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شیمی تجزیه مبحث: حلالیت رسوب ها

شیمی تجزیه مبحث: حلالیت رسوب ها. دکتر امید رجبی دانشیار گروه شیمی دارویی دانشکده داروسازی مشهد. AgCl(s) Ag + (aq) + Cl – (aq). K sp = [Ag + ][Cl – ]. The Solubility Product Constant, K sp. Many ionic compounds are only slightly soluble in water: ex. Ag salts, sulfides.

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شیمی تجزیه مبحث: حلالیت رسوب ها

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  1. شیمی تجزیهمبحث: حلالیت رسوب ها دکتر امید رجبی دانشیار گروه شیمی دارویی دانشکده داروسازی مشهد

  2. AgCl(s) Ag+(aq) + Cl–(aq) Ksp = [Ag+][Cl–] The Solubility ProductConstant, Ksp Many ionic compounds are onlyslightly solublein water: ex. Ag salts, sulfides Equations are written to represent the equilibriumbetween the compound and the ions present in asaturated aqueous solution

  3. Ksp’s (25 °C)

  4. Cl– Cl– Ag+ Ag+ + Ksp = x2 and [Ag+] = (Ksp)1/2 Ksp and Molar Solubility The solubility product constant is related to the solubility of an ionic solute Ksp = [Ag+][Cl–]; solubility given by [Ag+] From stoichiometry, the ion ratio is 1:1, so [Ag+] = [Cl–], both of which are unknown (x)

  5. Ag2SO4(s) 2 Ag+(aq) + SO4–2(aq) Solubility of Ag2SO4 if MgSO4 is added to solution The Common Ion Effect Le Châtelier’s principle is followed for the shift in concentration of products and reactants upon addition of either products or reactants to a solution The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution

  6. Common Ion Effect Illustrated

  7. Does Precipitation Occur? Precipitation should occur if Qip > Ksp Precipitation cannot occur if Qip < Ksp A solution is just saturated if Qip = Ksp Qip is the ion product reaction quotient and is based on initial conditions of the reaction

  8. Example If 1.00 mg of Na2CrO4 is added to 225 mL of 0.00015 M AgNO3, will a precipitate form? Ag2CrO4(s) 2 Ag+(aq) + CrO42–(aq) Ksp= 1.1 x 10–12

  9. A Conceptual Example Pictured here is the result of adding a few drops of concentrated KI(aq) to a dilute solution of Pb(NO3)2. What is the solid that first appears? Explain why it thendisappears.

  10. Example If 0.100 L of 0.0015 M MgCl2 and 0.200 L of 0.025 M NaF are mixed, should a precipitate of MgF2 form? MgF2(s) Mg2+(aq) + 2 F–(aq) Ksp= 3.7 x 10–8

  11. Example An aqueous solution that is 2.00 M in AgNO3 is slowly added from a buret to an aqueous solution that is 0.0100 M in Cl– and also 0.0100 M in I–. • Which ion, Cl– or I–, is the first to precipitate from solution? • When the second ion begins to precipitate, what is the remaining concentration of the first ion? • Is separation of the two ions by selective precipitation feasible? AgCl(s) Ag+(aq) + Cl–(aq) Ksp= 1.8 x 10–10 AgI(s) Ag+(aq) + I–(aq) Ksp= 8.5 x 10–17

  12. Selective Precipitation AgNO3 added to a mixture containing Cl– and I–

  13. CaF2(s) Ca2+(aq) + 2 F–(aq) AgCl(s) Ag+(aq) + Cl–(aq) Effect of pH on Solubility • If the anion of a precipitate is that of a weak acid, the precipitate will dissolve somewhat when the pH is lowered: Added H+ reacts with, and removes, F–; LeChâtelier’s principle says more F– forms. • If, however, the anion of the precipitate is that of a strong acid, lowering the pH will have no effect on the precipitate. H+ does not consume Cl– ; acid does not affect the equilibrium.

  14. Example What is the molar solubility of Mg(OH)2(s) in a buffer solution having [OH–] = 1.0 x 10–5 M, that is, pH = 9.00? Mg(OH)2(s) Mg2+(aq) + 2 OH–(aq) Ksp= 1.8 x 10–11 A Conceptual Example Without doing detailed calculations, determine in which of the following solutions Mg(OH)2(s) is most soluble: (a) 1.00 M NH3 (b) 1.00 M NH3 /1.00 M NH4+ (c) 1.00 M NH4Cl.

  15. Equilibria Involving Complex Ions Silver chloride becomes more soluble, not less soluble, in high concentrations of chloride ion.

  16. Ag+(aq) + 2 Cl–(aq) [AgCl2]–(aq) Complex Ion Formation • A complex ion consists of a central metal atom or ion, with other groups called ligands bonded to it. • The metal ion acts as a Lewis acid (accepts electron pairs). • Ligands act as Lewis bases (donate electron pairs). • The equilibrium involving a complex ion, the metal ion, and the ligands may be described through a formation constant, Kf: [AgCl2]– Kf = –––––––––– = 1.2 x 108 [Ag+][Cl–]2

  17. Complex Ion Formation Concentrated NH3 added to a solution of pale-blue Cu2+ … … forms deep-blue Cu(NH3)42+.

  18. Complex Ion Formationand Solubilities But if the concentration of NH3 is made high enough … … the AgCl forms the soluble [Ag(NH3)2]+ ion. AgCl is insoluble in water.

  19. Ag+(aq) + 2 NH3(aq) [Ag(NH3)2]+(aq) Kf= 1.6 x 107 AgBr(s) Ag+(aq) + Br–(aq) Example Calculate the concentration of free silver ion, [Ag+], in an aqueous solution prepared as 0.10 M AgNO3 and 3.0 M NH3. Example If 1.00 g KBr is added to 1.00 L of the solution described in Example 16.13, should any AgBr(s) precipitate from the solution? Ksp= 5.0 x 10–13

  20. حلالیت اگسالات کلسیم را در محلولی با غلظت یون هیدرونیوم 1.00x10-4 محاسبه کنید CaC2O4(S) Ca2+ + C2O42- C2O42- +H3O+ HC2O4- + H2O HC2O4- +H3O+ H2C2O4 + H2O

  21. Solubility = [Ca2+] = [C2O42-]+[HC2O4-]+[H2C2O4]

  22. Ksp=[Ca2+][C2O42-]=2.3x10-9 [Ca2+] = [C2O42-]+[HC2O4-]+[H2C2O4] Kb1=[C2O42-][H3O+]/[HC2O4-]=5.42x10-5 Kb2=[HC2O4-][H3O+]/[H2C2O4]=5.36x10-2 [H3O+]=1.0x10-4

  23. (1.0x10-4)[C2O42-]/ [HC2O4-]=5.42x10-5 [HC2O4-]=1.84[C2O42-] (1.0x10-4) x1.84[C2O42-]/ [H2C2O4-]=5.36x10-2 [H2C2O4-]= 0.0034[C2O42-]

  24. [Ca2+] = [C2O42-]+[HC2O4-]+[H2C2O4] [Ca2+]=[C2O42-]+ 1.84[C2O42-]+ 0.0034[C2O42-] [C2O42-]= [Ca2+]/2.84 Ksp=[Ca2+][C2O42-]=2.3x10-9 [Ca2+]=8.1x10-5 [Ca2+] [Ca2+]/2.84= 2.3x10-9 Solubility of CaC2O4=8.1x10-5

  25. محاسبات حلالیت در حالتی که غلظت یون هیدرونیوم متغیر است حلالیت PbCO3 را در آب محاسبه کنید:

  26. PbCO3 Pb2+ + CO32- CO32- + H2O HCO3- + OH- HCO3- + H2O H2CO3 + OH- 2H2O H3O+ + OH- Solubility= [Pb2+] =[CO32-]+[HCO3-]+[H2CO3]

  27. [Pb2+ ][CO32-]=Ksp=3.3x10-14 [Pb2+]=[CO32-]+[HCO3-]+[H2CO3] [HCO3-][OH-]/[CO32-]=K1b=1.0x10-14/4.7x10-11=2.13x10-4 [H2CO3][OH-]/[HCO3-]=K2b=1.0x10-14/4.45x10-7=2.25x10-8 [H3O+][OH-]=1.00x10-14 معادله موازنه بار یا Charge balance : 2[Pb2+ ]+[H3O+]=2[CO32-]+[HCO3-]+[OH-] 6 معادله و 6 مجهول داریم!

  28. [Pb2+]=[CO32-]+[HCO3-]+[H2CO3] 2[Pb2+ ]+[H3O+]=2[CO32-]+[HCO3-]+[OH-] 0=[OH-]+[HCO3-] [OH-] = [HCO3-] [HCO3-][OH-]/[CO32-] = K1b= 1.0x10-14/4.7x10-11 = 2.13x10-4 =[HCO3-]2/[CO32-] [HCO3-]=√K1b[CO32-] [Pb2+]=[CO32-]+√K1b[CO32-] [Pb2+]=[CO32-]+[HCO3-]

  29. [Pb2+ ][CO32-]=Ksp=3.3x10-14 [CO32-]=Ksp/ [Pb2+] [Pb2+]=[CO32-]+√K1b[CO32-] [Pb2+]= Ksp/ [Pb2+ ]+√K1b Ksp/ [Pb2+ ] [Pb2+] = 1.9x10-6

  30. حلالیت Fe(OH)3 را در آب محاسبه کنید: Ksp=[Fe3+][OH-]3=4x10-38 • Charge Balance: 3[Fe3+]+[H3O+]=[OH-] [Fe3+](3[Fe3+])3=4x10-38 [Fe3+]=2x10-10 [OH-]=6x10-10 فرض ما در مورد حذف H3O+ در مقابل Fe3+ اشتباه بوده است! [H3O+] = 1.7x10-5

  31. 3[Fe3+] « [H3O+] 3[Fe3+]+[H3O+]=[OH-] [Fe3+]=4x10-38/(1.00x10-7)3 = 4x10-17

  32. اثر غلظت الکترولیت روی حلالیت Effects of Ionic strength on Solubility

  33. Activity dilute solution

  34. Activity Ions in concentrated solution interact

  35. Activity • In concentrated solution, ions interaction makes it appear that there are fewer ions than there really are • aA = A [A]where:aA is activity of AA is the activity coefficient [A] is molarity (as usual)

  36. Significance • Ksp for BaSO4in water: 10-10in 0.01 M KNO3: 2.9 x 10-10 • Solubility has NOT really increased but K+ and NO3- act to shield Ba2+ and SO42+ from each other in solution

  37. Do All Ions Work Equally Well? • 2 factors important: • concentration of ions • charge on ions

  38. Ionic Strength •  = 0.5  ci Zi2where:ci is molarity of ith ionZi is charge on ith ion • Note: effect of square is to remove sign of charge so + and - don’t cancel out!

  39. Examples • Calculate and compare the ionic strength of 0.1 M NaCl, 0.1 M Na2SO4, and 0.1 M MgSO4 • 0.1 M NaCl = 0.10.1 M Na2SO4 = 0.30.1 M MgSO4 = 0.4

  40. Ksp=aAm.aBn =[A]m[B]n.γAm.γBn [A]m[B]n=Ksp/ γAm.γBn

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