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Ionic Equilibria III: The Solubility Product Principle

20. Ionic Equilibria III: The Solubility Product Principle. Chapter Goals. Solubility Product Constants 溶解度積常數 Determination of Solubility Product Constants 決定溶解度積常數 Uses of Solubility Product Constants Fractional Precipitation 分級沉澱

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Ionic Equilibria III: The Solubility Product Principle

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  1. 20 Ionic Equilibria III: The Solubility Product Principle

  2. Chapter Goals • Solubility Product Constants溶解度積常數 • Determination of Solubility Product Constants 決定溶解度積常數 • Uses of Solubility Product Constants • Fractional Precipitation分級沉澱 • Simultaneous Equilibria Involving Slightly Soluble Compounds • Dissolving Precipitates

  3. Aqueous Solutions: An Introduction Insoluble soluble

  4. Solubility Product Constants • Silver chloride, AgCl,is rather insoluble in water. • Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water. Ag+Cl-  Ag+(aq) + Cl-(aq) • The equilibrium constant expression for this dissolution is called a solubility product constant溶解度積常數. • Ksp = solubility product constant Ksp = [Ag+][Cl-] = 1.8x10-10 Appendix H 當溶液解離達平衡後，反應物以及生成物離子的濃度皆不改變，此離子濃度會達成一個固定的比例即稱溶解度積常數

  5. Solubility Product Constants • Consider the dissolution of silver sulfide in water. • The solubility product expression for Ag2S is: Ksp = [Ag+]2[S2-] = 1.0x10-49 H2O Ag2S(aq)2Ag++ S2- 100% Ksp的數值愈大，表示溶液中離子的濃度也愈大；也就是水溶解化合物時，化合物在水中分解成離子的數量也就愈多。

  6. Solubility Product Constants • The dissolution of solid calcium phosphate in water is represented as: • The solubility product constant expression is: Ksp = [Ca2+]3[PO43-]2 = 1.0x10-25 H2O Ca22+(PO43-)2(s)2Ca2++ 2PO43- 100%

  7. Solubility Product Constants • In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as: Ksp = [Ms+]r[Yr-]s H2O MrYs(s)rMs++ sYr- • 溶解度的另一表示方法: • The solubility of a compound molar solubility莫耳溶解度 • The number of moles that dissolve to one liter of saturated solution (M)每公升飽和水溶液中所含溶質的莫耳數 100%

  8. Solubility Product Constants • The same rules apply for compounds that have more than two kinds of ions. • One example of a compound that has more than two kinds of ions is calcium ammonium phosphate. H2O CaNH4PO4(s)Ca2+(aq) + NH4+(aq)+PO43-(aq) Ksp = [Ca2+][NH4+][PO43-] 100%

  9. Determination of Solubility Product Constants Example 20-1: One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25oC. Calculate the molar solubility of, and Ksp for, AgCl. The molar solubility can be easily calculated from the data: 0.00192g AgCl 143g AgCl 1.0L ? M AgCl= = 1.34x10-5 M AgCl(s) Ag++ Cl- 1.34x10-5 M 1.34x10-5 M 1.34x10-5 M Ksp = [Ag+][Cl-] = (1.34x10-5) x (1.34x10-5) = 1.8x10-10 M

  10. Determination of Solubility Product Constants Example 20-2: One liter of saturated calcium fluoride solution contains 0.0167 gram of CaF2 at 25oC. Calculate the molar solubility of, and Ksp for, CaF2. 0.0167g CaF2 ? M CaF2= = 2.14x10-4 M 78.1g 1.0L CaF2 Ca2++ 2F- 2.14x10-4 M 2.14x10-4 2x(2.14x10-4) Ksp = [Ca2+][F-]2 = (2.14x10-4) x (4.28x10-4)2 = 3.92x10-11 mol/L

  11. Uses of Solubility Product Constants Example 20-3: Calculate the molar solubility of barium sulfate, BaSO4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25oC. For barium sulfate, Ksp= 1.1 x 10-10. BaSO4 Ba2++ SO42- x M x M x M Ksp = [Ba2+][SO42-] = 1.1x10-10 (x)(x) = 1.1x10-10 x= 1.0x10-5 M [Ba2+]=[SO42-]= 1.0x10-5 M 1.0x10-5 mol ?g BaSO4= 1.0x10-5M x 1L = = 1.0x10-5 x 234 =2.34x10-3 g in 1L

  12. Uses of Solubility Product Constants Example 20-4: The solubility product constant for magnesium hydroxide, Mg(OH)2, is 1.5 x 10-11. Calculate the molar solubility of magnesium hydroxide and the pH of a saturated magnesium hydroxide solution at 25oC. Mg(OH)2 Mg2++ 2 OH- x M x M 2x M Ksp = [Mg2+][OH-]2 = 1.5x10-11 (x)(2x)2 = 1.5x10-11 4x3= 1.5x10-11 M x3= 3.75x10-12 M x= 1.6x10-4 M [OH-]= 2x (1.6x10-4) =3.2x10-4 M pOH=3.49, pH =10.51

  13. The Common Ion Effect in Solubility Calculations Example 20-5: Calculate the molar solubility of barium sulfate, BaSO4, in 0.010 M sodium sulfate, Na2SO4, solution at 25oC. Compare this to the solubility of BaSO4 in pure water. (Example 20-3). (What is the common ion? How was a common ion problem solved in Chapter 19?) Na2SO4 2Na++ SO42- Soluble ionic salt 完全解離 0.010 M 2x(0.010)M 0.010 M BaSO4 Ba2++ SO42- Slightly Soluble salt 溶解度很低 x M x M x M Ksp = [Ba2+][SO42-] = (x)(0.01+x) = 1.1x10-10 0.01+x 0.01 (0.01)x = 1.1x10-10 x = 1.1x10-8 Molar solubility of BaSO4

  14. The Common Ion Effect in Solubility Calculations • The molar solubility of BaSO4in 0.010 M Na2SO4 solution is 1.1 x 10-8M. • The molar solubility of BaSO4in pure water is 1.0 x 10-5M. • BaSO4 is 900 times more soluble in pure water than in 0.010 M sodium sulfate! • Adding sodium sulfate to a solution is a fantastic method to remove Ba2+ ions from solution! • If your drinking water were suspected to have lead ions in it, suggest a method to prove or disprove this suspicion.

  15. The Reaction Quotient in Precipitation Reactions The reaction quotient, Q, and the Ksp of a compound are used to calculate the concentration of ions in a solution and whether or not a precipitate will form. If Qsp< Ksp Forward process is favored(反應向右) No precipitation occurs; if solid is present, more solid can dissolve(不會產生沉澱) If Qsp= Ksp Solution is just saturated(正好飽和) Solid and solution are in equilibrium; neither forward nor reverse process is favored If Qsp> Ksp Reverse process is favored(反應向左) precipitation occurs to form more solid(產生沉澱)

  16. 100mL x 0.10M 100mL x 0.10M The Reaction Quotient in Precipitation Reactions Example 20-6: We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate form? MPb(NO3)2= =0.050 M Pb(NO3)2 100mL+100mL MK2SO4= =0.050 M K2SO4 100mL+100mL K2SO4 2K++ SO42- 0.05M 0.05x2M 0.05M  Will PbSO4 precipitate?? Pb(NO3)2 Pb2++ 2 NO3- 0.05M  0.05M 0.05x2M PbSO4 Pb2++ SO42- Qsp = [Pb2+][SO42-] = (0.050)(0.050) = 2.5x10-4 Ksp= 1.8x10-8 for PbSO4 Qsp> Ksp therefore solid forms

  17. The Reaction Quotient in Precipitation Reactions Example 20-7: Suppose we wish to remove mercury from an aqueous solution that contains a soluble mercury compound such as Hg(NO3)2. We can do this by precipitating mercury (II) ions as the insoluble compound HgS. What concentration of sulfide ions, from a soluble compound such as Na2S, is required to reduce the Hg2+ concentration to 1.0 x 10-8M? For HgS, Ksp=3.0 x 10-53. HgSHg2++ S2- Ksp = [Hg2+][S2-] = 3.0x10-53 = [S2-]x(1.0x10-8) [S2-] = 3.0x10-45 If enough S2-, in th form of Na2S, is added to just slightly exceed 3.0x10-45 M the mercury will precipitate

  18. 1 mol Hg2+ 1.0 L The Reaction Quotient in Precipitation Reactions Example 20-8: Refer to example 20-7. What volume of the solution (1.0 x 10-8M Hg2+ ) contains 1.0 g of mercury? ?L =1.0g Hg2+ x x 201g Hg2+ 1.0x10-8 mol Hg2+ ?L =5.0x105 L

  19. Fractional Precipitation • The method of precipitating some ions from solution while leaving others in solution is called fractional precipitation 分級沉澱. • If a solution contains Cu+, Ag+, and Au+, each ion can be precipitated as chlorides. CuClCu++ Cl- Ksp = [Cu+][Cl-]=1.9x10-7 AgClAg++ Cl- Ksp = [Ag+][Cl-]=1.8x10-10 AuClAu++ Cl- Ksp = [Au+][Cl-]=2.0x10-13

  20. Example 20-9: If solid sodium chloride is slowly added to a solution that is 0.010 M each in Cu+, Ag+, and Au+ ions, which compound precipitates first? Calculate the concentration of Cl- required to initiate precipitation of each of these metal chlorides. AuCl has the small Ksp, so it precipitates first By the same reasoning, CuCl will precipitate last • Calculate the concentration of Cl- required to precipate AuCl Ksp = [Au+][Cl-]=2.0x10-13 = (0.010)x[Cl-] [Cl-]=2.0x10-11 • Repeat the calculation for silver chloride. Ksp = [Ag+][Cl-]=1.8x10-10 = (0.010)x[Cl-] [Cl-]=1.8x10-8 • Finally, for copper (I) chloride to precipitate. Ksp = [Cu+][Cl-]=1.9x10-7 = (0.010)x[Cl-] [Cl-]=1.9x10-5

  21. Fractional Precipitation • These three calculations give the [Cl-] required to • precipitate AuCl ([Cl-] >2.0 x 10-11 M), • precipitate AgCl ([Cl-] >1.8 x 10-8 M), • precipitate CuCl ([Cl-] >1.9 x 10-5 M). • It is also possible to calculate the amount of Au+ precipitated before the Ag+ begins to precipitate, as well as the amounts of Au+ and Ag+ precipitated before the Cu+ begins to precipitate.

  22. [Au+]unprecipitated 1.1x10-5 Ksp = [Au+][Cl-]=2.0x10-13 Example 20-10: Calculate the percentage of Au+ ions that precipitate before AgCl begins to precipitate. • Use the [Cl-] from Example 20-9 to determine the [Au+] remaining in solution just before AgCl begins to precipitate. [Au+]x(1.8x10-8)=2.0x10-13 [Au+]=1.1x10-5 unprecipitated • The percent of Au+ ions unprecipitated just before AuCl precipitates is: % Au+unprecipitated= x100% [Au+]original = x100% 0.01 =0.1% unprecipitated • Therefore, 99.9% of the Au+ ions precipitates before AgCl begins to precipitate.

  23. [Ag+]unprecipitated 9.5x10-6 Fractional Precipitation • A similar calculation for the concentration of Ag+ ions unprecipitated before CuCl begins to precipitate is: Ksp = [Ag+][Cl-]=1.8x10-10 [Ag+]x(1.8x10-8)=1.9x10-5 [Ag+]=9.5x10-6 unprecipitated • The percent of Ag+ ions unprecipitated just before AgCl precipitates is: % Ag+unprecipitated= x100% [Ag+]original = x100% 0.01 =0.095% unprecipitated • Therefore, 99.905% of the Ag+ ions precipitates before CuCl begins to precipitate.

  24. Simultaneous Equilibria Involving Slightly Soluble Compounds • Many weak acids and bases react with many metal ions to form insoluble compounds.許多弱酸和弱鹼可以和金屬反應形成不可溶的化合物 • The weak acid or weak base equilibrium as well as the solubility equilibrium • Equlibria that simultaneously involve two or more different equilibrium constant expressions are simultaneous equilibria.

  25. [NH4+] [OH-] (x)(x) Simultaneous Equilibria Involving Slightly Soluble Compounds Example 20-12: If 0.10 mole of ammonia and 0.010 mole of magnesium nitrate, Mg(NO3)2, are added to enough water to make one liter of solution, will magnesium hydroxide precipitate from the solution? • For Mg(OH)2, Ksp = 1.5 x 10-11. Kb for NH3 = 1.8 x 10-5. Calculate Qsp for Mg(OH)2 and compare it to Ksp. • Mg(NO3)2 is a soluble ionic compound so [Mg2+] = 0.010 M. • Aqueous ammonia is a weak base that we can calculate [OH-]. NH3 + H2O NH4++ OH- 0.1M -x +x +x x2=1.8x10-6 Kb= x=1.3x10-3 =[OH-] = = 1.8x10-5 [NH3] (0.10-x) 可溶性鹽類 Mg(NO3) 2Mg2++ 2NO3- 0.01M +0.01M +0.01x2M Qsp = [Mg2+][OH-]2 = (0.010)(1.3x10-3)2 = 1.7x10-8 Qsp> Ksp thus Mg(OH)2 will precipitate

  26. [base] (x) Example 20-13: How many moles of solid ammonium chloride, NH4Cl, must be used to prevent precipitation of Mg(OH)2 in one liter of solution that is 0.10 M in aqueous ammonia and 0.010 M in magnesium nitrate, Mg(NO3)2 ? (Note the similarity between this problem and Example 20-12.) Calculate the maximum [OH-] that can exist in a solution that is 0.010 M in Mg2+. Ksp = [Mg2+][OH-]2=1.5x10-11 [OH-]2x(0.010)=1.5x10-11 [OH-]  3.9x10-5 M [OH-]2=1.5x10-9 Using the maximum [OH-] that can exist in solution, determine the number of moles of NH4Cl required to buffer 0.10 M aqueous ammonia so that the [OH-] does not exceed 3.9 x 10-5M. NH4Cl →NH4++ Cl- xM xM xM Buffer solution NH3 + H2O NH4++ OH- (0.1M) (3.9x10-5) [OH-]= Kb x =1.8x10-5 x [salt] (0.10) Because there is 1.0L of solution There are 0.046 mol NH4Cl x= 0.046 =[NH4Cl]

  27. Simultaneous Equilibria Involving Slightly Soluble Compounds • Check these values by calculating Qsp for Mg(OH)2. Qsp = [Mg2+][OH-]2 = (0.010)(3.9x10-5)2 = 1.5x10-11 Qsp = Ksp Thus this system is at equilibrium • Use the ion product for water to calculate the [H+] and the pH of the solution. [H+][OH-] =1.0x10-14 [H+]x(3.9x10-5)=1.0x10-14 [H+] =2.6x10-10 pH=9.59

  28. Dissolving Precipitates • For an insoluble solid, if the ion concentrations (of either the cation or anion) are decreased, a solid precipitate can be dissolved.(在不可溶固體中,若任一離子濃度降低,則此沉澱的固體會被再溶解) • The trick is to make Qsp < Ksp. • One method is to convert the ions into weak electrolytes.採用的方法乃將離子轉成弱電解質 • Make these ions more water soluble. 讓此離子更水溶

  29. Dissolving Precipitates • If insoluble metal hydroxides are dissolved in strong acids, they form soluble salts and water.(不可溶金屬氫氧化合物可溶於強酸而形成可溶性鹽類及水) • For example, look at the dissolution of Mg(OH)2 in HCl. Mg(OH)2(s) + 2 HCl(aq) MgCl2(aq) + 2H2O(l) or Mg(OH)2(s) + 2 H+(aq) Mg2+(aq) + 2H2O(l) Notice that the insoluble Mg(OH)2 is converted to the more soluble MgCl2

  30. Dissolving Precipitates • A second method is to dissolve insoluble metal carbonates in strong acids.(另一方法是將不可溶金屬碳酸化合物溶於強酸中) • The carbonates will form soluble salts, carbon dioxide, and water. CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + CO2(g) + 2H2O(l) or CaCO3(s) + 2 H+(aq)  Ca2+(aq) + CO2(g) + 2H2O(l) Notice that the formation of carbon dioxide removes ions and makes the carbonate dissolve

  31. Dissolving Precipitates • A third method is to convert an ion to another species by an oxidation-reduction reaction.(第三種方法乃利用氧化還原反應改變離子態) • For example, the dissolution of insoluble metal sulfides in hot nitric acid causes the sulfide ions to be oxidized to elemental sulfur. 3PbS(s) + 8H++ 2NO3- 3Pb2+ + 3S0(s) +2NO(g) + 4H2O(l) [Pb2+][S2-]=8.4x10-28=Ksp The NO3- oxidizes S2- to S0 making the Qsp<Ksp

  32. Dissolving Precipitates • A fourth method is complex ion formation.(第四種方法是形成錯離子) • The cations in many slightly soluble compounds will form complex ions. • This is the method used to dissolve unreacted AgBr and AgCl on photographic film. • Photographic “hypo” is Na2S2O3. h 2AgBr(s) 2Ag(s) + Br2(l) In the unexposed portion of the film AgBr(s) is left AgBr(s) + 2 Na2S3O2(aq)  Na3Ag(S2O3)2 (aq) + NaBr(aq) AgBr(s) + 2 Na2S32- [Ag(S2O3)2]3-(aq) + Br-(aq)

  33. Dissolving Precipitates • Copper(II) hydroxide, which is light blue colored, dissolves in aqueous ammonia to form dark blue, [Cu(NH3)4]2+. Cu(OH)2(s) + 4NH3(aq)  [Cu(NH3)4]2+(aq) + 2OH- Cu(OH)2(s)  Cu2+(aq) + 2OH-(aq) Cu2+(aq) + 4NH3  [Cu(NH3)4]2+(aq)

  34. Complex Ion Equilibria • A metal ion coordinated to several neutral molecules or anions forms compounds called complex ions. • Familiar examples of complex ions include: [Ag(NH3)2]+ [Cu(NH3)4]2+ Co(NH3)2Cl2 [Pt(NH3)2]+

  35. [Cu2+] [NH3]4 [Ag+] [NH3]2 Complex Ion Equilibria • The dissociation of complex ions can be represented similarly to equilibria. • For example: [Cu(NH3)4]2+  Cu2+ + 4NH3 Kd= [[Cu(NH3)4]2+] [Ag(NH3)2]+  Ag+ + 2NH3 Kd= [[Ag(NH3)2]+] Complex ion equilibrium constants are called dissociation constants Kd.

  36. [Ag+] [NH3]2 (x)(2x)2 Complex Ion Equilibria Example 20-14: Calculate the concentration of silver ions in a solution that is 0.010 M in [Ag(NH3)2]+. Kd = 6.3 x 10-8 Write the dissociation reaction and equilibrium concentrations. [Ag(NH3)2]+  Ag+ + 2NH3 (0.01-x)M (x)M (2x)M Substitute the algebraic quantities into the dissociation expression. Kd= = = 6.3x10-8 [[Ag(NH3)2]+] (0.010-x) 4x3= 6.3x10-10 x3= 1.6x10-10 x= 5.4x10-4 M

  37. [Ag+] [NH3]2 [Ag+] [NH3]2 (3.6x10-8)[NH3]2 ? M AgCl = 0.01mol/2.0L = 5x10-3 M AgCl(s) + 2NH3 [Ag(NH3)2]++ Cl- Example 20-15: How many moles of ammonia must be added to 2.00 L of water so that it will just dissolve 0.010 mole of silver chloride, AgCl? 5x10-3 M 2(5x10-3) M 5x10-3 M 5x10-3 M AgCl  Ag+ + Cl- Ksp=[Ag+][ Cl-] =1.8x10-10 [Ag(NH3)2]+  Ag+ + 2NH3 Kd= =6.3x10-10 [[Ag(NH3)2]+] The [Ag+] in the solution must satisfy both equilibrium constant expressions. Because the [Cl-] is known, the equilibrium concentration of Ag+ can be calculated from Ksp for AgCl. Ksp=[Ag+][ Cl-] =1.8x10-10 =[Ag+](5x10-3) Which is the maximum [Ag+] possible [Ag+] =3.6x10-8 = =6.3x10-10 Kd= (5x10-3) [[Ag(NH3)2]+] [NH3]= 0.094 M [NH3]2= 8.75x10-3 [NH3]total= 0.094 M + 2x(5.0x10-3) =0.104M ? mol NH3= 2.0L x 0.104M = 0.21

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