Physical Chemistry
26. Liquid-Solid Solutions

Ruifeng ZHOU

Faculty of Science &Institute for the Advancement of Higher Education
Hokkaido UniversityTable of Content

1. Nonelectrolytes
2. Electrolytes
3. Colligative properties21. Nonelectrolytes

3• In a binary solution, if the fraction of component 1 ismuch greater than that of component 2 (𝑥𝑥1 ≫ 𝑥𝑥2 ),component 1 is called the solvent; component 2 iscalled the solute.
• In this chapter, we will discuss solutions withvolatile liquid solvents and a non-volatile solidsolutes.
• There are many ways to describe the concentrationof the solute.
• Amount concentration (molarity, 𝑐𝑐) is the amount ofsubstance of solute per volume of solution.

1. Nonelectrolytes

𝑛𝑛2𝑐𝑐 ≡𝑉𝑉
• Molality (𝑏𝑏) is the amount of substance of soluteper mass of solvent.
𝑛𝑛2𝑏𝑏 ≡𝑚𝑚1
• It is more convenient to use the dimensionlessmolality (𝕞𝕞).
𝑏𝑏𝕞𝕞 ≡𝑏𝑏°• 𝑏𝑏° = 1mol � kg −1 is the standard molality.

41. Nonelectrolytes

5• The amount of substance of the solvent is
𝑚𝑚1
𝑛𝑛1 =𝑀𝑀1• The mole fraction of the solute is
𝑛𝑛2 𝑏𝑏𝑚𝑚1 𝑏𝑏𝑚𝑚1 𝑥𝑥1= 𝑛𝑛 = 𝑚𝑚 = 𝑥𝑥1 𝑀𝑀1 𝑏𝑏𝑥𝑥2 =11𝑛𝑛𝑥𝑥1
𝑀𝑀1• The activity of the solvent can be calculated fromthe partial pressure of its vapour (nearly ideal gas).
𝑃𝑃1𝑎𝑎1 = ∗𝑃𝑃1

61. Nonelectrolytes• Then, 𝜇𝜇1 can found.
𝜇𝜇1 = 𝜇𝜇1l + 𝑅𝑅𝑅𝑅 ln 𝑎𝑎1• However, 𝜇𝜇2 cannot be calculated from 𝑃𝑃2 ≈ 0.
• 𝜇𝜇2 can be obtained from 𝜇𝜇1 , because they are
related by Gibbs-Duhem equation.
𝑥𝑥1 𝑑𝑑𝜇𝜇1 + 𝑥𝑥2 𝑑𝑑𝜇𝜇2 = 0• When the solute has low concentration, the solvent
obeys Raoult’s law.
lim ln 𝑎𝑎1 = ln 𝑥𝑥1 = lim ln 1 − 𝑥𝑥2 = −𝑥𝑥2 = −𝑥𝑥1 𝑀𝑀1 𝑏𝑏
𝑥𝑥1 →1≈ −𝑀𝑀1 𝑏𝑏

𝑥𝑥2 →01. Nonelectrolytes

7• When the solute has high concentration, thisrelation is corrected by the osmotic coefficient (𝜙𝜙).
ln 𝑎𝑎1 = −𝜙𝜙𝑀𝑀1 𝑏𝑏• Thus, 𝜙𝜙 can be calculated by 𝑃𝑃1 from experiment .
𝑑𝑑𝜇𝜇1 = 𝑅𝑅𝑅𝑅𝑑𝑑 ln 𝑎𝑎1 = 𝑅𝑅𝑅𝑅𝑑𝑑 −𝜙𝜙𝑀𝑀1 𝑏𝑏= −𝑅𝑅𝑅𝑅𝑀𝑀1 𝜙𝜙𝑑𝑑𝑏𝑏 + 𝑏𝑏𝑑𝑑𝜙𝜙• On the other hand, 𝜇𝜇2 can be expressed as𝜇𝜇2 = 𝜇𝜇2 ° + 𝑅𝑅𝑅𝑅 ln 𝑎𝑎2• 𝜇𝜇2 ° is the 𝜇𝜇 of the hypothetical standard state ofsolute at 𝑏𝑏 = 𝑏𝑏° but exhibiting infinite-dilution
behaviour. lim 𝛾𝛾2 = 1.
𝑏𝑏→0

81. Nonelectrolytes

𝑎𝑎2 = 𝛾𝛾2 𝕞𝕞
𝑑𝑑𝜇𝜇2 = 𝑅𝑅𝑅𝑅𝑑𝑑 ln 𝑎𝑎2 = 𝑅𝑅𝑅𝑅 𝑑𝑑 ln 𝛾𝛾2 + 𝑑𝑑 ln 𝕞𝕞• Insert 𝑑𝑑𝜇𝜇1 and 𝑑𝑑𝜇𝜇2 into the Gibbs-Duhem equation.
𝑑𝑑𝕞𝕞=0−𝑥𝑥1 𝑅𝑅𝑅𝑅𝑀𝑀1 𝜙𝜙𝑑𝑑𝑏𝑏 + 𝑏𝑏𝑑𝑑𝜙𝜙 + 𝑥𝑥2 𝑅𝑅𝑅𝑅 𝑑𝑑 ln 𝛾𝛾2 +𝕞𝕞• Divided the equation by 𝑅𝑅𝑅𝑅𝑥𝑥1 𝑀𝑀1 𝑏𝑏 = 𝑅𝑅𝑅𝑅𝑥𝑥2 .
𝑑𝑑𝑏𝑏
𝑑𝑑𝕞𝕞+ 𝑑𝑑𝜙𝜙 + 𝑑𝑑 ln 𝛾𝛾2 +=0− 𝜙𝜙𝑏𝑏𝕞𝕞• Note that,𝑑𝑑𝑏𝑏 𝑑𝑑𝕞𝕞=𝑏𝑏𝕞𝕞

1. Nonelectrolytes• Thus,𝜙𝜙 − 1𝑑𝑑𝕞𝕞 + 𝑑𝑑𝜙𝜙
𝑑𝑑 ln 𝛾𝛾2 =𝕞𝕞• Integrate both sides from 0 to 𝕞𝕞.
𝕞𝕞𝕞𝕞𝕞𝕞
𝜙𝜙 𝕞𝕞′ − 1ln 𝛾𝛾2 � = �𝑑𝑑𝕞𝕞′ + 𝜙𝜙�
𝕞𝕞′000• When 𝕞𝕞 = 0, 𝛾𝛾2 0 = 1 and 𝜙𝜙 0 = 1.
𝕞𝕞
𝜙𝜙 − 1𝑑𝑑𝕞𝕞′ln 𝛾𝛾2 = 𝜙𝜙 − 1 + �𝕞𝕞′0• 𝛾𝛾2 can be calculated by 𝜙𝜙 𝕞𝕞 from experiment.

9101. Nonelectrolytes• Example: water-sucrose solution at 25℃.
𝑃𝑃1ln ∗𝑃𝑃1
𝜙𝜙 = −𝑀𝑀1 𝑏𝑏
24.0

1.00

23.5

0.98

23.0

0.96

1.5

3.5

1.4

3.0

1.3

2.5

21.5

Ideal

21.0

0.90
0.88

20.5

0.86

Real0123m

45

6

γ2

φ

1.2

2.0
1.5

1.1
1.0

0

123m

45

6

1.0

γ20.92

φ22.0

20.0

4.0

0.94

a1

P1(torr)

22.5

1.6

1. Nonelectrolytes• 𝜙𝜙 of water - sucrose solution at 25℃ can be fitted
by the following polynomial.
𝜙𝜙= 1 + 7.349 × 10−2 𝕞𝕞 + 1.9783 × 10−5 𝕞𝕞2 − 5.688
× 10−5 𝕞𝕞3 + 6.036 × 10−4 𝕞𝕞4 − 2.517 × 10−5 𝕞𝕞5• When 𝕞𝕞 = 1, 𝜙𝜙 = 1.08816.
1
𝜙𝜙 − 1
�𝑑𝑑𝕞𝕞′ = 0.08163
𝕞𝕞′
0ln 𝛾𝛾2 = 1.08816 − 1 + 0.08163 = 0.1698
𝛾𝛾2 = 1.185

112. Electrolytes

12• An electrolyte has cations (+) and anions (-).
• The cations and anions form the solute of anelectrolyte solution.
• In an electrolyte solution, the cations and anions
CANNOT be separated.
• The ion-ion interaction (∝ 𝑟𝑟 −1 ) is much stronger
than the dipole-dipole (Keesom) interaction (∝ 𝑟𝑟 −3 )and the (London) dispersion force (∝ 𝑟𝑟 −6 ) betweenmolecules.
• The behaviours of electrolytes are different fromnonelectrolytes.

2. Electrolytes• Consider a strong electrolyte, C𝜈𝜈+ A𝜈𝜈− , which
dissociates completely when dissolved.
C𝜈𝜈+ A𝜈𝜈− → 𝜈𝜈+ C𝑧𝑧+ + 𝜈𝜈− A𝑧𝑧−

• The amounts of solvent, cation and anion are𝑚𝑚1
𝑛𝑛1 =𝑀𝑀1
𝑛𝑛+ = 𝜈𝜈+ 𝑏𝑏𝑚𝑚1• 𝑏𝑏 is the molality of C𝜈𝜈+ A𝜈𝜈− .
𝑥𝑥+ 𝑛𝑛+ 𝜈𝜈+ 𝑏𝑏𝑚𝑚1== 𝑚𝑚= 𝜈𝜈+ 𝑏𝑏𝑀𝑀1
1
𝑥𝑥1 𝑛𝑛1𝑀𝑀113142. Electrolytes• The activities of the ions cannot be measured
separately.
• The Gibbs-Duhem equation is𝑥𝑥1 𝑑𝑑𝜇𝜇1 + 𝑥𝑥+ 𝑑𝑑𝜇𝜇+ + 𝑥𝑥− 𝑑𝑑𝜇𝜇− = 0−𝑑𝑑𝜇𝜇1 = 𝑏𝑏𝑀𝑀1 𝜈𝜈+ 𝑑𝑑𝜇𝜇+ + 𝜈𝜈− 𝑑𝑑𝜇𝜇−• Since 𝜇𝜇1 = 𝜇𝜇1l + 𝑅𝑅𝑅𝑅 ln 𝑎𝑎1 ,𝑑𝑑𝜇𝜇1 = 𝑅𝑅𝑅𝑅𝑅𝑅 ln 𝑎𝑎1• When the concentration is low, the solvent is ideal.
lim ln 𝑎𝑎1 = ln 𝑥𝑥1 = lim ln 1 − 𝑥𝑥+ − 𝑥𝑥− = − 𝑥𝑥+ + 𝑥𝑥−
𝑥𝑥1 →1𝑥𝑥2 →0= − 𝜈𝜈+ + 𝜈𝜈− 𝑏𝑏𝑀𝑀1 = −𝜈𝜈𝑏𝑏𝑀𝑀12. Electrolytes𝜈𝜈 ≡ 𝜈𝜈+ + 𝜈𝜈−• At higher concentrations,ln 𝑎𝑎1 = −𝜙𝜙𝜙𝜙𝑀𝑀1 𝑏𝑏• Therefore,𝑑𝑑𝜇𝜇1 = −𝜈𝜈𝑀𝑀1 𝑅𝑅𝑅𝑅 𝜙𝜙𝑑𝑑𝑏𝑏 + 𝑏𝑏𝑑𝑑𝜙𝜙• As for the electrolyte, since 𝜇𝜇+ = 𝜇𝜇+ ° + 𝑅𝑅𝑅𝑅 ln 𝑎𝑎+ ,𝑑𝑑𝜇𝜇+ = 𝑅𝑅𝑅𝑅𝑑𝑑 ln 𝑎𝑎+• Thus,𝜈𝜈+ 𝑑𝑑𝜇𝜇+ + 𝜈𝜈− 𝑑𝑑𝜇𝜇− = 𝑅𝑅𝑅𝑅𝜈𝜈+ 𝑑𝑑 ln 𝑎𝑎+ + 𝑅𝑅𝑅𝑅𝜈𝜈− 𝑑𝑑 ln 𝑎𝑎−= 𝑅𝑅𝑅𝑅𝑅𝑅 ln 𝑎𝑎+ 𝜈𝜈+ 𝑎𝑎− 𝜈𝜈−15

2. Electrolytes• The activities of ions can be expressed as𝑎𝑎+ = 𝛾𝛾+ 𝕞𝕞+• The mean ionic molality (𝕞𝕞) is𝕞𝕞𝜈𝜈 ≡ 𝕞𝕞+ 𝜈𝜈+ 𝕞𝕞− 𝜈𝜈−• The mean ionic activity coefficient (𝛾𝛾)̅ is𝛾𝛾̅ 𝜈𝜈 ≡ 𝛾𝛾+ 𝜈𝜈+ 𝛾𝛾− 𝜈𝜈−• The mean ionic activity (𝑎𝑎)
� is𝑎𝑎� 𝜈𝜈 ≡ 𝑎𝑎+ 𝜈𝜈+ 𝑎𝑎− 𝜈𝜈− = 𝛾𝛾+ 𝜈𝜈+ 𝕞𝕞+ 𝜈𝜈+ 𝛾𝛾− 𝜈𝜈− 𝕞𝕞− 𝜈𝜈− = 𝛾𝛾̅ 𝜈𝜈 𝕞𝕞𝜈𝜈
• Thus,̅𝑎𝑎� = 𝛾𝛾𝕞𝕞

16

2. Electrolytes• Example: NaCl. 𝜈𝜈+ = 𝜈𝜈− = 1. 𝜈𝜈 = 2.
𝕞𝕞+ = 𝕞𝕞− = 𝕞𝕞𝕞𝕞𝜈𝜈 = 𝕞𝕞2 = 𝕞𝕞𝕞𝕞 = 𝕞𝕞2• Example: CaCl2. 𝜈𝜈+ = 1, 𝜈𝜈− = 2. 𝜈𝜈 = 3.
𝕞𝕞+ = 𝕞𝕞𝕞𝕞− = 2𝕞𝕞𝕞𝕞𝜈𝜈 = 𝕞𝕞3 = 𝕞𝕞 2𝕞𝕞 2 = 4𝕞𝕞3
• Example: Na3Fe(CN)6. 𝜈𝜈+ = 3, 𝜈𝜈− = 1. 𝜈𝜈 = 4.
𝕞𝕞+ = 3𝕞𝕞𝕞𝕞− = 𝕞𝕞𝕞𝕞𝜈𝜈 = 𝕞𝕞4 = 3𝕞𝕞 3 𝕞𝕞 = 27𝕞𝕞417

2. Electrolytes• Therefore,𝜈𝜈+ 𝑑𝑑𝜇𝜇+ + 𝜈𝜈− 𝑑𝑑𝜇𝜇− = 𝑅𝑅𝑅𝑅𝑅𝑅 ln 𝛾𝛾̅ 𝜈𝜈 𝕞𝕞𝜈𝜈= 𝜈𝜈𝑅𝑅𝑅𝑅 𝑑𝑑 ln 𝕞𝕞 + 𝑑𝑑 ln 𝛾𝛾̅• The Gibbs-Duhem equation becomes𝜈𝜈𝑀𝑀1 𝑅𝑅𝑅𝑅 𝜙𝜙𝑑𝑑𝑏𝑏 + 𝑏𝑏𝑑𝑑𝜙𝜙 = 𝑏𝑏𝑀𝑀1 𝜈𝜈𝑅𝑅𝑅𝑅 𝑑𝑑 ln 𝕞𝕞 + 𝑑𝑑 ln 𝛾𝛾̅
• Divide it by 𝑏𝑏𝑀𝑀1 𝜈𝜈𝑅𝑅𝑅𝑅.
𝑑𝑑𝑏𝑏
𝑑𝑑𝕞𝕞+ 𝑑𝑑𝜙𝜙 =𝜙𝜙+ 𝑑𝑑 ln 𝛾𝛾̅
𝕞𝕞𝑏𝑏• It has the same form as nonelectrolyte.
𝕞𝕞
𝜙𝜙 − 1𝑑𝑑𝕞𝕞′ln 𝛾𝛾̅ = 𝜙𝜙 − 1 + �𝕞𝕞′018

192. Electrolytes• Example: water-NaCl solution at 25℃.
𝑑𝑑𝜙𝜙
=∞
lim
𝕞𝕞→0 𝑑𝑑𝕞𝕞
24

1.8

1.15

1.6

1.10

1.4

1.05

1.2

1.00

1.0

0.95

0.8

21

Ideal
20

19Real
0

12

34

0.90

5m

0∞

1𝜙𝜙 = 1 + 𝑐𝑐1 𝕞𝕞 + � 𝑐𝑐𝑖𝑖 𝕞𝕞𝑖𝑖
2𝑖𝑖=123m

450.6γ

22φP1(torr)

23

1.20

2. Electrolytes

20• Debye-Hückel theory can explain the 𝛾𝛾̅ at low
concertation.
• The change of internal energy due to theelectrostatic energy of ions (𝑈𝑈 el ) is calculated,which contributes to the change of 𝜇𝜇2 .
• The mean electrostatic potential at a distance (𝑟𝑟)from an ion (𝑞𝑞) is (considering other ions),𝑞𝑞𝑒𝑒 −𝜅𝜅𝜅𝜅12� 𝐶𝐶𝑖𝑖 𝑞𝑞𝑖𝑖 2𝜅𝜅 =𝜑𝜑 =𝜀𝜀𝑘𝑘B 𝑇𝑇
4𝜋𝜋𝜋𝜋𝜋𝜋𝑖𝑖• 𝜀𝜀 = 𝜀𝜀0 𝜀𝜀𝑟𝑟 is the electric permittivity. 𝑘𝑘B is Boltzmann
constant. 𝐶𝐶 is the number concentration of ion.

212. Electrolytes• An ion is surrounded by an ionic atmosphere of theopposite charges.
• Aqueous 1mM NaCl.
• 𝑇𝑇 = 25℃, 𝜀𝜀r = 78.54.
0.10electrolytevaccumphi (V)

0.08

0.06

0.04

0.02

0.00

02

4

6r (nm)810

222. Electrolytes• The ionic strength is• And,1𝐼𝐼 ≡ � 𝑧𝑧𝑖𝑖 2 𝑐𝑐𝑖𝑖
2𝑖𝑖

𝜅𝜅𝑞𝑞+ 𝑞𝑞−𝑧𝑧+ 𝑧𝑧− e32𝑁𝑁A=−ln 𝛾𝛾̅ =𝐼𝐼38𝜋𝜋𝜋𝜋𝑘𝑘B 𝑇𝑇
8𝜋𝜋𝜀𝜀 3 𝑘𝑘B 𝑇𝑇 3

• Debye-Hückel theory predicts that ln 𝛾𝛾̅ is
proportional to 𝑐𝑐 at low concentrations, typicallyless than 1mM.

2. Electrolytes• For example,• Dissociation degree, size and shape of ion andsolvation should be considered at high 𝑐𝑐.

23

3. Colligative properties

24• Colligative properties are solvent properties which
depend on the number of particles of solute rather
than the types of them.
• The colligative properties are the results of thechange of 𝜇𝜇1 due to the solute.
• For example, the melting (freezing) point (𝑇𝑇m ) of thesolvent changes due to the solute.
• At the melting point,
𝜇𝜇1s = 𝜇𝜇1sln = 𝜇𝜇1l + 𝑅𝑅𝑇𝑇m ln 𝑎𝑎1
𝜇𝜇1s − 𝜇𝜇1l
∆fus 𝐺𝐺ln 𝑎𝑎1 ==−𝑅𝑅𝑇𝑇m
𝑅𝑅𝑇𝑇m

25

3. Colligative properties• Differentiate both side wrt 𝑇𝑇f at constant 𝑃𝑃.
∆fu𝑠𝑠 𝐺𝐺
𝜕𝜕 ln 𝑎𝑎1
1 𝜕𝜕 𝑇𝑇m=−
𝜕𝜕𝑇𝑇m 𝑃𝑃𝜕𝜕𝑇𝑇m𝑅𝑅𝑃𝑃• According to the Gibbs-Helmholtz equation,𝐺𝐺𝜕𝜕 ln 𝑎𝑎1
∆fus 𝐻𝐻𝜕𝜕
𝑇𝑇 = − 𝐻𝐻=>0𝜕𝜕𝜕𝜕 𝑃𝑃𝑇𝑇 2
𝜕𝜕𝑇𝑇m 𝑃𝑃 𝑅𝑅𝑇𝑇m 2• The addition of solute lowers 𝑎𝑎1 , which in turnlowers the melting point.

3. Colligative properties• When 𝕞𝕞 → 0,ln 𝑎𝑎1 ≈ −𝑏𝑏𝑀𝑀1
𝜕𝜕 ln 𝑎𝑎1
𝜕𝜕𝑏𝑏∆fus 𝐻𝐻= −𝑀𝑀1=
𝜕𝜕𝑇𝑇m 𝑃𝑃𝜕𝜕𝑇𝑇m 𝑃𝑃 𝑅𝑅𝑇𝑇m 2
𝜕𝜕𝑇𝑇m𝑏𝑏°𝑀𝑀1 𝑅𝑅𝑇𝑇m 2≡ −𝐾𝐾m=−𝜕𝜕𝕞𝕞 𝑃𝑃∆fus 𝐻𝐻• The change of melting point is proportional to 𝕞𝕞.
∆𝑇𝑇m = −𝐾𝐾m 𝕞𝕞

26

3. Colligative properties• For example, cyclohexane, 𝑀𝑀1 = 84.16g � mol−1 .
• At 𝑃𝑃 = 𝑃𝑃°, 𝑇𝑇m = 279.6K, ∆fus 𝐻𝐻 = 2.68kJ � mol−1 .
• Calculate 𝐾𝐾m .
𝑏𝑏°𝑀𝑀1 𝑅𝑅𝑇𝑇 2= 20.4K
𝐾𝐾m =∆fus 𝐻𝐻• If 0.2mol � kg −1 hexane is added to cyclohexane,∆𝑇𝑇m = −𝐾𝐾m 𝕞𝕞 = −20.4K × 0.2 = −4.08K• The melting point lowers to 275.5K.

27

3. Colligative properties• A similar relation can be derived for boiling point(𝑇𝑇b ).
𝜕𝜕𝑇𝑇b𝑏𝑏°𝑀𝑀1 𝑅𝑅𝑇𝑇b 2≡ 𝐾𝐾b > 0=𝜕𝜕𝕞𝕞 𝑃𝑃∆vap 𝐻𝐻

28• The addition of solute raises the boiling point.
• When 𝕞𝕞 is low, the change of 𝑇𝑇b is proportional to𝕞𝕞.
∆𝑇𝑇b = 𝐾𝐾b 𝕞𝕞• The change of melting and boiling points dependon 𝕞𝕞, but independent of the type of the solute.

29

3. Colligative properties
Boiling pointChemical potential

Melting point

Non-volatile, nonfreeze solute

SolidLiquid/SolventTemperatureVapour

30

3. Colligative properties• 𝜇𝜇1 decreases when solute is added.
• Osmosis: the solvent diffuses form high 𝜇𝜇1 phase tolow 𝜇𝜇1 phase through a semipermeable membrane.
• The diffusion is stopped by the osmotic pressure (Π)
which increases 𝜇𝜇1 to 𝜇𝜇1l .

BAB

AB

A

31

3. Colligative properties• When the 2 sides are in equilibrium,𝜇𝜇1l 𝑃𝑃B + 𝑅𝑅𝑅𝑅 ln 𝑎𝑎1 = 𝜇𝜇1l 𝑃𝑃B − 𝑅𝑅𝑅𝑅𝜙𝜙𝑀𝑀1 𝑏𝑏 = 𝜇𝜇1l 𝑃𝑃A• On the other hand, at constant 𝑇𝑇,
𝜇𝜇1l 𝑃𝑃B − 𝜇𝜇1l 𝑃𝑃A = 𝑉𝑉�1l 𝑃𝑃B − 𝑃𝑃A = 𝑉𝑉�1l Π• Therefore,

𝜙𝜙𝑅𝑅𝑅𝑅𝑀𝑀1 𝑏𝑏
Π=𝑉𝑉�1l𝑑𝑑𝑑𝑑 = −𝑆𝑆𝑆𝑆𝑆𝑆 + 𝑉𝑉𝑉𝑉𝑉𝑉• At low concentration, 𝜙𝜙 ≈ 1, 𝑀𝑀1 𝑏𝑏 ≈ 𝑥𝑥2 .
𝑅𝑅𝑅𝑅
𝑅𝑅𝑅𝑅 𝑛𝑛2𝑛𝑛2Π ≈ l 𝑥𝑥2 ≈ l≈ 𝑅𝑅𝑅𝑅 l ≈ 𝑐𝑐𝑐𝑐𝑐𝑐
𝑉𝑉1
𝑉𝑉�1𝑉𝑉�1 𝑛𝑛1

3. Colligative properties

32• Osmotic pressures can be used to calculate themolar mass of polymers.
• Example: 2.2g of a certain polymer is dissolved inwater to make a 300ml solution. It has an osmoticpressure of 7.45torr at 20℃. Determine the molar
mass of the polymer.
Π103 Pa−3==0.4mol
𝑐𝑐 =�m
𝑅𝑅𝑅𝑅 8.314J � mol−1 � K −1 × 293K𝑚𝑚
2.2g
=
𝑀𝑀 =
=−3−4−3𝑐𝑐𝑐𝑐 0.4mol � m × 3 × 10 m
≈ 1.8 × 104 g � mol−1

Appendix: Debye-Hückel theory

33• First, the electrostatic potential and charge density
are related by the Poisson’s equation.
𝜌𝜌2∇ 𝜑𝜑 = −𝜀𝜀• 𝜑𝜑 is the electrostatic potential.
• 𝜌𝜌 is the charge density.
• 𝜀𝜀 = 𝜀𝜀0 𝜀𝜀𝑟𝑟 is the electric permittivity.
• For an ion at the origin of coordinate, 𝜑𝜑 and 𝜌𝜌 arefunctions of 𝑟𝑟 only (spherical symmetry).
1 𝑑𝑑𝑑𝑑𝜑𝜑
2 𝜑𝜑 =2𝑟𝑟∇
2𝑟𝑟 𝑑𝑑𝑟𝑟
𝑑𝑑𝑟𝑟

Appendix: Debye-Hückel theory34• According to the Boltzmann distribution, the charge
density of the 𝑖𝑖th ions around a charge 𝑞𝑞 is𝑞𝑞 𝜑𝜑− 𝑖𝑖 𝑇𝑇𝜌𝜌𝑖𝑖 = 𝑞𝑞𝑖𝑖 𝐶𝐶𝑖𝑖 𝑒𝑒 𝑘𝑘B• 𝐶𝐶𝑖𝑖 and 𝑞𝑞𝑖𝑖 are the overall number concentration andcharge of the 𝑖𝑖 th ions.
𝑁𝑁𝑖𝑖 𝑁𝑁𝐴𝐴 𝑛𝑛𝑖𝑖== 𝑁𝑁𝐴𝐴 𝑐𝑐𝑖𝑖𝐶𝐶𝑖𝑖 =𝑉𝑉𝑉𝑉• The normalization is already included (check later).
𝑞𝑞 𝜑𝜑− 𝑖𝑖𝜌𝜌 = � 𝑞𝑞𝑖𝑖 𝐶𝐶𝑖𝑖 𝑒𝑒 𝑘𝑘B 𝑇𝑇𝑖𝑖

35Appendix: Debye-Hückel theory• It yields the Poisson-Boltzmann equation.
𝑞𝑞𝑖𝑖 𝜑𝜑
1 𝑑𝑑𝑑𝑑𝜑𝜑1−2𝑘𝑘B 𝑇𝑇=
�𝑟𝑟−
𝑞𝑞𝐶𝐶𝑒𝑒𝑖𝑖𝑖𝑖
𝑟𝑟 2 𝑑𝑑𝑟𝑟
𝑑𝑑𝑟𝑟
𝜀𝜀𝑖𝑖

• It does not have analytical solution.
• Expand the exponential function to the first order.
𝑞𝑞𝑖𝑖 𝜑𝜑𝜌𝜌 = � 𝑞𝑞𝑖𝑖 𝐶𝐶𝑖𝑖 1 −𝑘𝑘B 𝑇𝑇𝑖𝑖

• The solution is charge neutral, so� 𝑞𝑞𝑖𝑖 𝐶𝐶𝑖𝑖 = 0𝑖𝑖36Appendix: Debye-Hückel theory• Define 𝜅𝜅 2 .

𝜑𝜑
� 𝐶𝐶𝑖𝑖 𝑞𝑞𝑖𝑖 2𝜌𝜌 = −𝑘𝑘B 𝑇𝑇𝑖𝑖

𝜌𝜌1
� 𝐶𝐶𝑖𝑖 𝑞𝑞𝑖𝑖 2 = 𝜅𝜅 2 𝜑𝜑− = 𝜑𝜑𝜀𝜀𝜀𝜀𝑘𝑘B 𝑇𝑇𝑖𝑖

• The Poisson equation becomes
1 𝑑𝑑𝑑𝑑𝜑𝜑22=𝜅𝜅𝑟𝑟
𝜑𝜑2𝑟𝑟 𝑑𝑑𝑟𝑟𝑑𝑑𝑟𝑟

Appendix: Debye-Hückel theory• It is a Helmholtz equation. The solutions are𝑒𝑒 −𝜅𝜅𝜅𝜅𝑒𝑒 𝜅𝜅𝜅𝜅
𝜑𝜑 = 𝐴𝐴+ 𝐵𝐵𝑟𝑟
𝑟𝑟
• Where 𝜅𝜅 > 0.
• Consider the boundary condition𝑒𝑒 𝜅𝜅𝜅𝜅lim 𝜑𝜑 = 𝐵𝐵=0𝑟𝑟→∞𝑟𝑟∴ 𝐵𝐵 = 0• The only realizable solution is𝑒𝑒 −𝜅𝜅𝜅𝜅
𝜑𝜑 = 𝐴𝐴𝑟𝑟

37

Appendix: Debye-Hückel theory38• When the concentration approaches 0, 𝜅𝜅 = 0. The
electric potential is from a single charged particle.
𝐴𝐴𝑞𝑞
lim 𝜑𝜑 = =𝑐𝑐→0𝑟𝑟 4𝜋𝜋𝜋𝜋𝜋𝜋• So,𝑞𝑞𝑒𝑒 −𝜅𝜅𝜅𝜅
𝜑𝜑 =4𝜋𝜋𝜋𝜋𝜋𝜋39

Appendix: Debye-Hückel theory• Now, check the normalization.
• The total charge except the origin is+∞𝑞𝑞𝑞 = �04𝜋𝜋𝑟𝑟 2 𝜌𝜌𝜌𝜌𝜌𝜌 = −4𝜋𝜋𝜋𝜋𝜅𝜅 2 �= −𝜅𝜅 2 𝑞𝑞 �• Let 𝜅𝜅𝑟𝑟 = 𝑥𝑥+∞0𝑟𝑟𝑒𝑒 −𝜅𝜅𝜅𝜅 𝑑𝑑𝑑𝑑+∞

𝑞𝑞′ = −𝑞𝑞 �0+∞0𝑟𝑟 2 𝜑𝜑𝑑𝑑𝑑𝑑

𝑥𝑥𝑒𝑒 −𝑥𝑥 𝑑𝑑𝑥𝑥 = −𝑞𝑞Γ 2 = −𝑞𝑞

40

Appendix: Debye-Hückel theory

distance, which is thetypical size of the ionicatmosphere.
• The charge in a thinshell at 𝑟𝑟 of thickness
𝑑𝑑𝑑𝑑 is𝑑𝑑𝑞𝑞𝑞 = 4𝜋𝜋𝑟𝑟 2 𝑑𝑑𝑑𝑑𝜌𝜌𝑑𝑑𝑞𝑞𝑞= −𝑞𝑞𝜅𝜅 2 𝑒𝑒 −𝜅𝜅𝜅𝜅 𝑟𝑟
𝑑𝑑𝑑𝑑1𝑟𝑟 =𝜅𝜅500000400000-dQ/dr (e/m)•1
has the unit of𝜅𝜅

300000200000100000002040r (nm)

6080

41

Appendix: Debye-Hückel theory• The total electrostatic energy between the central
ion and all the other ions is+∞+∞𝑞𝑞�
𝜌𝜌4𝜋𝜋𝑟𝑟 2 𝑑𝑑𝑑𝑑 = −𝑞𝑞 � 𝑟𝑟𝜅𝜅 2 𝜑𝜑𝑑𝑑𝑑𝑑4𝜋𝜋𝜋𝜋𝜋𝜋00
𝜅𝜅 2 𝑞𝑞2 +∞ −𝜅𝜅𝜅𝜅𝜅𝜅𝑞𝑞2
� 𝑒𝑒 𝑑𝑑𝑑𝑑 = −=−4𝜋𝜋𝜋𝜋 04𝜋𝜋𝜋𝜋• The total electrostatic internal energy of all the ionsis231𝜅𝜅𝑞𝑞
𝜅𝜅
𝑉𝑉𝑘𝑘
𝑇𝑇𝜅𝜅
𝑖𝑖
B=−
� 𝐶𝐶𝑖𝑖 𝑉𝑉𝑞𝑞𝑖𝑖 2 = −𝑈𝑈 el = − � 𝑁𝑁𝑖𝑖
4𝜋𝜋𝜋𝜋8𝜋𝜋28𝜋𝜋𝜋𝜋𝑖𝑖

𝑖𝑖

Appendix: Debye-Hückel theory• Consider the correction of 𝜇𝜇 due to electrostaticenergy.
elelel𝜕𝜕𝐴𝐴𝜕𝜕𝑈𝑈
𝜕𝜕𝑆𝑆=− 𝑇𝑇𝜇𝜇𝑖𝑖el =𝜕𝜕𝑛𝑛𝑖𝑖 𝑇𝑇,𝑉𝑉𝜕𝜕𝑛𝑛𝑖𝑖 𝑇𝑇,𝑉𝑉𝜕𝜕𝑛𝑛𝑖𝑖 𝑇𝑇,𝑉𝑉• On one hand,
𝜕𝜕𝑈𝑈 el
𝑉𝑉𝑘𝑘B 𝑇𝑇 𝑑𝑑𝜅𝜅 3 𝜕𝜕𝜅𝜅 2=−
𝜕𝜕𝑛𝑛𝑖𝑖 𝑇𝑇,𝑉𝑉
8𝜋𝜋 𝑑𝑑𝜅𝜅 2 𝜕𝜕𝑛𝑛𝑖𝑖 𝑇𝑇,𝑉𝑉𝑉𝑉𝑘𝑘B 𝑇𝑇 3
1 𝑁𝑁A 𝑞𝑞𝑖𝑖 23𝑁𝑁A 𝜅𝜅𝑞𝑞𝑖𝑖 2=−𝜅𝜅=−
16𝜋𝜋𝜋𝜋8𝜋𝜋 2 𝜀𝜀𝑘𝑘B 𝑇𝑇 𝑉𝑉

42

43

Appendix: Debye-Hückel theory• On the other hand,𝜕𝜕𝑆𝑆 el𝜕𝜕 𝜕𝜕𝐴𝐴el=−𝜕𝜕𝑛𝑛𝑖𝑖 𝑇𝑇,𝑉𝑉𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑇𝑇 𝑛𝑛 ,𝑉𝑉𝜕𝜕 𝜕𝜕𝐴𝐴el=−𝜕𝜕𝑇𝑇 𝜕𝜕𝑛𝑛𝑖𝑖 𝑇𝑇,𝑉𝑉• Therefore,𝑛𝑛𝑖𝑖 ,𝑉𝑉=−𝑖𝑖𝑇𝑇,𝑉𝑉𝜕𝜕𝜇𝜇𝑗𝑗el𝜕𝜕𝑇𝑇𝑛𝑛𝑖𝑖 ,𝑉𝑉el23𝑁𝑁
𝜅𝜅𝑞𝑞𝜕𝜕𝜇𝜇𝐴𝐴𝑖𝑖𝑖𝑖el+ 𝑇𝑇𝜇𝜇𝑖𝑖 = −16𝜋𝜋𝜋𝜋𝜕𝜕𝑇𝑇 𝑛𝑛 ,𝑉𝑉𝑖𝑖

Appendix: Debye-Hückel theory• It is an OED of the form:
𝑑𝑑𝑑𝑑 𝑦𝑦− + 𝑓𝑓 𝑥𝑥 = 0
𝑑𝑑𝑑𝑑 𝑥𝑥• Where 𝜇𝜇𝑖𝑖el → 𝑦𝑦, 𝑇𝑇 → 𝑥𝑥. The solution is𝑓𝑓 𝑥𝑥
𝑑𝑑𝑥𝑥
𝑦𝑦 = 𝑥𝑥 �𝑥𝑥

• So,

44

1
∑𝑖𝑖 𝐶𝐶𝑖𝑖 𝑞𝑞𝑖𝑖 2223𝑁𝑁A 𝜅𝜅𝑞𝑞𝑖𝑖
3𝑁𝑁A 𝑞𝑞𝑖𝑖
𝜀𝜀𝑘𝑘B 𝑇𝑇′el𝑇𝑇 �𝜇𝜇𝑖𝑖 = 𝑇𝑇 �𝑑𝑑𝑇𝑇 =𝑑𝑑𝑇𝑇′2216𝜋𝜋𝜋𝜋𝑇𝑇16𝜋𝜋𝜋𝜋𝑇𝑇′

45

Appendix: Debye-Hückel theory253𝑁𝑁𝑞𝑞
1𝐴𝐴𝑖𝑖−𝑇𝑇𝜇𝜇𝑖𝑖el =� 𝐶𝐶𝑖𝑖 𝑞𝑞𝑖𝑖 2 � 𝑇𝑇′ 2 𝑑𝑑𝑇𝑇′
16𝜋𝜋𝜋𝜋𝜀𝜀𝑘𝑘B𝑖𝑖

33𝑁𝑁𝐴𝐴 𝑞𝑞𝑖𝑖 212−𝑇𝑇=
� 𝐶𝐶𝑖𝑖 𝑞𝑞𝑖𝑖 2 − 𝑇𝑇 216𝜋𝜋𝜋𝜋𝜀𝜀𝑘𝑘B
3𝑖𝑖2 𝜅𝜅𝑁𝑁𝐴𝐴 𝑞𝑞𝑖𝑖 21𝑁𝑁𝑞𝑞𝐴𝐴 𝑖𝑖
2
� 𝐶𝐶𝑖𝑖 𝑞𝑞𝑖𝑖 = −=−8𝜋𝜋𝜋𝜋 𝜀𝜀𝑘𝑘B 𝑇𝑇
8𝜋𝜋𝜋𝜋𝑖𝑖• Suppose electrostatic energy is the only correction
to chemical potential.

46Appendix: Debye-Hückel theory

𝜇𝜇𝑖𝑖el = 𝑅𝑅𝑅𝑅 ln 𝛾𝛾𝑖𝑖
𝜈𝜈+ ln 𝛾𝛾+ + 𝜈𝜈− ln 𝛾𝛾−𝜅𝜅𝑁𝑁A 𝜈𝜈+ 𝑞𝑞+ 2 + 𝜈𝜈− 𝑞𝑞− 2=−ln 𝛾𝛾̅ =𝜈𝜈
8𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜈𝜈• Use 𝜈𝜈+ 𝑞𝑞+ + 𝜈𝜈− 𝑞𝑞− = 0。
𝜅𝜅−𝜈𝜈− 𝑞𝑞− 𝑞𝑞+ − 𝜈𝜈+ 𝑞𝑞+ 𝑞𝑞−
𝜅𝜅𝑞𝑞+ 𝑞𝑞−
=ln 𝛾𝛾̅ = −𝜈𝜈
8𝜋𝜋𝜋𝜋𝑘𝑘B 𝑇𝑇
8𝜋𝜋𝜋𝜋𝑘𝑘B 𝑇𝑇• Define the ionic strength.
1𝐼𝐼 ≡ � 𝑧𝑧𝑖𝑖 2 𝑐𝑐𝑖𝑖
2𝑖𝑖

47

Appendix: Debye-Hückel theory𝜅𝜅 2 =112
� 𝐶𝐶𝑖𝑖 𝑞𝑞𝑖𝑖 =� 𝑁𝑁A 𝑐𝑐𝑖𝑖 𝑧𝑧𝑖𝑖 e 2𝜀𝜀𝑘𝑘B 𝑇𝑇
𝜀𝜀𝑘𝑘B 𝑇𝑇𝑖𝑖

21𝑁𝑁eA2
� 𝑧𝑧𝑖𝑖 𝑐𝑐𝑖𝑖 = 2𝐼𝐼=
𝜀𝜀𝑘𝑘B 𝑇𝑇
𝜀𝜀𝑘𝑘B 𝑇𝑇

• Finally,𝑖𝑖𝑖𝑖

𝜅𝜅𝑞𝑞+ 𝑞𝑞−𝑧𝑧+ 𝑧𝑧− e32𝑁𝑁A=−ln 𝛾𝛾̅ =𝐼𝐼38𝜋𝜋𝜋𝜋𝑘𝑘B 𝑇𝑇
8𝜋𝜋𝜀𝜀 3 𝑘𝑘B 𝑇𝑇 3