1 / 58

Chapter 4 Acid-Base Equilibria Three issues ： 溶液理论 Acid-Base Theory Acid-Base Equilibria

Chapter 4 Acid-Base Equilibria Three issues ： 溶液理论 Acid-Base Theory Acid-Base Equilibria. 1. Solutions 1.1 Concentration units. (1) Molarity (M). (2) Molality (m). (3) Mole fraction (X). X A +X B + … = 1. 1-2 Colligative properties of nonelectrolyte solutions.

normap
Download Presentation

Chapter 4 Acid-Base Equilibria Three issues ： 溶液理论 Acid-Base Theory Acid-Base Equilibria

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. Chapter 4 Acid-Base Equilibria Three issues： 溶液理论 Acid-Base Theory Acid-Base Equilibria

  2. 1. Solutions1.1 Concentration units (1)Molarity (M) (2) Molality (m)

  3. (3) Mole fraction (X) XA +XB + … = 1

  4. 1-2 Colligative properties of nonelectrolyte solutions • Those solution properties which depend primarily on the concentration of solute particles rather than their nature are called colligative properties(依数性). Four physical properties of solutions are the same for all nonvolatile solutes: • vapor pressure lowering • boiling point elevation • freezing point depression • osmotic pressure

  5. (1) Vapor pressure lowering P = PA0XA = PA0 (1-XB) “在一定温度下，稀溶液的蒸气压等与纯溶液的蒸气压与溶剂摩尔数的乘积。”

  6. (2) Boiling point elevation ΔTb= T溶液-T溶剂=Kbm m:溶液的质量摩尔浓度单位：mol·kg-1, Kb:溶剂的沸点上升常数,单位K·kg·mol-1,是溶质的质量摩尔浓度为1mol·kg-1时所引起的溶液沸点上升的数值。

  7. (3) Freezing point depression ΔTf=Tf溶剂-Tf溶液=Kf  m Kf：溶剂的摩尔凝固点下降常数，单位K·kg·mol-1, 即m= 1mol·kg-1时所引起的溶液凝固点的下降值。例如：Kf（H2O）=1.68 K·kg·mol-1

  8. 例：汽车的散热冷却水，冬季加入乙二醇(或甲醇⁄甘油），可防止结冻，冻裂水箱。求：在1kg水中，加入甘油989g，可使溶液的凝固点下降多少度？（甘油的分子量：MC3H8O3 = 92g /mol） • 解：ΔTf = Kf m • Kf（H2O）=1.86 K·kg·mol-1 • m= (989 92) ⁄ 1kg =10.75mol·kg-1 •  ΔTf =1.86  10.75=20(K) • ΔTf =273- Tf溶液 • Tf溶液=253K

  9. （4）Osmotic pressure

  10. Л= n RT ⁄ V= CRT P = n RT ⁄ V

  11. 1-3 强电解质溶液互吸理论 问题：电解质溶液“依数性”反常 解释：完全电离，且离子互吸

  12. 强电解质溶液的互吸理论要点： （1）强电解质在水中完全电离，电离度а=100%. （2）（只考虑正 负离子静电作用，不考虑离子的水合，离子溶剂作用）形成“离子氛” 每个离子周围异号离子多于同号离子，净结果是在每个离子周围形成球形对称的“离子氛”，离子氛的电荷在数值上等于中心离子的电荷，符号相反。

  13. （3）活度：溶液中实际发挥作用的离子浓度，叫有效浓度，或称活度。（3）活度：溶液中实际发挥作用的离子浓度，叫有效浓度，或称活度。 （4）“活度”的根源在于离子间作用力 （m、 Z越大，作用力越大） 用离子强度来衡量 单位：mol/kg

  14. （5）25℃,稀溶液（I  0.01m ） 1923年，德拜-休克尔的极限公式 (Debye –Hückel’s limiting law)

  15. 例1：0.1mol·kg-1盐酸和0.1molkg-1CaCl2溶液等体积混和，计算该溶液的I及a H+。 解：mH+=0.05mol·kg-1 mca2+=0.05 mol·kg-1 mcl-=0.15 mol·kg-1 ∴I=1／2∑mizi2=1／2(0.05×12+0.05×22+0.15×12)=0.2 mol·kg-1 ㏒f±(HCl)=-0.509∣1×1∣0.2.1／2 = -0.2276 f±=0.519=f+=f- ∴aH+ = f+ mH+／m0=0.59×0.05= 0.0295，误差较大。 ㏒f±(HCl)= -0.509∣1×1∣0.2.1／2 / （1+0.2.1／2）= -0.1573 f±= 0.70 ∴aH+ = 0.70×0.05= 0.035

  16. 例2：I=1.0×10-4mol·kg-1,计算NaCl溶液，MgSO4溶液的f±。例2：I=1.0×10-4mol·kg-1,计算NaCl溶液，MgSO4溶液的f±。 解： 一价离子 NaCl ㏒f±=-0.059∣1×1∣(1.0×10 –4) 1／2 = - 0.00509 , f±=0.99 二价离子 MgSO4 ㏒f±=-0.059∣2×2∣(1.0×10 –4) 1／2 = -0.02036, f±=0.95 三价离子㏒f±=-0.059∣3×3∣(1.0×10 –4) 1／2 =-0.0458 , f±=0.90 四价离子㏒f±=-0.059∣4×4∣(1.0×10 –4) 1／ 2= -0.0814, f±=0.83 结论：I一定时，Z越小，f越大1.

  17. 例3：分别计算浓度为1.0×10-2mol·kg-1，1.0×10-3 mol·kg-1和1.0×10-4mol·kg-1NaCl溶液中Na+,Cl-的活度系数及活度。 解：I=Σ0.5miZi2=0.5×0.01×12 + 0.5×0.01×12 = 0.01 mol·kg-1 logf±= -0.509∣1×1∣×0.01½ = -0.0509 ∴f±= 0.89 = f+ = f- aNa+ = aCl- = 0.89×0.01= 0.0089 同样可得，m=1.0×10-3 mol·kg-1 即I=1.0×10-3 mol·kg-1时，f±=0.96 m=1.0×10-4 mol·kg-1时， I=1.0×10-4 mol·kg-1，f±=0.99 结果表明，m越稀， I越小，(I=1.0×10-4 mol·kg-1)， f1.

  18. 2 Acid-Base Theory2-1 Arrhenius Definition of Acids and Bases Acids are compounds that all their cations in aqueous solutions are protons (H+). Eg. HCl, HAc. Bases are compounds that all their anions in aqueous solutions are hydroxides (OH-). Eg. NaOH, KOH

  19. The acid/ base strength can be scaled by ionization extent(α). If α of an acid equals to 100%, it is called strong acid, its ionization is completed and it is called strong acid. If α of an acid is much less than 100%, its ionization is partial and it is called weak acid.

  20. The essence of acids and bases reaction is the neutralization reaction: H++ OH- = H2O △rHmº =-55.84KJ/mol

  21. It can easily explain that the heat of neutralization reactions of all strong acids and strong bases is equal to -55.84KJ/mol and the heat of neutralization reactions of all weak acids and weak bases is less than 55.84KJ/mol The heat of mixing dilute salt solutions is zero.

  22. In sum, it make the aqueous solution chemistry systematical and theoretical. However, it limited the concept of acids and bases in aqueous solutions and it cannot explain the same kind of problems in non- aqueous solution or non- solvent system.

  23. 2-2 Bronsted-Lowry Definition of Acids and Bases Acids are species that donate a proton (H+). Bases are species that accept a proton. According to this definition, acid and base have conjugated base and acid. They are conjugated acid-base pair. Acid = Base + H+ HCl = Cl- + H+ NH4+ =NH3 + H+

  24. Comments • Acid /Base can be a molecule , a cation and an anion. • Acid and base can be transformed. • A compound that can act as either an acid or a base is called amphiprotic. • There is no concept of salts.

  25. The Bronsted-Lowry definition of acids and bases does not encompass all chemical compounds that exhibit acidic and basic properties.

  26. 2-3 Lewis Acids and Bases A Lewis acid is an electron-pair acceptor . A Lewis base is an electron-pair donor.

  27. Lewis acid + Lewis base = Acid- base complex Eg. H+ + OH- = H2O H+ +NH3 = NH4+ Ag+ + 2 NH3 = Ag(NH3)2+ Cu + 4NH3 = Cu(NH3)42+ BF3 + F- = BF4- Al3++ 6 H2O = Al(H2O)63+ Ag+ + Cl- =AgCl HCl + NH3 = NH4Cl

  28. 3.Acid-Base Equilibria—— The Proton Transfer in Aqueous Solution Key issue：Calculation of [H+]

  29. 3-1 The ion product for water---- Kw H2O  H+ + OH- rHm = 55.84KJ/mol Kw = [H+][OH-] T, Kw

  30. Kw at different temperature

  31. acidic solution : [H+] > [OH-] neutral solution: [H+] = [OH-] basic solution: [H+] < [OH-] The pH scale: pH = -log [H+]

  32. 3-2 Weak -acid/base Equilibria(1)The Weak -acid/base HAc + H2O  H3O+ +Ac-rHm = -0.46KJ/mol HAc  H+ +Ac- t0 c0 0 0 te c0 - x x x Ka = x2/( c0 - x)

  33. If c0 / Ka > 400, c0 – x  c0 , [H+] =  Ka c0  =  Ka /c0 As the initial acid concentration decreases, the percent dissociation of the acid increases. (Dilution Law) HAc + H2O  H3O+ +Ac- As for weak base, [OH-] ==  Kb c0

  34. Summary • Ka /Kb stands for the dissociation extent • of acid/base, Ka , the strength of the acid. • Since the rHm is small, Ka /Kb change • very little with the temperature. The main factor that affect acid-base equilibria is concentration. • Common Ion Effect • Salt Effect

  35. Example 1: Calculate the [H+] and  of the following solutions: (a)0.1M HAc (b)1.0 10-5M HAc (c)Add NaAc(s) to 0.1M HAc ,keeping [NaAc] =0.20M (d)Add NaCl(s) to 0.1M HAc ,keeping [NaCl] =0.20M Solution: [H+]=1.3 10-3M, = 1.3% [H+]=7.16 10-6M, = 71.6% (need to be calculated exactly) [H+]=9.0 10-6M, = 0.009% [H+]=1.9 10-3M, = 1.9%

  36. (2) The Dissociation of Polyprotic Acids Acids with more than one ionizable proton are polyprotic acids. Diprotic acid: H2S, H2SO3,H2CO3,H2C2O4 Triprotic acid : H3PO4

  37. H2S  HS- + H+ Ka1 = 1.310-7 HS-  S2- + H+ Ka2 = 7.110-15 Ka1 > Ka2 , Successive acid dissociation constants typically differ by several orders of magnitude. This fact simplifies pH calculations involving polyprotic acids， because we can usually neglect the H+ coming from the subsequent dissociations.

  38. Example 2:Calculate the saturated H2S aqueous solution(0.1M). Solution: H2S  HS- + H+ t0 0.1 0 0 te 0.1- x 0.1 x x x = [H+] = [HS-] =  Ka c0 =1.14 10-4M HS-  S2- + H+ t0 1.14 10-4 0 1.14 10-4 te 1.14 10-4- y y 1.14 10-4+y 1.14 10-41.14 10-4  Ka2 = y = 7.110-15M H2O  H+ + OH- x+y+[OH-]x [OH-] [OH-]= Kw /x = 8.810-11M

  39. Conclusion: [H+] =  Ka1 c0

  40. 3-3. Hydrolysis of Salts When salts dissolve in water, the pH of the solution is affected. NaCl , NaAc, NH4Cl, NH4Ac, NaHCO3

  41. (1) NaAc Ac- + H2O  HAc + OH- t0 c0 0 0 te c0 - x x x Kh = Kb = [HAc][OH-][H+]/[Ac-][H+] = Kw / Ka = 5.610-10 x = [OH-] =  Kh c0 =  Kw c0 / Ka h = [OH-]/ c0 =  Kw / Ka c0

  42. Example 3: Calculate the pH and h of the following salt solutions: 0.010M NaAc (Ka = 1.810-5 ) 0.010M NaCN (Ka = 4.9310-10 ) Solution: (a) Ac- + H2O  HAc + OH- t0 0.01 0 0 te 0.01- x 0.01 x x x = [OH-] =  Kh c0 = Kw c0 / Ka = 2.3710-6 M [H+] = 4.2310-9M, pH= 8.37 h= [OH-]/ c0 = 0.0237% pH =10.65, h = 4.5% 结论：Ka越小（酸越弱），[OH-]和h越大。

  43. NaAc [OH-] =  Kh c0 = Kw c0 / Ka Na2CO3 [OH-] =  Kh1 c0 = Kw c0 / Ka2 NH4Cl [H+] =  Kh c0 = Kw c0 / Kb NH4Ac [H+] =  Kw Ka / Kb

  44. (2) NaHCO3 NaHCO3(c0) Na+ (c0)+HCO3-( c0) Two trends: HCO3- H+ + CO32- Ka2 = 5.610-11 HCO3- + H2O OH- + H2CO3 Kh2 = Kw / Ka1 = 2.410-8 [OH-] = [H2CO3] – [CO32-] Kw /[ H+]= [H+][ HCO3-]/ Ka1 - Ka2[ HCO3-]/[H+] [H+]2[ HCO3-]/ Ka1= Kw + Ka2[ HCO3-] [H+]2= Ka1(Kw + Ka2[ HCO3-])/[ HCO3-] [H+] =  Ka1(Kw + Ka2 c0)/ c0 If Ka2 c0>> Kw , [H+] =  Ka1 Ka2

  45. Example 4: Calculate the pH of 0.1M solutions: (a)NaHCO3 (b)NaH2PO4 (c)Na2HPO4 Solution: (a) [H+] =  Ka1 Ka2 (b) [H+] =  Ka1 Ka2 (c) [H+] =  Ka2 Ka3 =1.6610-10 pH =9.78

  46. Practice: Rank the following salts in order of increasing pH of their aqueous solutions: (a)KNO3, K2SO3, K2S, Fe (NO3) 2 (b) NH4NO3, NaHSO4, NaHCO3 , Na 2CO3

  47. (3)The factors that affect the hydrolysis of salts: • C: h = [OH-]/ c0 =  Kh / c0 • c0, h • T: rHm > 0 • T, Kh Applications: 可通过控制溶液pH来控制水解。 Eg. SnCl2/concentrated HCl KCN/KOH

More Related