Inorganic Chemistry

无机化学
Ming GUO
郭明

1

 Ming GUO
Tel：+86-13634183205

Email：guoming@zafu.edu.cn

 2 2
 Learning and Examination
Learning methods: Preview - review -
summary.
Examination mode: Closed book
independent examination.
Total score = 40% of the usual results
(experiments + homework + classroom
exercises + performance) + 60% of the final
exam score.
Homework: Regular release on Xuexi Tong
3
APP. 

3 3
Learning requirements - Observe, Listen,
Remember, Practice

Observe --- Read and review the PPT courseware

carefully;
Listen --- Pay attention in class: listening to the teacher’s

explanation has a twice-as-effective result as self-study;

Remember --- Take notes as much as possible: ensure

there is a reference to consult;

 Practice --- Homework: consolidate the results of listening
in class - Very important!!

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
44
 Chapter 1
Dispersed Systems

 Solutions

 Colligative properties
 5
 Learning objectives

1. Understand the classification of dispersed

systems.

2. Master the definition of solution concentration

and their mutual conversion.

3. Understand the colligative properties of dilute

solutions.
 6
1.1 Classification of Dispersed Systems

1. Definition:
A dispersed system is a system formed when one or
several substances are dispersed in another
substance.

dispersed phase
The substance that is being dispersed

dispersion medium
The substance that causes the dispersion
 7
Dispersion systems

 8
 2. Classification
1) By the state of aggregation
Dispersed Dispersing Examples
substance medium
Gas Gas air
liquid Gas cloud, fog
solid Gas smoke, dust
Gas liquid soda, foam
liquid liquid milk
solid liquid mud, solution
Gas solid foam plastic, bread
liquid solid pearl, gelatin
solid solid alloy, colored glass  9
2) By the diameter of the dispersed particles

 10
 1.2 Solutions
1. Solutions
A dispersed system in which the solute is
evenly dispersed in the solvent as small
molecules, ions, or atoms.
Amount of substance and its unit：
1) Amount of substance n:
A physical quantity that represents the
number of basic units of a substance.
2) Unit: mol
 11
3) Basic unit: The fundamental components of a
substance in a system, which can be molecules,
atoms, ions, electrons, or other specific
combinations of particles.

4) Molar mass ：MB = mB /nB

 12
  molar mass calculation
Example: Calculate the molar mass of carbon
dioxide (CO₂).
 Find the relative atomic masses: The relative atomic
mass of carbon (C) is approximately 12, and the relative
atomic mass of oxygen (O) is approximately 16.
Calculate the relative molecular mass: Relative
molecular mass = (1 × the relative atomic mass of carbon)
+ (2 × the relative atomic mass of oxygen) = 12 + (2 ×
16) = 12 + 32 = 44.
Determine the molar mass: The molar mass of carbon
dioxide is 44 g/mol. 13 
 Calculate the molar mass

 Exercise 1: sodium hydroxide (NaOH)

MNa=23 g/mol, MO=16 g/mol, MH=1 g/mol

MNaOH = 23 + 16 + 1 = 40 g/mol

 Exercise 2: copper(II) sulfate (CuSO₄)

MCu=63.5 g/mol, MS=32 g/mol, MO=16 g/mol

MCuSO4 = 63.5 + 32 + (4 ×16) = 159.5 g/mol

14

2. Methods for expressing solution concentration

1) Molar concentration c(B)

① Definition: The amount of substance of solute

B per unit volume of solution

② Formula:

③ Unit: mol·L-1 (mol·dm-3)

 15
Example
Dissolve 4.6 g of ethanol in water to prepare
1.0 L of solution. Calculate the molar
concentration of ethanol.
Given: M(ethanol)=46 g/mol


c= n/V= (m/M)/V
=(4.6/46)/1.000
=0.1 mol/L
16
 2) Mole fraction xi

① Definition:
The ratio of the amount of substance of component
B to the total amount of substance of the solution
② Formula:

③ Dimension： 1
 17
Example
Dissolve 4.6 g of ethanol in 180 g of
water to prepare a solution. Calculate
the mole fraction of ethanol.
 Given: M(ethanol)=46 g/mol, M(H2O)=18 g/mol

x = n(ethanol)/[(n(ethanol)+n(H2O)]
= (4.6/46)/[(4.6/46)+(180/18)]
= 0.01
 18
 3) Mass Molarity b(B)
① Definition: The amount of substance of solute B
per unit mass of solvent
n( B )
② Formula: b( B ) 
m( A)
③ Unit: mol·kg-1
④ Feature: Independent of temperature;
very dilute aqueous solutions
c(B) ≈ b(B)  19
Example : 17.1g of sucrose is dissolved in 500g
of water. Calculate the mass molarity of the
sucrose solution. Msucrose=342 g/mol.

Answer:

 20
 4) Mass fraction ω
① Definition : The ratio of the mass of solute B
to the mass of the solution ，ω

② Formula : w (B) m(B) / m(total) *100%

③ Dimension : 1

 21
Example
Dissolve 4.6 g of ethanol in 180 g of
water to prepare a solution. Calculate
the mass fraction of ethanol.

w = m(ethanol)/m(total)
=4.6/(4.6+180)
=2.5%
22

6) Conversion between concentrations

Example 3: A 48% sulfuric acid solution

has a density of 1.38 g·mL-1. Calculate the
following for this solution: (1) Molar
concentration; (2) Mass molarity; (3) Mole
fraction. MH2SO4=98 g/mol

 23
Answer
(1)

 24
(2)

(3)

 25
1.3Numerical dependence of dilute solutions

1. Colligative Property：
Relates only to the number of solute
particles, not to the nature of the solute.

1) vapor pressure lowering

2) Boiling point elevation
3) Freezing point lowering
4) Osmotic pressure
26

1. Decrease in the vapor pressure of the solution
vapor pressure
At a certain temperature, the pressure at which
the liquid phase (solid phase) is in equilibrium with its
gas phase is called the saturated vapor pressure of the
liquid (solid) at that temperature, or vapor pressure
for short. vapor molecule
liquid molecule

vaporization

liquid vapor
coagulate

27
Reasons for a decrease in the vapor pressure of a solution

After a solute that is difficult to volatilize is dissolved in a solvent,

the number of solvent molecules on the surface of the solution
decreases due to the presence of solute plasmas, so that the number
of solvent molecules evaporated from the solution at the same
temperature is less than that of the pure solvent, the vapor pressure
of the solution is lower than the vapor pressure of the pure
vapor molecule
solvent. liquid molecule
solute molecule

28 2024/11/13
 Drying capacity of common desiccant
P2O5 > KOH > 浓H2SO4 > NaOH > CaCl2

Strong water absorption makes it easy to deliquesce

into saturated aqueous solution in the air, because its
vapor pressure is lower than that of water vapor in the
air, thus causing water vapor in the air to continuously
condense into solution.

29 2024/11/13
 Quantitative Relationship of Vapor Pressure Decrease
Raoult's law：
At a given temperature,the vapor pressure of a dilute solution of a non-volatile
non-electrolyte is equal to the product of the vapor pressure of the pure solvent
and the mole fraction of solvent in solution.

p = pA*· xA
purified water

30 2024/11/13
 xA + xB = 1
p = pA*· xA

Δp = pA*- p =pA*- pA* · (1-xB) = pA* · xB

Another formulation of Raoult's law is that the decrease in

vapor pressure of a dilute solution of a non-volatile
non-electrolyte at a given temperature is directly proportional
to the substance fraction of the solute.
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When the solution is very dilute：nA 》nB，XB = nB/nA+nB ≈nB/nA

Raoult's lawcan also be stated as follows: the decrease in

vapor pressure of a dilute solution of a difficult-to-volatilize
non-electrolyte is approximately proportional to the mass
molar concentration of the solution at a given temperature,
independent of the nature of the solute.
2. The boiling point of the solution increases

101. 325
a’ b’

liquid

Tf  Tf*  Tf  Kf bB
solution
a Tb  Tb  Tb*  K b bB
0. 611
gas
solid b ΔT
c ΔT f b

t凝 0 100 t沸
t/℃
Variation of the vapor pressure of water, ice and aqueous
33
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solutions with temperature
2. The boiling point of the solution increases
purified water
p
fructose
P外 The boiling point of a
liquid is the temperature
at which the vapor
pressure of the liquid is
equal to the outside
atmospheric pressure.

Tb * Tb T

 34 2024/11/13
 Tb  Tb  Tb*  K b bB

ΔTb：The boiling point of the solution increases

Tb*, Tb ：Boiling points of pure solvents and solutions
bB ：Mass molar concentration in mol·kg–1

Kb : Molar boiling point elevation constant, different

solvents have different Kb values in K·kg·mol-1
Kb :Depends on the properties of the pure solvent
and not on the solute properties
35 2024/11/13
 3)Application of boiling point elevation
(1) Calculate the boiling point of a solution
For example 3. Dissolve 4.56 g of urea{ CO(NH2)2 }in 100
g of water and calculate the boiling point of this solution at
101.3 kPa.

373.15+ 373.54
36
 2024/11/13
(2) Determination of the molar mass of non-
volatile non-electrolytes
For example 4. Dissolve 0.40 g of glucose) C6H12O6 (in
20.0 g of water, the boiling point of the solution is
measured to be 100.056℃ ,calculate the molar mass
of glucose.
0.40/M glucose C6H12O6

20/1000
Tb = 0.056 = Kb· b glucose C6H12O6

M glucose C6H12O6 = 182.9 g·mol-1

37 2024/11/13
 3. Decrease in the freezing point of the solution

p purified water The temperature

at which the vapor
fructose
ice pressure of a liquid
0.611 kPa is equal to the
vapor pressure of a
solid and the two
phases coexist in
Tf T*
f = 273.15 K T equilibrium -
freezing point

38
 2024/11/13
1)Quantitative relationship of freezing point depression

Tf  Tf*  Tf  K f bB

ΔTf : Decrease in the freezing point of the solution

Tf*, Tf : Freezing points of pure solvents and solutions
bB : Mass molar concentration in mol·kg–1
Kf : Molar freezing point depression constant in K·kg·mol-1

Kf ：Depends on the properties of the pure solvent

and not on the solute properties
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 2) Causes of increase in boiling point and
decrease in freezing point:

Decrease in the vapor pressure of the solution

3) Application of freezing point depression:：

(1)Explain phenomena such as drought resistance and
cold tolerance in plants;
(2)Calculate the freezing point of a solution;
(3)Determination of the molar mass of difficult-to-
volatilize non-electrolytes;
40 2024/11/13
Eg5. Dissolve 0.40 g of glucose in 20.0 g of water and
measure the freezing point of the solution to be - 0.207℃ .
Calculate the molar mass of glucose.
Answer:

41
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1. Decrease in the vapor pressure of the solution

2. The boiling point of the solution increases

Tb  Tb  T  K bbB
b
*

3. Decrease in the freezing point of the solution

Note: 1) electrolyte
2)hard evaporate
3)dilute solution
42
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4. Osmotic pressure of solutions
Osmotic phenomenon
(osmosis): the process of
automatic d iffusion o f
ПV= nB×R×T solvent molecules into
solution through a
semipermeable membrane.
solution purified
water

semi-permeable membrane

Simple device showing the phenomenon of osmotic pressure

43
 2024/11/13
 Conditions for osmosis

①Presence of a semi-permeable membrane

② There is a concentration difference between the

solutions on both sides of the membrane.

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44

Hypertonic, hypotonic, isotonic solutions
Cells of plants

45
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45
Quantitative relationship of osmotic pressure

n
Π  cRT  RT
V

or ΠV  nRT

Note: unit
c：mol/m3 п：Pa
c：mol/L п：kPa
46 2024/11/13
46
 Application of osmolality
① Explain the phenomena of life in plants and
animals
② Calculate the molar mass of a polymer
compound

47 2024/11/13
47
 Eg6. An aqueous solution containing 5g·L-1of a
soluble polysaccharide has an osmotic pressure of
3.24 kPa at 278 K. Find the molar mass of the
polysaccharide.
Answer: n
Π  cRT  RT
V

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48

 Thinking Questions：
1.How does a tree get water from the soil？
2.Can plants burn if you fertilize too much？
3.Can a freshwater fish swim into saltwater and live？

2024/11/13
49
 Reverse osmosis
If an external pressure greater than the osmotic pressure is applied to
one side of the solution, the solvent permeates from the solution side
through the semipermeable membrane in the direction of the pure
solvent or low concentration, a phenomenon known as reverse osmosis.
Applications: desalination and wastewater treatment

solution purified
water
50 2024/11/13
50
 Numerical dependence of dilute solutions

Lower vapor pressure

Depends on the number of solute
Decrease in freezing point
primes
Increased boiling point
Dependent on the number of
osmotic pressure
solute particles

dilute solution Number of particles of solute

value of dependent variable

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51

Arrange the freezing point( Tf )and boiling point
(Tb )of the following solutions in descending order
1) 0.001mol/L C6H12O6 solution
2) 0.001mol/L sucrose solution
3) 0.001mol/L Na2SO4 solution
4) 0.001mol/L NaCl solution
5) 0.001mol/L HAc solution
6) 0.01mol/L C6H12O6 solution

Tf: 1) = 2) > 5) > 4) > 3) > 6)

Tb: 6) > 3) > 4) > 5) > 1) = 2)
2024/11/13
52

 At larger solution concentrations, the qualitative relationships
remain, but the quantitative relationships are complex.

It is generally believed that the ionization of a strong

electrolyte is 100%, but it has been experimentally demonstrated
that the ionization of a strong electrolyte in solution is less than
100%, and this ionization is called the apparent ionization of the
strong electrolyte.

— — +
+ — +
— +
+ — + — +
+ —
— + — +
—
+ —
53 2024/11/13
Ionic Atmosphere Schematic 
 Theory of ionic mutual attraction in strong
electrolyte solutions
ionic atmosphere
This is an important concept in the
Debye-Shocker theory.They believed
that in a solution, every ion is
surrounded by counter sign ions,
which are unevenly distributed due
to the interaction of positive and
negative ions.
If the center ion is positive and surrounded by more negative ions,
some of the charges cancel each other out, but the remaining
r
charges form a spherical negative ion atmosphere at a distance from
the center ion; and vice versa.An ion can be both the center ion and
a member of another ion atmosphere.
54 2024/11/13
 Debye-Shocker's limiting law
Based on the concept of ionic atmosphere and introducing a
number of assumptions, Debye-Shocker derived a formula for
calculating the ionic activity coefficient i in dilute solutions of
strong electrolytes, known as Debye-Shocker's limit law.

lg i   Azi I 2

where zi is the charge of the i ion , I is the i onic

strength, and A is a constant related to temperature.

Since the activity coefficients of individual ions

cannot be verified by experimental measurements, this
formula is of little use. 55  2024/11/13
 Debye-Shocker's limiting law
A common representation of the Debye-
Shocker's limiting law :
lg     A | z z | I
This formula applies only to dilute solutions of strong
electrolytes, systems where ions can be handled as point
charges.where   is the average ionic activity coefficient
obtained from this formula ,   is the theoretically calculated
value.With the electric potential method,   can be determined
by the experimental value, used to test the scope of application
of the theoretical calculations.
56 2024/11/13
 For systems with large ionic radii that cannot be
treated as point charges, the Debye-Huckle limit law
equation is modified to:
 A | z z | I
lg   
1  aB I

where a is the average effective diameter of the ion ,

3.5 1010 m approximately , and B is a temperature-
and solvent-dependent constant, in an aqueous
solution at 298 K.
1 1 1
B  0.33 10 (mol  kg) m aB  1(mol  kg ) 2
10 1
2 -1

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1.4 Colloidal Solution
1.Basic Characteristics of Sols
1) Sols are multi-phase systems. S0 = A / V
2) High dispersion, large specific surface.
3)Unstable,easy to agglomerate and sink.
L=1cm n=1 S=6cm-2
L=0.5cm n=8 S=12cm-2
L=0.1cm n=1000 S=60cm-2
L=100nm=10-5cm n=1015 S=6·105cm-2
L=1nm =10-7cm n=1021 S=6·107cm-2
58 2024/11/13
2.Surface properties
1) High surface energy
The surface particles
Schematic diagram
have more energy than
of solid interface
the internal particles,
and the higher energy
is called surface energy.

total surface area

Specific surface energy
(surface tension)
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 Reduced surface energy
(Stabilize the system)

Reduce S0 Reduced surface tension

Small particles Saturation of residual

merge into large gravity by adsorption
particles
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 2024/11/13
2) Absorbent

Adsorption
The process by which one substance automatically
aggregates at the interface of another substance.

① molecular adsorption
Adsorption of solid adsorbents in non-electrolyte or
weakly electrolyte solutions.

Characteristics: Similarity attracts

61 2024/11/13
 ② Ion adsorption
Adsorption of ions by adsorbents in strong
electrolyte solutions.
A. Ion selective adsorption
Law: Preferential adsorption of ions related to composition
Characteristics: Positive absorption with positive,
negative absorption with negative.
eg AgNO3+KBr==AgBr(collosol)+KNO3
B. ion exchange adsorption
The exchange of ions between the adsorbent and the
solution is called ion exchange adsorption.
62 2024/11/13
Eg: Clay particles are negatively charged, with
exchangeable cations Ca2+(Mg2+.K+.Na+)on the
surface, when ammonium nitrogen fertilizer is
applied ,NH4+ is exchanged with Ca2+ on the
surface of the colloidal particles：

 
2 NH4 NH4 2
Ca   Ca
NH4 NH4

Characteristics: Reversible process

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 2024/11/13
 2.Properties of sols
1)Optical properties-Tyndall phenomenon（solute-specific）
A beam of light is shone onto a transparent sol, and a
bright column of light is observed in the vertical
direction of the light.
2) Kinetic Properties - Brown Motion
Brown movement is the dispersant molecules with different
sizes and different directions of force on the colloidal particles
constantly impact and produce, due to the imbalance of the
force, so continuously in different directions, different speeds
for irregular movement.
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 3) Electrical property
①Electrophoresis - Directional movement of a dispersant
in a dispersant under an external electric field.
- +

clay
Negative: The water rises and clears.
Positive: water drops, cloudy, clay particles are
negatively charged. 65
 2024/11/13
②Electroosmosis - The solid phase is immobile
and the dispersant moves directionally under an
external electric field.

3.Reasons for sol-gel particles to be electrically

charged
Adsorption Charged - selectively adsorbs ions
related to their composition.

66 2024/11/13
Eg1.Iron hydroxide sol is made by hydrolysis
of ferric chloride in boiling water.
FeCl 3  3H 2O
FeCl3+3H2O== Fe(OH)3+3HCl
Fe(OH)3+ HCl==FeOCl+2H2O
FeOCl==FeO++Cl-
FeO+is adsorbed on the surface of Fe(OH)3 colloidal
particles making them positively charged.

67
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Eg2. Arsenic sulfide sols are prepared by passing
hydrogen sulfide gas through a saturated arsenic
acid solution.
Answer
2H3AsO3+3H2S=As2S3+6H2O
H2S=H++HS-
Arsenic sulfide sols are negatively charged
by preferentially adsorbing HS-.

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Dissociative Charging - The dissociation of the
sol surface, which charges the gel particles.

Eg :Silica gel dissociation：

HSiO3-,SiO32-remain on the surface of

the gelatinous particles to make them
negatively charged.
69 2024/11/13
 4. colloidal structure
diffusion layer colloidal particles
Cl-
Cl-
Cl-
Cl-
FeO+ FeO FeO+
+
Cl-
FeO+ [Fe(OH) ] FeO+
Cl- 3 m Cl -
FeO+ FeO+
Cl -

FeO FeO+
+ FeO +

Cl -
Cl- colloid
Cl -
Cl -
adsorbent layer
 70 nucleus 2024/11/13
Eg1. An As2s3 sol is prepared from H2S
and H3AsO3 with an excess of H2S.Write
the structure of the colloid.
Answer：

71 2024/11/13
Eg 2. A sol is prepared from the reaction between
BaCl2and k2SO4solution, if the k2SO4solution is in excess,
write the structural formula of the colloid of the sol.

Answer：
BaCl2+K2SO4=BaSO4 溶胶+2KCl
[( BaSO4 ) m · n SO42- · 2(n-x)K+]2x- ·2xK+
or [( BaSO4 ) m · n SO42- · (2n-x)K+]x-·xK+
The electrolyte k2SO4 is the stabilizer of the sol.

72 2024/11/13
Eg 3. To prepare a negatively charged AgIsol,
how much 0.005mol.L-1AgNO3solution should
be added to 25mL 0.016mol.L-1KI solution?
Answer:AgI sol is negatively charged, then
KI is in excess of：

That is, by adding less than80mL of 0.005mol.L-

1 AgNO , the structural formula of the colloid is
3

73 2024/11/13
5 . Stabilization and coagulation of sols

1)Causes of sol-gel stabilization

Brown Campaign
Repulsion of the same charge
Solvation

74
 2024/11/13
 2)Methods of sol-gel coagulation

Factors in sol-gel coagulation

The sol is heated Interconden Addition of

for a long time sation of sols electrolyte
(greatest effect on
solutes)
75
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 ① coagulation value

The minimum concentration at which the sol starts

to coagulate within a certain period of time.
Unit: mmol
The smaller the coagulation value, the greater the
coagulation ability.
The coagulation value of NaCl on As2S3 sol is 51
The polymerization value of MgCl2 on As2S3 sol is
.0.72
The polymerization value of AlCl3 for As2S3 sol is
.0.093 76
 2024/11/13
Factors affecting the magnitude of the
condensation value
A、The higher the valence, the lower the
coagulation value.The higher the value, the lower
the coagulation value.
For example, for negative solvation：Al3+>Mg2+>K+

For orthosols：PO43->SO42->Cl-
B.The larger the hydration ion radius, the higher the
coagulation value and the lower the aggregation and
sinking capacity.
77 2024/11/13
Example:Ag2CrO4sol was prepared by mixing
equal volumes of 0.008 mol.L-1 AgNO3 solution
and 0.005 mol.L-1 K2CrO4 solution.Write the
structural formula of the colloid of this
sol.There are three electrolytes MgSO4 ,
K3[Fe(CN)6] , [Co(NH3)6]Cl3, what kind of
ions are involved in the coagulation of this
sol?
What is the order of coagulation value of the
three electrolytes on the sol?
78 2024/11/13
解:

79
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 1.5 Emulsion
1.surface active reagent
1)Definition：
Substances that significantly reduce the specific
surface energy of water are called surface-active
reagent.
2) Structural features: amphiphilic
molecule
Polar portion, hydrophilic group
Nonpolar fraction, hydrophobic clusters

3)Functions: Wetting, emulsifying, etc.

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4) Reasons for lowering the specific
surface energy of water

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 82