医学基础化学（ 1 ）

刘洛生

82378sheng7ed@126.com

CHAPTER 1 Solution ( 4h) 2 Electrolyte solution (5h) 3 Buffer solution (4h) 5 Rates of chemical reaction(4h) 6 Electrochemistry (6h) 10 Titrimetry (4h) 11 Spectrophotometry(2h) 12 The colloidal system and surface phenomena (5h) 7 Atomic Structure (4h) 8 Molecular Structure (4h) 9 Coordination Compounds (6h)

课本（英文、中文 共 48 学时）

顺序

• 先中文，再英文 记忆英文的词汇和相关定义

• 正确理解基本概念、基本原理及适用范围

抽象→联系实际，勤思考，理解，推理。

预习自学 记笔记，及时复习、消化、答疑。

重视作业 适当做习题。

每章要求的作业： summary （上交）

学习方法：

CHAPTER ONE

SOLUTION

1-1 Basic Terms of Solution

1-2 Solubility of Solution

1-3 Solution Concentration

1-4 Colligative Properties of Solutions

问题： 生理盐水？ 等渗溶液？ 肾脏衰竭的治疗方法——血液透析原理？

溶液浓度、渗透及渗透压概念有关

（ 5%GS 、 50%GS ）

（ 0.9%NaCl ）

1-1      Basic Terms of Solution

• A solution is a homogeneous mixture of two or more substances evenly distributed in each other.

• When one of these substances is a liquid and the others are so1ids or gases, we tend to think of the liquid as the substance that dissolves the solid or gas. The liquid is referred to as the solvent and the solids and gases as the solutes.

• Solvent is the substance in a solution that dissolves another substance; it is the substance in the solution that is in greater amount.

• Solute is either a substance in a solution considered to have been dissolved by a solvent, or else a substance in a solution that is in smaller amount.

• Two liquids that mix completely in one another to form a solution, no matter what the proportions of liquids, are said to be miscible.

• Two liquids that do not mix together or dissolve in one another in significant amounts, but rather tend to separate into two distinct layers, are said to be immiscible .

Water – alcohol

Water – oil

Solution Process • What happens when a substance like NaCl

dissolves in water?

When NaCl is added to water, the water molecules orient themselves on the surface of the NaCl crystals.

The positive end of the water dipole is oriented towards the Cl- ions, while the negative end of the water dipole is oriented toward the Na+ ions.

• The dipole-ion( 偶极离子 ) attractions between water molecules and Na+ and Cl- ions are sufficiently strong to pull these ions from their positions in the crystal.

• Once removed from the crystal, the Na+ and Cl- ions are surrounded by water molecules.

• Such interactions between solute and solvent molecules is known as solvation( 溶剂化 ).

• When the solvent is water it is known as hydration. （水合）

1-2 Solubility of Solution• Solubility is the maximum amount of substance

that dissolves in a given volume of solvent at a specified temperature.

• Suppose you add 60 g of sodium thiosulfate to 100 ml of water contained in a flask at 20℃.

• After stirring vigorously, you find that 50g of the crystalline compound dissolves, leaving 10g at the bottom of the flask.

• Thus, the solubility of sodium thiosulfate at 20℃ is 50 g in 100 ml of water.

 The solution you obtain after you have dissolved the maximum amount of solute in a solvent at a given temperature is said to be a saturated solution.

• Thus, the solution of 100ml water containing 50g of sodium thiosulfate at 20℃ is a saturated solution of sodium thiosulfate.

• Any solution that contains less solute in a given volume of solvent than is contained in the saturated solution is said to be an unsaturated solution.

• supersaturated solution, which is a solution that contains more of a solute than is contained in the saturated solution.

• How do you know whether a solution is saturated or supersaturated? (P5)

Factors That Affect Solubility

1. 1. Nature of Solute / SolventNature of Solute / Solvent.

2. Temperature2. Temperature - -i) Solids/Liquids: Solubility increases with temperature

increase .

ii) Gas : Solubility decreases with temperature increase .

3. Pressure Factor3. Pressure Factor - -i) Solids/Liquids : Very little effect

Solids and Liquids are already close together, extra pressure will not increase solubility.

ii) Gas: Solubility increases with Pressure.

Solubilities of Solids vs Temperature

MOST salts have greater solubility in hot water.

A few salts have negative heat of solution, (exothermic( 放热 ) process) and they become less soluble with increasing temperature.

Temperature & the Solubility of GasesThe solubility of gases DECREASES at higher temperatures

1-3 Solution Concentration

1-3.1 Amount-of-substance and Molar Mass

1-3.2 Mass Percent of Solute

1-3.3 Mole fraction (Amount-of-substance

Fraction)

1-3.4 Molality

1-3.5 Amount-of-substance Concentration

Mass percent: (mass solute / mass of solution) x 100%

Molarity(c): moles solute / Liter solution

Molality* (mB) ： moles solute / Kg solvent

Mole Fraction(x) : moles solute / total moles solution

* Note that molality is the only concentration unit in which denominator （分母） contains only solvent information rather than solution.

ways of expressing concentration:ways of expressing concentration:

% (m/m) =

% (m/v) =

% (v/v) =

1-3.2 Mass percent of solute (p11)

100xsolutionmasssolutemass

100xsolutionvolumesolutemass

100xsolutionvolumesolutevolume

Example 3.5 g of CoCl2 is dissolved in 100mL solution.

Assuming the density of the solution is 1.0 g/mL, what is concentration of the solution in % mass?

% = 3.5 g CoCl2

100g H2O

= 3.5% (m/m)

1-3.3 Mole fraction (Amount-of-substance Fraction) Moles of A (Amount-of-substance of A) Mole fraction of A = Sum of moles of all components

or nA

xA = nA + nB

Example1-1 A chemist prepared a solution by adding 200.4g of pure ethanol (C2H5OH) to 143.9g of water. Calculate the

mole fraction of these two components? The mol masses of ethanol and water are 46.02g and 18.02g, respectively.

Solution:

200.4g C2H5OH Amount-of-substance of C2H5OH = = 4.355mol 46.02g/mol  143.9g H2O Amount-of-substance of H2O = = 7.986 mol 18.02g/mol  Using equation we can write the mole fraction of ethanol and water as 4.355mol xC2H5OH = = 0.3529 (4.355mol+7.986mol)  7.986mol xH2O = = 0.6471 (4.355mol+7.986mol)

The Amount-of-substance of C2H5OH and H2O present are :

1-3.4 Molality _____moles solute / kg solvent

• Molality is a concentration unit equal to the amount-of-substance of solute dissolved in a kilogram of solvent.

amount-of-substance of solute (nB)Molality (mB)= kilograms of solvent (mA)

nB or mB = mA

若将 58.5gNaCl 溶于 1kg 水，所得溶液的质量摩尔浓度是 1.00 mo1·kg ―1

Molarity (c): _____moles solute / Liter solution

If 0.435 g of KMnO4 is dissolved in enough water to give 250. mL of solution, what is the molarity of KMnO4?

Now that the number of moles of substance is known, this can be combined with the volume of solution — which must be in liters — to give the molarity. Because 250. mL is equivalent to 0.250 L .

As is almost always the case, the first step is to convert

the mass of material to moles.

0.435 g KMnO4 • 1 mol KMnO4 = 0.00275 mol KMnO4

158.0 g KMnO4

Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110mol/L

0.250 L solution

1-3.5Amount-of-substance Concentration

• Amount-of-substance Concentration is a concentration unit equal to the amount-of- substance of solute dissolved in a liter of solution

Amount-of-substance of solute Amount-of-substance Concentration = liters of solution

nB

or cB=

V

实验室常用——物质的量浓度• 如 50ml ， 18 mo1· L ―1 的浓硫酸中所含

H2SO4 的物质量 (mol) 。

90.01000

5018

42SOHn (mol)

Solutions By “Parts”• Parts per million is the number of particles of

solute per one-million particles of solution. Fluoridated drinking water contains ~ 1 ppm of fluoride ion.

1 ppm = 1 mg/L (10-3)• Parts per billion （十亿） is the number of

particles of solute per one-billion particles of solution.

1 ppb = 1 µg/L (10-6)• Parts per trillion （万亿） is the number of

particles of solute per one-trillion particles of solution.

1 ppt = 1 ng/L (10-9)

等物质的量的反应规则

• 关于浓度，细读中文书 p8-10

aA + tT dD + eE

)()()()( eEndDntTnaAn

1-4 Colligative Properties of Solutions

1-4.1 Vapor Pressure Lowering

1-4.2 Boiling Point Elevation

1-4.3 Freezing Point Depression

1-4.4 Osmosis and Osmotic Pressure

1-4.5 The Colligative Properties of Electrolyte Solution

1-4.6 Isotonic( 等 ), Hypotonic ( 低 ), Hypertonic ( 高 ) Solution

1-4.7 Osmolarity

1-4.8 Crystalloid Osmotic Pressure and Colloidal

Osmotic Pressure

•The properties of a solution are differ from the properties of solute and solvent

The first depends on the nature of solute eg. color 、 electric and the volume change

The second not depend on the nature of solute is called

colligative property

eg. vapor pressure 、 boiling point 、 freezing point and

osmotic pressure

1-4 Colligative Properties of Solutions

•Two kinds of properties :

• colligative property: All these properties depend on the collective effect of the number of dissolved particles (molecules or ions) rather than on the nature of the particular particles involved

• (1) the lowering of the vapor pressure of the solution relative to that of the pure solvent, •(2) the elevation of the boiling point, •(3) the depression of the freezing point,•(4) the phenomenon of osmotic pressure.

• Evaporation is the process by which water is converted from its liquid form to its vapor form.

1-4.1 Lowering of Vapor Pressure

Evaporation is a surface phenomenon - some molecules have enough kinetic energy to escape.

1-4.1 Lowering of Vapor Pressure

Condensation is the change of water from its gaseous form (water vapor) into liquid water.

If the container is

closed, an equilibrium

is reached where

an equal number of

molecules return to

the surface.

Saturated Vapor PressureEquilibrium: V evaporation = V condensation .

• At this point the vapor is said to be saturated, and the pressure of that vapor (usually expressed in mmHg) is called the saturated vapor pressure.

• The pressure of this equilibrium is called the saturation vapor pressure .

Lowering of Vapor Pressure

Solvent moleculesolute molecule

(nonvolatile)

(a)vapor pressure of pure solvent (b) vapor pressure of solutionpo p

• The presence of a non-volatile （非挥发性的） solute means that fewer solvent particles are at the solution’s surface, so less solvent evaporates!

Δp = po- p

• Vapor pressure of a dilute solution that containing nonvolatile, nonelectrolyte is directly proportional to the amount-of-substance fraction of the solvent.

p∝ xA

. p is the vapor pressure of solution

xA is amount-of-substance fraction of solvent The constant is the vapor pressure of pure solvent p°

　 　　　　　　 　 　 p = p°xA …… （经验公式）

比例系数

xA+ xB =1

p= po xA = po(1- xB)= po - po xB

∴ po- p = po xB

　 Δp = po xB

Raoult’s law : Vapor pressure lowing of a dilute solution that containing nonvolatile, nonelectrolyte is directly proportional to the amount-of-substance fraction of the solute and has nothing to do with the nature of solute.

Δp Vapor Pressure Lowering

Δp≥0

For a dilute solution nA>> nB ， (nA+nB≈nA)

Δp= K bB (1-9)

A

B

BA

BB n

n

nn

nx

AA M

gn

1000

BA

B nM

pxpp1000

Suppose: solvent is 1000g:

Δp= K bB

mB

mB

若以水为溶剂， 1kgH2O 中含有 nB

在数值上 mB≈nB

• Raoult’s law :

• Vapor pressure lowing of a dilute solution that containing nonvolatile, nonelectrolyte is directly proportional to the molality of the solute and has

nothing to do with the nature of solute.

Δ p = K mB

Raoult’s law : Δ p = K　mB

T

p pure solvent

solution

solution

[ 例1-6] 已知 20℃ 时水的饱和蒸气压为 2.33kPa ，将 17.1g 蔗糖(C12H22O11) 与3.00g 尿素 [CO(NH2)]2] 分别溶于 100g 水。计算这两种溶液的蒸气压各是多少？

解：蔗糖的摩尔质量 M ＝342 g·mol － 1 ，所以溶液的 质量摩尔浓度

H2O 的摩尔分数

蔗糖溶液的蒸气压

)( 500.0100

1000

342

1.17 1 kgmolmB

991.05.05.55

5.55

500.00.18

10000.18

1000

2

OHx

(kPa) 31.2991.033.2 22

OHOH xpp

00

0

• 尿素的摩尔质量 M ＝ 60.0 g·mol － 1 ，溶液的 质量摩尔浓度

所以尿素溶液中蒸气压 p ＝ 2.31kPa ，

)kg(mol 500.0100

1000

0.60

00.3 1-Bm

这两种溶液的质量分数不同，但溶剂的摩尔分数相同，蒸气压也就相等。

991.05.05.55

5.55

500.00.18

10000.18

1000

2

OHx

Application of Vapor Pressure Lowering• Describe what is happening in the pictures below.• Use the concept of vapor pressure lowering to

explain this phenomenon.

1-4.2 Elevation of Boiling Point

•Why is responsible for increasing the boiling point of water with the addition of anti-freeze（防冻剂） (or salt)?

•Evaporation occurs when molecules have sufficient energy to escape the interface of a liquid substance.

Normal Boiling Point:Normal Boiling Point:

Temperature at which the vapor Temperature at which the vapor pressure of the liquid is equal to pressure of the liquid is equal to the atmospheric pressure.the atmospheric pressure.

PPH2OH2O==PP 外外 =101.3 =101.3 kkPPaa

T=100T=10000CC

When When a solutea solute contaminates the contaminates the

solvent, the surface area is solvent, the surface area is

reduced because of the attraction reduced because of the attraction

between solute and solvent, thus the physical between solute and solvent, thus the physical properties (i.e., vapor pressure) are properties (i.e., vapor pressure) are alteredaltered..

•Boiling Point Elevation Δ Tb = (Tb -Tb°) Tb°_ b.p. of pure solvent Tb _ b.p. solvent in solution Δ Tb _ b.p. Elevation

ΔTb= Kb mB

Kb　is different for different solvent

for water Kb = 0.512

Raoult’s law : Boiling Point Elevation of a dilute solution that containing nonvolatile, nonelectrolyte is directly proportional to the amount-of-substance fraction of the solute and has nothing to do with the nature of solute.

•Table 1- 1 Boiling-Point elevation and Freezing- Paint Depression Constants

Solvent Formula Tb(℃) Kb Tf(℃) Kf

Acetic acid CH3COOH 118.1 3.07 17 3.9

Benzene C6H6 80.1 2.53 5.5 4.9

Carbon tetrachloride CCl4 76.7 5.03 -22.9 32

Diethyl ether C4H 10O 34.7 2.02 - 116.2 1.8

Ethanol C2H5OH 78.4 1.22 - 114.7 -

Naphthalene C10H8 - - 80.5 6.8

Water H2O 100.0 0.512 0.0 1.86

萘

二乙醚

Example 1-3 （ P17 ）

• SolutionBecause the molar mass of biphenyl is 154 g ·mol-1, 5.50

g of biphenyl contains 5.50g /154g ·mol-1 =0.0357 mol biphenylThe molality of the biphenyl in the solution is 0.0357 mol mB = ---------------- = 0.357 mol ·kg-1 0. 1000 kg ΔTb 0.903 K Kb = -------- = ----------------- - =2.53 K · kg ·mol-1 mB 0.357 mol /kg

• When 5.50 g of biphenyl (C12H 10), a nonvolatile compound, is dissolved in 100.0 g of benzene, the boiling point of the solution exceeds the boiling point of pure benzene by 0.903 . What is K℃ b for benzene?

Example 1- 4 （ p9 ）

• When 6.30 g of a nonvolatile

hydrocarbon （烃） of unknown molar

mass is dissolved in 150.0 g of

benzene, the boiling point of the

solution is 80.696 . The boiling point ℃of pure benzene is 80.099 . What is ℃the molar mass of the hydrocarbon?

Exercise

1. What would the boiling point of a solution be if 60 g of NaCl were dissolved on 1000g of H2O?

1-4.3 Depression of Freezing Point

• Freezing Point, temperature at which a liquid congeals( 凝结 ) into the solid state at a given pressure.

• The temperature at which the solid and liquid states of the substance are in equilibrium.

凝固点 是物质的固相与它的液相平衡共存的温度。

Water•

0 0C P 固相 = P 液相（固 - 液两相共存）

0 0C 以上 P 固相 〉 P 液相 （冰溶化）

0 0C 以下 P 固相 〈 P 液相 （水转化为冰）

Freezing point depression

• The depression of the freezing point of a solution with respect to the pure solvent is analogous （相似的） to boiling point elevation.

• Here, we consider only the case that only solvent crystallizes out of solution.

凝固点 是溶剂的固相与液相平衡 共存的温度。

solution ： 0 0C P 溶液 〈 P 冰

（冰吸收热量融化，使体系温度降低）

Since the solution vapor pressure curve lies below that of the pure liquid solvent due to Raoult‘s law, the point of intersection （交点） gives rise to a lower freezing point for the solution.

• ΔTf = Kf mB

for water Kf = 1.86

(1-11)

Tf is subtracted from the normal freezing point of the solvent.

Kf is the molal freezing point depression constant

of the solvent. Kf is different for different solvent

[ 例 1-9] 乙二醇 CH2(OH)CH2(OH) 是一种常用的汽车防冻剂，它溶于水并完全是非挥发性的（ b.p.197℃ ）。 计算在 2505g 水中溶解 651g 该物质的溶液的凝固点。 夏天能否将它用于汽车散热器中？ 乙二醇的摩尔质量是 62.01g/mol 。

解：溶液的质量摩尔浓度

kgmolmB /19.42505

1000

01.62

651

ΔTf ＝ Kf·mB ＝ 1.86×4.19 ＝ 7.79(K)

因为纯水的凝固点是 273K ， 该溶液凝固点 Tf ＝ 273-7.79 ＝ 265.21K

沸点升高 ΔTb ＝ 0.52×4.19 ＝ 2.2 （ K ）此溶液在 375.2K 沸腾，所以夏天它能用于汽车散热器中防止溶液沸腾。

Determining Molecular Weight (p21)

•Dissolves 2.06 g of an unknown solid

substance in 164 g of cyclohexane (C6H12 环己

烷 ) and determines that the freezing point of

this solution is 4.76 .The freezing point of ℃

pure cyclohexane is 6.47 ; the ℃ Kf is 20.0K

kg/mol. What is the molecular weight of the

unknown solid, (Assume that it is nonionizable

and nonvolatile.)

Solution:The freezing-point depression is

ΔTf =freezing point of pure solvent - freezing point of solution

= 6.47℃ - 4.76℃ = 1.71 ℃according to ΔTf = Kf mB

mB =ΔTf / Kf = 1.71 / 20.0 =0.0855 (mol / kg)Amount-of-substance of solute

(nB)= 0.164kg×0.0855 mol / kg =0.0140 mol n = m/M Molar mass unknown =2.06g /0.0140mol = 147 g/mol

1-4.4 Osmotic Pressure

• Osmosis: • the process of solvent flow

through a semipermeable membrane from a pure solvent or from a dilute solution to a more concentrated solution in order to equalize the concentrations of solutes on the two sides of the membrane.

Figure : An experiment in osmosis.

• Osmosis is the spontaneous movement ( 自发运动 )of water across a semi-permeable membrane from an area of low solute concentration to an area of high solute concentration

达平衡状态时，半透膜两侧的水位差所表示的静压大小反映出溶液渗透能力的大小，溶液的渗透能力用渗透压表示。

1-4.4 Osmotic Pressure

•Osmotic

Pressure - The

Pressure that

must be applied

to stop osmosis.

渗透压的定义是：将纯溶剂与溶液以半透膜隔开时，为了阻止溶剂分子向溶液渗透所需要加给溶液的额外压力。

一定温度下，溶液越浓， 渗透压也越大。

• 实验证明：在一定温度下，溶液的渗透压与它的浓度成正比；在一定浓度时，溶液的渗透压与绝对温度成正比。

1886年荷兰物理化学家范托夫（ van’t Hoff ）综合上述实验结果，指出了非电解质的稀溶液的渗透压与温度、浓度的关系：

• πV ＝ nRT

• osmotic pressure:

Π= cRT • For a dilute solution, the amount-of-substance

concentration is approximately equal to the molality of solution,

so

Π= mBRT

公式的意义在于：一定温度下，溶液的渗透压与溶液的质量摩尔浓度成正比，也就是说，与溶液中所含溶质质点的数目成正比，而与溶质的本性无关。

Problem• What would the boiling point of a solution be if

18.0g of glucose were dissolved on 1000g of H2O?

freezing point ? Osmotic Pressure？

• What would the boiling point of a solution be if 5.85g of NaCl were dissolved on 1000g of H2O?

1-4.5 The Colligative Properties of Electrolyte Solution

• Δ p = i K　　mB

• ΔTb = i Kb mB

• ΔTf = i Kf mB i - Van’t Hoff factor

• Π　= i RT mB

i For sucrose, i =1; for KNO3 , i =2 ; for MgCl2, i = 3

1- 4.6 Isotonic, Hypotonic, Hypertonic Solution

Osmosis —is important in many biological processes

A cell might be thought of as an aqueous solution surrounded by a semipermeable membrane.

The solutions surrounding cells must have

the same osmotic pressure.

Otherwise, water will either leave the cell,

dehydrating （脱水） it ,

or enter the cell, causing it to burst （溶胀） .

Osmosis and Blood Cells(a) A cell placed in an isotonic solution. The net movement of water in and out of the cell is zero because the concentration of solutes inside and outside the cell is the same. (5%C6H12O6)

(b) In a hypertonic solution, the concentration of solutes outside the cell is greater than that inside. There is a net flow of water out of the cell, causing the cell to dehydrate, shrink, and perhaps die. (50% C6H12O6)

(c) In a hypotonic solution, the concentration of solutes outside of the cell is less than that inside. There is a net flow of water into the cell, causing the cell to swell and perhaps to

burst. (1 % C6H12O6)

IsotonicIsotonic HypertonicHypertonic HypotonicHypotonic

水、电解质紊乱水、电解质紊乱

1-4.7 Osmolarity （渗透物质的量浓度）

Osmolarity （ osmol/L or Osm/L).

is the measure of solute concentration,

defined as the number of osmoles

(Osm) of solute per liter of solution

渗量浓度 __ 指 1L 溶液中能产生渗透效应的各种溶质 微粒（分子或离子）的总物质的量。

1-4.7 Osmolarity number of particles• osmolarity = molarity×------------------------- = molarity × i molecule of solute

the unit of osmolarity is the Osm/L

• For nonionizable solute: Osm/L ≈ mol/L ,

but for the electrolyte: Osm/L ≈ mol/L × i

•For example: 1 mole of glucose dissolved in 1 litre of water has an osmolarity of 1 osmole (Osm/L ). If 1 mole of another sugar, such as sucrose （蔗糖） were added to the same litre of water, the osmolarity would be 1 Osm/L.

For example:

If 1 mole of NaCl were dissolved in 1 litre of water it would produce a 1 mol/LNaCl solution with an osmolarity 2 osm/L because NaCl dissociates into Na+ and Cl- (two particles)

in solution.

                        Na2SO4 i=3

•Body fluids have an osmolarity of about 300 mOsm/L.

(280-320 mOsm/L)

For dilute solutions, osmolarity is expressed in

milliosmols per liter (m Osm/L).

• What is the osmolarity of a 0.9% NaCl solution? osmolarity = molarity × i 1000 = 0.9%×----------------- ×2 = 0.308 Osm/L 100×M(58.5)

( 308mOsm/L)

• What is the osmolarity of a 5% glucose solution?

Solution:• Glucose is a nonelectrolyte; number of particlesosmolarity = molarity ×------------------------- = molarity × i molecule of solute

1000 =5%×----------------- × 1 = 0.277 Osm/L 100×M(180)

= 277mOsm/L

1-4.8 Crystalloid Osmotic Pressure and Colloidal Osmotic Pressure

• Crystalloid Osmotic Pressure (772KPa)

by NaCl, NaHCO3, glucose,0.75% regulate

balance of fluid and electrolyte on the two side of cell wall

Colloidal Osmotic Pressure (2.93~4.00KPa)

by large molecular colloidal substances (Protein),7% regulate balance of fluid and electrolyte on the two side of blood capillary wall.

（（ICFICF））

（（ECFECF））

（（Interstitial fluidInterstitial fluid））

（（P

lasm

aP

lasm

a））

水、电解质紊乱水、电解质紊乱

Crystalloid Osmotic PressurColloidal Osmotic Pressure ?

Crystalloid Osmotic Pressure If lack of water in the body, the outside fluids of cells become hypertonic, can lead to dehydratio

If water intake is too much, the outside fluids of cells become hypotonic, can lead to hemolysis,

（（ICFICF））

（（ECFECF））

（（Interstitial fluidInterstitial fluid））

（（P

lasm

aP

lasm

a））

水、电解质紊乱水、电解质紊乱

If the protein in plasma become less because some

disease, Colloidal Osmotic Pressure become

less,small crystal molecules osmosis to the outside

of blood capillary wall,so the edema （水肿） occur

•

edema occur

Dialysis• Process of cleaning blood

透析是指溶质通过半透膜，从高浓度溶液向低浓度方向运动。

血液透析包括溶质的移动和水的移动，即血液和透析液在透析器（人工肾）内借半透膜接触和浓度梯度进行物质交换，使血液中的代谢废物和过多的电解质向透析液移动，透析液中的钙离子、碱基等向血液中移动。

A solution, called the dialysate （ 透 析 液 ） , circulates outside of the cellophane tube. This dialysate is very carefully prepared not only to be isotonic with blood but also to have the same concentrations of all the essential substances that should be left in solution in the blood.

When these concentrations match, the rate at which such solutes migrate( 迁移 ) out of the blood equals the rate at which they return. In this way several key equilibria are maintained, and there is no net removal of essential components.

Process of cleaning blood