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def. c os. n os / V. Content review. Osmolarity ( 渗透浓度 ). 2. isotonic, hypertonic and hypotonic sulutions ( 等渗、高渗与低渗溶液 ). hypotonic. hypertonic. isotonic. 280 mmol. L -1. 320 mmol. L -1. Burst, Hemolysis. Shrivel Crenation. normal. Exercises Choice:

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Content review

def

cos

nos/ V

Content review

  • Osmolarity (渗透浓度)

2. isotonic, hypertonic and hypotonic sulutions (等渗、高渗与低渗溶液)

hypotonic

hypertonic

isotonic

280 mmol.L-1

320mmol.L-1

Burst, Hemolysis

Shrivel

Crenation

normal


Content review

Exercises

Choice:

What is the osmolarity of 0.0350 mol.L-1 (NH4)2SO4?

A 0.0350 mol.L-1 B 0.0700 mol.L-1

C 0.105 mol.L-1 D 0.140 mol.L-1

2. The osmolarity of Na+ in the physiological saline is ?

A 77 mmol.L-1 B 190 mmol.L-1

C 154 mmol.L-1 D 391 mmol.L-1

C

B


Content review

True or false:

1. When a cell is placed in a concentrated sodium chloride solution, water enters the cell, causing the cell to swell.

F

2. When red blood cells are placed in a hypotonic solution, they swell up and finally burst; this process is called hemolysis.

3. It is osmolality within body fluid or blood serum level of 280-320 mmol.L-1.

4. Red blood cells will remain the same size when placed in 0.10 mol. L-1 CaCl2 solution.

T

T

T


Content review

Calculation

When 18 g of glucose (葡萄糖,C6H12O6) and 2.34 g of NaCl are dissolved in 100 mL of water, What is the osmolarity of the final solution?is it a isotonic, hypotonic or hypertonic solution?

solution:

So, it is a hypertonic solution.


Content review

Case

Chapter 3 Electrolyte Solutions

Conclusion:

acidosis (酸毒症)


Content review

3.2 Brønsted-Lowry Acid-Base Theory

3.4 Solving Problems Involving Acid-Base Equilibria

Chapter 3 Electrolyte Solutions


Content review

aims

  • Identify conjugate acid-base pairs

  • Determine the relationship between Kb for a base and the Ka of its conjugate acid

  • Calculate the concentrations of the species present in a solution of a weak acid or base


Content review

New words

  • Acid, acidic酸 suan

  • Base, alkaline碱 jian

  • alkalescent 弱碱性的

  • conjugate acid-base pair 共轭酸碱对

    gong e suan jian dui

  • amphiprotic 两性liang xing

  • cation阳离子 yang li zi

  • anion 阴离子 yin li zi

  • acid-ionization Constant 酸电离常数,酸常数, Ka

    suan chang shu

  • base-ionization Constant 碱常数, Kb

  • monoprotic acid一元酸

  • Binary 二元的 er yuan

  • step- ionization分步电离

    fen bu dian li


Content review

布朗斯特

3.2 Brønsted-Lowry Acid-Base Theory

1. definition

acid(酸):An acid is a proton (H+)donor

base(碱):A base is a proton (H+) acceptor

Notice:

Only focus on the proton

base

acid

base

acid


Content review

example

布朗斯特

There are many electrolyte ions in blood plasma, such asNa+, K+ , Mg2+,Ca2+,Cl-, HCO3-, HPO42- andSO42-. Which is the acid and which is the base?

Notice:

1. Acids and bases can be ions as well as molecular substances.

2. Some species that can act as either acids or bases are called amphiprotic (两性物质) such as water.

3. The proton-containing negative ions are amphiprotic.


Content review

HAc H+ + Ac-

NH4+ H+ + NH3

H3PO4 H+ + H2PO4-

H2PO4-H+ + HPO42-

HPO42-H+ + PO43-

Acid Proton + Conjugate base

When a acids loses a proton, the species formed is base. The two species are conjugate acid-base pair. Every Brønsted -Lowry acid has a conjugate base, and every base has a conjugate acid.


Content review

Example:Write down the conjugate bases of these acids.

H2O, +NH3CH2COO-, NH4+, [Al(H2O)6]3+

OH-, NH2CH2COO-, NH3, [Al(OH)(H2O)5]2+

H2CO3, HCO3-, H3PO4, H2PO4-, HPO42-

HCO3- , CO32- , H2PO4-, HPO42-, PO43-

分析:

Write the formula of the conjugate base of

any acid simply by removing a proton.


Content review

2. the acid-base reaction

Essence:

An acid-base reaction is the transfer of a proton from

a proton donor (acid) to a proton acceptor (base).


Content review

Notice:

For a given pair of reactants, write the equation for the

transfer of only one proton from one species to the other.

A Brønsted -Lowry acid-base reaction produces a conjugate

acid and base.


Content review

The direction for an acid-base reaction is from the stronger acid and base to the weaker acid and base.

HB1 + B2- B1- + HB2

stronger stronger weaker weaker

acid base base acid

Notice:

The strongest acids have the weakest conjugate bases, and

the strongest bases have the weakest conjugate acids.

Example 4 In the following equations, label each species as an acid or a base. Show the conjugate acid-base pairs.

HCO3-+ HF H2CO3+ F-


Content review

HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)

HNO3 + HAc H2Ac+ +NO3-

HNO3 + HC1H2NO3+ + Cl -

3. Relative Strengths of Acids and Bases

The terms stronger and weaker are used here only in acomparative sense. Because it relates to the types of solvent system.

For example:


Content review

  • Notice

  • The strength of the acid is a measure of its tendency to liberate proton.

  • When you say an acid loses its proton readily, you can also say that its conjugate base does not hold the proton very tightly. The stronger the acid, the weaker is its conjugate base.

  • The direction for an acid-base reaction is from the stronger acid and base to the weaker acid and base.


Content review

3.4 Solving Problems Involving

Acid-Base Equilibria

1. Self-Ionization of Water

H2O(l) + H2O (l) H3O+ (aq) + OH-(aq)

Kc =[H3O+ ][OH-]=constant= Kw


Content review

  • Notice

  • Kw is the ion-product constant for water. At 25℃, the value of Kw is 1.0 × 10-14.

  • Like any equilibrium constant, Kwvaries with temperature.

    0℃ 1.10×10-15,25℃ 1.00×10-14,100℃ 5.50×10-13

    ALL CALCULATIONS WILL BE AT 298.15 K , UNLESS OTHERWISE INDICATED


Content review

  • Kw is applicable to all dilute aqueous solutions.

    If you add an acid or a base to water, the concentrations of H+ and OH- will no longer be equal. The equilibrium-constant equation Kw =[H3O+ ][OH-] will still ho1d. Example: 0.1mol/L HCl

  • In an acidic solution, [H+] > 1.0 × 10-7 M.

  • In a neutral solution, [H+] = 1.0 × 10-7 M.

  • In a basic solution, [H+] < 1.0 × 10-7 M.

  • Equation indicates an inverse relationship between [H +] and [OH -]; if we know either the hydrogen or hydroxide ion concentration, we can, by Equation, find the other.


Content review

Example: please calculate the concentrations of H+ and OH- for 0.1 M HCl, pure water and 0.1 M NaOH. The equilibrium-constant equation Kw =[H3O+ ][OH-]

  • Solution:

  • 0.1 M HCl

  • [H3O+ ] = 0.1 M

  • [OH-]=Kw/ [H3O+ ] = 10-14/0.1=10-13 M

  • (2) pure water

  • [H3O+ ] = [OH-]=10-7 M

  • 0.1 M NaOH

  • [OH-] = 0.1 M

  • [H3O+]=Kw/ [H3O+ ] = 10-14/0.1=10-13 M


Content review

  • 2. The pH of A Solution

  • Definition: pH is the negative logarithm of the molar hydrogen ion concentration.

  • pH = -log[H+]

  • Remember, a high pH (greater than 7) means basic and a low pH (less than 7) means acidic. The higher the pH, the lower the acidity, and vice versa.

  • If [H+] = l0-x, then pH = x.

  • [H+] = antilog(-pH) = 10-pH


Content review

Example: please calculate the concentrations of H+ and OH- for 0.1 M HCl, pure water and 0.1 M NaOH. The equilibrium-constant equation Kw =[H3O+ ][OH-]

  • Solution:

  • 0.1 M HCl

  • [H3O+ ] = 0.1 MpH= - log0.1= 1

  • [OH-]=Kw/ [H3O+ ] = 10-14/0.1=10-13 M

  • (2) pure water

  • [H3O+ ] = [OH-]=10-7 M pH= - log 10-7 = 7

  • 0.1 M NaOH

  • [OH-] = 0.1 M

  • [H3O+]=Kw/ [H3O+ ] = 10-14/0.1=10-13 M pH= - log 10-13 = 13


Content review

pOH = -log[OH-]

  • Relationship between pH and pOH

  • [H+][OH-] = Kw

  • (-log [H+]) + (-log [OH-]) = -log Kw

  • pH + pOH = pKw=14

  • Just as we know that the product of the hydrogen ion and hydroxide ion concentrations is 1.0 × 10-14 , the sum of the pH and pOH must always be 14.00 (at 25℃).


Content review

pH Values for Some Common Solutions

  • Substance pH

  • household ammonia 11.9

  • milk of magnesia 10.5

  • household detergent 9.2

  • Sea water 8.5

  • blood 7.4

  • pure water 7.0

  • milk 6.9

  • urine 6.0

  • black coffee 5.0

  • tomato juice 4.1

  • wine 3.5

  • carbonated beverage 3.0

  • vinegar 2.9

  • lemon juice 2.2

  • human stomach acid 1.7


Content review

Example :

A sample of orange juice has a hydrogen-ion concentration of 2.9×10-4 M. What is the [H3O+ ] , [OH-], pH and pOH?

Solution:

[H3O+ ]= 2.9×10-4M

pH = - log(2.9×10-4) = 3.54

pOH =14-pH=14-3.54=10.46

[OH-]= 10-10.46M

or [OH-]= Kw/ [H3O+ ] = 10-14/(2.9×10-4)=3.45×10-11 M


Content review

  • 3. The pH of Monoprotic Acids or Bases

  • Acid-ionization Constant and Base-ionization Constant

  • acid: HB+H2O B- +H3O+

  • base: B-+H2O HB +OH-

  • If Ka is large, the acid is strong. If Kb is large, the base is strong.

  • Like any equilibrium constant, Ka (or Kb) varies with temperature.


Content review

The Relationship between Ka and Kb for Conjugate Acid-Base Pairs

Ka · Kb = Kw


Content review

Ka · Kb = Kw

  • Notice

  • This relationship is general and shows that the product of acid- and base-ionization constants in aqueous solution for conjugate acid-base pairs equals the ion-product constant for water, Kw.

  • As Ka decreases, Kb for the conjugate base must increase, since Kw is a constant. We see that the weaker an acid (the smaller Ka ), the stronger its conjugate base (the larger Kb ).

  • Obtain Ka from Kb or Kb from Ka .


Content review

Example: At 25℃, the Ka ofHAc is 1.8×10-5 and the Kb ofNH3 is 1.8×10-5,too. Calculate the following : (a) Kb for Ac-; (b) Ka for NH4+.

Solution

(a) The conjugate acid of Ac- is HAc, whose Ka is 1.8×10-5 . Hence,

Kb= Kw/Ka = 1.0×10–14/1.8×10-5 = 5.6×10–10

(b) The conjugate base of NH4+ is NH3, whose Kb is 1.8×10-5 . Hence,

Ka = Kw/Kb = 1.0×10–14/1.8×10-5 = 5.6×10-10


Content review

  • Calculating Concentrations of Species in a Weak Acid Solution (Approximation Method)

  • Procedure

  • Write the equilibrium equation.

  • Set up a table in which you write the starting, change, and equilibrium values of each substance under the balanced equation.

  • Substitute the equilibrium-constant equation for the equilibrium concentrations.

  • Solve the equilibrium-constant equation for the equilibrium concentrations.


Content review

HB(aq) H+ (aq) + B-(aq)

Starting ca 0 0

Change - x + x + x

Equilibrium ca- xxx

If ca/Ka≥500 (or α< 5% )

simplifying assumption


Content review

Example A 0.25 M solution of HCN has a pH of 5.00. What is Ka? What is αof Hydrocyanic acid in this solution?


Content review

HCN (aq) H+ (aq)+ CN -(aq)

Solution:

[H+] = antilog (-5.00) = 10-5

Starting 0.25 0 0

Change -10-5 +10-5 +10-5

Equilibrium 0.25- 10-5 10-5 10-5


Content review

Example: What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0.10M nicotinic acid, HC6H4NO2, at 25℃? What is the pH of the solution? What is the degree of ionization of nicotinic acid? The acid-ionization constant, Ka, is 1.4×10-5.

[HA]= c - x ≈ ca

pH= -log[H+ ]


Content review

Solution HNic (aq) H+ (aq)+ Nic -(aq)

Starting 0.10 0 0

Change -x +x +x

Equilibrium 0.10- x x x

[HNic]= 0.10 - x ≈ ca=0.1 mol/L

pH= -log[H+ ] = 2.92


Content review

Summary


Content review

Example

Morphine, C17H19NO3, is administered medically to relieve pain. It is a naturally occurring base, or alkaloid. What is the pH of a 0.0075 mol·L-1 solution of morphine at 25 ℃? The base-ionization constant, Kb, is 1.6×10-6 at 25 ℃.

Solution


Content review

  • Procedure

  • Identify a salt solution is acidic or basic.

  • Obtain Ka or Kb for the ion that hydrolyzes by using KaKb = Kw .

  • Calculation of the concentrations of species present follows that for solutions of weak acids or bases.

  • basic→Kb →

  • acidic→Ka →


Content review

  • 4. Polyprotic Acids

  • Step- ionization

  • When polyprotic acids ionize, they do so by steps, releasing one hydrogen ion with each step.

  • 1st stage H2CO3 H+ + HCO3- Ka1

  • 2st stage HCO3- H+ + CO32- Ka2

  • Each succeeding acid is weaker than the one before. For any diprotic acid, Ka2 < Ka1. This is always true because it is more difficult to remove a proton from a negatively charged species (HA-) than from an uncharged molecule (H2A).


Content review

1st stage H2CO3 H+ + HCO3- Ka1

2st stage HCO3- H+ + CO32- Ka2

CO32- + H2O HCO3- + OH- Kb1

HCO3- + H2O H2CO3 + OH- Kb2

Ionization Constants for Acids (Ka) and Base (Kb)

H2CO3~HCO3- Ka1 Kb2 = KW

HCO3-~CO32- Ka2 Kb1 = KW


Content review

Key points

共轭酸碱对:

Ka · Kb = Kw

弱电解质的解离度:

一元弱酸弱碱pH:


Content review

第四章 电解质溶液

【教学目标】

1、掌握酸碱质子理论的基本内容(质子酸碱的概念、共轭关系、酸碱反应的实质、酸碱强度);

2、掌握同离子效应及有关计算;

3、掌握一元弱酸弱碱pH计算。

【教学重难点】

酸碱质子理论、同离子效应、一元弱酸弱碱溶液

【课后作业】

课后习题 p45 T5(4)(5),T8,T9


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