1. Electrolyte solutions conduct electricity due to the presence of ions. Strong electrolytes fully dissociate into ions in solution, while weak electrolytes only partially dissociate. 2. Ion transport in electrolyte solutions occurs through diffusion due to concentration gradients and migration due to applied electric fields. Conductivity and molar conductivity describe how well solutions conduct, and depend on factors like ion type and concentration. 3. At infinite dilution, the limiting molar conductivity of electrolytes can be calculated from the ion mobilities. For concentrated solutions, effects like ion-ion interactions cause conductivity to decrease with concentration.

1. Physical Chemistry II (InCh3107)
Chapter 1
Electrolyte Solutions
1

2. 1.1 Introduction
Electrolytic solutions: Solutions which conduct electricity.
Types of electrolytes
a) True Electrolytes: Composed of ions in pure state.
E.g. NaCl
b) Potential Electrolytes: Form ions in solutions.
consist of uncharged molecules in the pure state.
E.g. 1. HCl – strong electrolyte
HCl + H2O  H3O+ + Cl- (aq)
E.g. 2. CH3COOH – weak electrolyte
2
CH3COOH + H2O CH3COO-
+ H3O+

3. 1.2. Transport Properties
 Ions move in electrolyte solutions between anode and cathode.
 Solvated ions move at different velocities, depending on their size
and charge.
The movement is of two types.
a) Diffusion: Movement of ions due to concentration gradient.
b) Migration: Movement of ions due to electric field which is
applied between two electrodes immersed in an electrolyte
solution.
 Diffusion is described by Fick’s first law.
𝐉i = −𝐃
𝝏𝑪i
𝝏𝒙
Where Ji = flux of species i of concentration Ci in the x-direction,
𝜕𝐶i
𝜕𝑥
= Concentration gradient of i, D = Diffusion coefficient of i
3

4.  In the presence of an applied electric field, the flux of ions is
described by:
𝐉i = −𝐃
𝝏𝑪i
𝝏𝒙
− 𝒁i𝑪i
𝑭
𝑹𝑻
𝑬
Where the second term is due to migration effect.
1.2.1. Conductance and Conductivity
a) Conductance (G): The conductance of a solution is the inverse of
its resistance.
𝑮 =
𝟏
𝑹
Unit: Siemens (S). 1S = 1 -1 = 1CV-1s-1
 The resistance of a sample increases with its length (l) and
decreases with its cross-sectional area (A).
Hence,
𝑮 =
𝑨
𝒍
Where  = conductivity
4

5. b) Conductivity (): It is described by the equation:
 = 𝑮
𝒍
𝑨
Unit: Sm-1
 The conductivity of a sample is measured in a conductivity cell.
𝑙
𝐴
= 𝑐𝑒𝑙𝑙 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
 = cell constant  G = cell constant/R
 The conductivity of a solution depends on the number of ions
present.
c) Molar conductivity (m): It is the conductivity of a solution which
contains one molar of the electrolyte.
m =

𝒄
where c = molar concentration of the electrolyte.
5

6. Unit: Sm2mol-1
Range: 10-2Sm2mol-1
E.g.
The resistance of a conductivity cell filled with 0.01 M KCl solution at
25oC is 747.5 . Its conductivity is 0.14 -1m-1.
If the resistance of a 0.005 M CaCl2 is 876 . Calculate
a) the cell constant b) the conductivity of CaCl2 solution c) the molar
conductivity of the CaCl2 solution.
Solution:
a) Cell constant =   R = 0.14-1 m-1  747.5  = 104.6 m-1
b)  𝐶𝑎𝐶𝑙2 =
𝑐𝑒𝑙𝑙 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑅
=
104.65𝑚−1
876
= 0.11946−1𝑚−1
= 1.1946𝑚𝑆𝑐𝑚−1
c) 0.005M = 0.005 mol dm-3 = 5 mol m-3
m =

𝑐
=
0.11946𝑆𝑚−1
5𝑚𝑜𝑙𝑚−3
= 0.023892𝑆𝑚2𝑚𝑜𝑙−1 = 238.92𝑆𝑐𝑚2𝑚𝑜𝑙−1
6

7. Exercise: The conductivity of a 1.0 mol dm-3 aqueous KCl solution at
25oC is 0.112-1 cm-1.
Find its molar conductivity. (Answer: 112 Scm2 mol-1)
 Molar conductivity depends on:
1) Type of solvent - E.g. m (LiCl), (water) > m (LiCl), (propanone)
2) Type of electrolyte – E.g. m (HCl) > m (KCl)
3) Concentration: Molar conductivity is expected to be independent
of concentration.
In reality, as concentration increases:
i. For strong electrolytes: the interaction between ions increases.
 Therefore, molar conductivity decreases with increasing
concentration.
7

8. ii. For weak electrolytes: The degree of dissociation decreases.
Therefore, molar conductivity decreases.
d) Limiting Molar Conductivity (o
m)
 It is the molar conductivity of a solution at infinite dilution (c  0)
Law of Independent migration of ions:
 At infinite dilution, the conductivity of an electrolyte is a result
of the independent contributions of individual ions.
8
CH3COOH
HCl
KCl

C
CH3COOH
KCl
HCl
m
C

9. o
m = ++ + --
where + , - = limiting ionic conductivity of + ve and – ve ions
+, - = number of + ve and – ve ions per formula unit
 The law holds for strong electrolytes.
Examples of limiting ionic conductivities:
9

10. Example:
Determine the limiting molar conductivity of Na2SO4.
Solution
o
m = ++ + -- = 2Na
+ + SO4
2- = (25.01 + 116) mSm2mol-1
= 26.02 mSm2mol-1
 The law of independent migration of ions is important to
determine the limiting molar conductance of weak electrolytes.
Example:
o
m of NH4Cl, NaCl and NaOH in aqueous solution at 25oC are
0.01497, 0.01265 and 0.02478 Sm2mol-1, respectively. Determine o
m
of NH4OH which is a weak electrolyte.
Solution
o
m (NH4OH) = o
m (NH4Cl) + o
m (NaOH) - o
m (NaCl) = o
m (NH4
+) + o
m (Cl-) + o
m
(Na+) + o
m (OH-) - o
m (Na+) - o
m (Cl-) = o
m (NH4
+) + o
m (OH-)
10

11. Therefore,
o
m (NH4OH) = (0.01497 + 0.02478 - 0.01265) Sm2mol-1 = 0.02710 Sm2mol-1
e) Effect of concentration on molar conductivity
i. For strong electrolytes: At low concentrations, the molar
conductivities of strong electrolytes vary linearly with the square
root of the concentration.
m = o
m − 𝒄½ Kohlrausch’s law
 The c½ dependence arises from interactions between ions that
retard the ion’s progress.
ii. For weak electrolytes: The concentration dependence arises from
the decrease in the degree of dissociation () with increasing
concentration due to the displacement of the equilibrium:
11

12. For a weak acid, HA:
𝐾a =
𝐻3𝑂+ 𝐴−
(aq)
[𝐻𝐴]
At equilibrium, [H3O+] = [A-] = c where  is the degree of ionization.
[HA] = (1 - ) c
Therefore,
𝐾a =
𝑐22
1− 𝑐
𝐾a =
𝑐2
1−
Rearrangement yields,
1

= 1 +
𝑐
𝐾a
When the degree of dissociation is , m becomes:
m = o
m
Hence,
 =
m
o
m
12
H
A + H
2O A
-
+ H
3O+

13. Substituting for  yields,
Ostwald dilution law
 o
m can be determined by extrapolation to zero concentration in
the plot of 1/m vs. mc.
Example 1: The resistance of a 0.025M HAc measured in a cell (cell
constant = 0.367 cm-1) was found to be 444 . Calculate Ka. (o
m
(HAc ) = 390.7Scm2mol-1)
Solution:
𝐾a =
𝑐2
1−
1dm-3 = 10-3cm-3
13

14. Example 2: The molar conductivity of 0.01 M HAc at 298 K is m =
1.65 mSm2mol-1 and its limiting molar conductivity is 39.05 mSm2mol-
1. Calculate the acidity constant of the acid.
Solution
14

15. 1.2.2. The mobilities of Ions and the Transport Number
 These properties help to understand the role of ions in electrical
conduction.
a) Drift speed and Mobility
 When the potential difference between two electrodes a distance
l apart is  , the ions in the solution between them experience a
uniform electric field of magnitude, E.
An ion of charge Ze experiences a force of magnitude,
15

16.  As a result, the cation accelerates towards the negative electrode
and the anion accelerates towards the positive electrode.
 As the ion moves through the solvent, it experiences a frictional
retarding force, Ffri, which is proportional to its speed, s.
Ffri = fs
Where,
f = 6a
a = radius of the solvated ion,  = viscosity of the solvent (kgm-1s-1)
Drift Speed: The speed of an ion when the accelerating force is
balanced by the viscous drug.
F = Ffri ZeE = 6as
𝒔 =
𝒁𝒆𝑬
𝟔𝒂
16

17.  Accordingly, the drift speed of an ion is proportional to the
strength of the applied field.
s  E, hence
s = uE where u is ionic mobility.
Substitution yields:
Examples of ionic mobilities in water at 298 K
17
ion u (10-8 m2V-1s-1)
H+ 36.23
Na+ 5.19
K+ 7.62
OH- 20.64
Cl- 7.91
Br- 8.09

18. b) Mobility and Molar conductivity
 The molar conductivity of an ion is proportional to the ion’s
mobility.
 = 𝒖𝒁𝑭
Where F = Faraday’s constant = Nae
 The equation applies to cations and anions.
+ = 𝒖+𝒁+𝑭 − = 𝒖−𝒁−𝑭
 At infinite dilution the limiting molar conductivity is given by:
o
m = +𝒖+𝒁+ + −𝒖−𝒁− 𝑭
 For a symmetrical z : z electrolyte (E.g. CuSO4, 2 : 2), the above
equation simplifies to:
o
m = 𝒁(𝒖+ + 𝒖−)F
Where Z = +Z+ = -Z-
18
,

19. c) Mobility and Diffusion Coefficient (D)
Example
If u+ = 5  10-8m2V-1s-1, at 25 oC,
d) Transport number
 It is the fraction of total current carried by the ions of a specified
type.
Note: t+ + t- = 1 19

20.  The limiting transport number, to, of an ion is related to the
mobility of the ion by:
+Z+ = -Z- - Electrical neutrality
Therefore,
Using the relationship between mobility and ionic conductivity,
Therefore,
20

21. Measuring Transport Number
 It is measured using a moving boundary method.
Experiment
Apply a direct current to a solution.
The solution consists of;
- indicator solution E.g. a salt NX which is denser
- leading solution – E.g. a salt of interest – MX
21
K+ Cl-
Cd2+
Cl-
Initial Cd2+ boundary
Final Cd2+ boundary
Cathode
Anode
e-
e-

22. KCl – leading solution, CdCl2 – indicator solution
 The mobility of K+ > Cd2+
The total charge passed across a cross – sectional area is:
Q = It
No. of K+ in the shaded region = NAcV = NAclA
Total charge of K+ in the shaded region in time t:
Q+ = Z+ eNAcV = Z+ FcV
Therefore, total charge transferred when a current I flows for time t
is:
 By measuring the distance moved by the indicator solution, t, 
and u of the ions can be determined.
22

23. Example:
In a moving boundary experiment a 0.1 M KCl solution was layered
above a CdCl2 solution in a tube of cross sectional area 0.276 cm2.
When 6.25  10-3 A was passed through for 4000 s the boundary
moved a distance of 4.6 cm.
Calculate the transport number of K+.
Solution
23

24. 1.3. Activity and Activity Coefficients
 For an ideal solution, the chemical potential of a solute is given by:
Where, mB = molality of B
o
B = standard chemical potential – a chemical potential when
mB = 1molkg-1= mo
B
 The equation holds for very dilute solutions (less than 10-3 molkg-1)
o
m = (+𝑢+𝑍+ + −𝑢−𝑍−)
 For a real solution,
Where,
aB = activity of B, B = activity coefficient of B.
 Note: B  1 asmB  0 24

25. 1.3.1. Activities of ions in solution:
 The total molar Gibbs energy of the ions in ideal solution is the
sum of the chemical potential of the positive and the negative
ions.
Gm
ideal = +
ideal + -
ideal
 For a real solution,
Gm = ++ - = +
ideal + -
ideal + RT ln + + RT ln- = Gm
ideal + RT ln+-
 Gm becomes a function of the activities of the ions.
1.3.2. Mean activity coefficient (±)
 It is the mean of the individual activity coefficients.
 For a 1 : 1 electrolyte of univalent ions,
± = (+-)½
25

26. Accordingly,
+ = +
ideal + RTln± , -= -
ideal + RTln±
For a compound MpXq,
Gm = p++ q- = Gm
ideal + pRTln ++ qRTln-
= Gm
ideal + RTln +
p + RTln-
q
± = (+
p-
q)1/p + q
I = i
ideal + RTln±
26

27. 1.3.3. Ionic strength
 It is a measure of the non-ideality of an electrolyte solution which
is imposed by a solution containing ions.
Where Zi is charge of the ion, bi = molality, bo = 1molkg-1
Example
Calculate the ionic strength of a solution that contains 0.3 molal
NaOH and 0.1 molal Na3PO4.
Solution
bOH
- = 0.3 m, bPO4
3- = 0.1 m, bNa
+ = 0.3 + 3  0.1= 0.6 m
27

28. 1.3.4. The Debye – Hückel limiting law
 The mean activity coefficient of an ionic solution depends on
a) ionic strength
b) Temperature
c) dielectric constant of the solvent
d) charge of the ions (Zi) given by the expression:
Where A = 0.509 for aqueous solution at 25 oC.
 It is used to calculate the mean activity coefficient of an ionic
solution and holds for very low concentrations.
Note: The mean activity coefficient depends on the concentrations
of all ions in the solution.
28

29. Example: Calculate a) the ionic strength b) the mean activity
coefficient c) the activity of 1.00 mmolkg-1 CaCl2 (aq) at 25oC.
Solution
a)
ZCa
2+ = 2, ZCl
- = -1, bCa
2+ = 0.001molkg-1, bCl
- = 0.002 molkg-1
I = ½ (22 × 0.001 + -12 × 0.002) = 0.003
b) log = -0.509 Z+Z-I½ = -0.509 × 2 × (0.003)½ = -0.05599,  = 0.879
c)
aA+B- = (m) = 
m+
+-
- where,  = + + -
aCaCl2 = 
3m3 × 1 x 4 = (0.879)3 × (0.001)3 × 4 = 2.7166 × 10-6
29