1 / 43

Content review

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:

johnlpeterson
Download Presentation

Content review

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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

  2. 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

  3. 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

  4. 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.

  5. Case Chapter 3 Electrolyte Solutions Conclusion： acidosis (酸毒症)

  6. 3.2 Brønsted-Lowry Acid-Base Theory 3.4 Solving Problems Involving Acid-Base Equilibria Chapter 3 Electrolyte Solutions

  7. 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

  8. 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

  9. 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

  10. 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.

  11. 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.

  12. 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.

  13. 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).

  14. 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.

  15. 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-

  16. 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:

  17. 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.

  18. 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

  19. 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

  20. 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.

  21. 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

  22. 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

  23. 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 ？ ？ ？

  24. 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℃).

  25. 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

  26. 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

  27. 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.

  28. The Relationship between Ka and Kb for Conjugate Acid-Base Pairs Ka · Kb ＝ Kw

  29. 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 .

  30. 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

  31. 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.

  32. 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

  33. 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?

  34. 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

  35. 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+ ]

  36. 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

  37. Summary

  38. 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

  39. 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 →

  40. 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).

  41. 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

  42. 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：

  43. 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

More Related