
.
The equation accounts for some
activity coefficients over a moderate
range of dilute solution (up to about
0.1 mol kg-2); nevertheless it remains
very poor near 1.0 mol kg-1.
.