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Solubility Equilibria

Solubility Equilibria. AP Chemistry Chapter 15b. Solubility. Substances are soluble in varying degrees. Salt and sugar are both soluble CaSO 4 is less soluble in hot water. It forms protective coating for tubes in boilers.

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Solubility Equilibria

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  1. Solubility Equilibria AP Chemistry Chapter 15b

  2. Solubility • Substances are soluble in varying degrees. • Salt and sugar are both soluble • CaSO4 is less soluble in hot water. It forms protective coating for tubes in boilers. • BaSO4 has very low solubility. It is used in gastrointestinal x-rays. • Nearly all salts that dissolve in water can be considered to dissociate 100%

  3.  Ca2+ + 2F- CaF2  CaF2 Ca2+ + 2F-  As it begins to dissolve, the ions are formed. When a solid is first added to water, no ions are present.  Ca2+ + 2F- CaF2 As concentrations of ions increase, it is more and more likely that these will collide and re-form the solid. Ultimately, a state of dynamic equilibrium will be reached. The solution is said to be saturated.

  4. Molar Solubility • The amount of a solid which dissolves is expressed in moles per liter. • Calculations involving Ksp are divided into 3 categories: • Calculate Ksp from solubility data • Calculate solubility from Ksp • Problems with precipitation.

  5. K = [Ca2+][F-]2 [CaF2]  CaF2(s) Ca2+ + 2F-  For the saturated solution with CaF2 and a given temperature, the solubility product constant, Ksp, is: Ksp = [Ca2+][F-]2

  6. CuBr(s) Cu+ + Br-  Lots of solid 2.0x10-4 2.0x10-4 15.12 Copper(I) bromide has a measured solubility of 2.0x10-4 mol/L at 25oC. Calculate its Ksp value. Solid 0 0 -2.0x10-4 +2.0x10-4 +2.0x10-4 Ksp = [Cu+][Br-] Ksp = [2.0x10-4][2.0x10-4] Ksp = 4.0x10-8

  7. Bi2S3(s) 2Bi3+ + 3S2-  Lots of solid 2.0x10-15 3.0x10-15 15.13 Calculate the Ksp value for bismuth sulfide (Bi2S3) which has a solubility of 1.0x10-15 mol/L at 25oC. Solid 0 0 -1.0x10-15 +2.0x10-15 +3.0x10-15 Ksp = [Bi3+]2[S2-]3 Ksp = [2.0x10-15]2[3.0x10-15]3 Ksp = 1.1x10-73

  8. Cu(IO3)2(s) Cu2+ + 2 IO3-  Lots of solid x 2x 15.14 The Ksp value for copper(II) iodate, Cu(IO3)2, is 1.4x10-7 at 25oC. Calculate its solubility at that temperature. Solid 0 0 -x +x +2x Ksp = [Cu2+][IO3-]2 1.4x10-7 = x(2x)2 1.4x10-7 = 4x3 x = 3.3x10-3 mol/L

  9. Relative Solubilities We must be careful using Ksp values in predicting relative solubilities. Salts producing same # ions: AgI Ksp = 1.5x10-16 CuI Ksp = 5.0x10-12 CaSO4 Ksp = 6.1x10-5 List most soluble to least…. CaSO4 > CuI > AgI Salts producing different # ions: Sol in mol/L CuS Ksp = 8.5x10-45 Ag2S Ksp = 1.6x10-49 Bi2S3 Ksp = 1.1x10-73 9.2x10-23 5.0x10-17 1.0x10-15 List most soluble to least…. Bi2S3 > Ag2S > CuS

  10. Ag2CrO4(s) 2 Ag+ + CrO42-  Lots of solid .100+2x x Common Ion Effect Find the solubility of Ag2CrO4 (Ksp 9.0x10-12) in solution with .100 M AgNO3. Solid .100 0 -x +2x +x Ksp = [Ag+]2[CrO42-] 9.0x10-12 = (.100+2x)2 x x = 9.0x10-10 mol/L (in pure water, solubility is 1.3x10-4 mol/L)

  11. CaF2(s) Ca2+ + 2 F-  15.15 Calculate the solubility of solid CaF2 (Ksp = 4.0x10-11) in a 0.025 M NaF solution. Lots of solid x .025 + 2x Ksp = [Ca2+][F-]2 4.0x10-11 = (.025+2x)2 x x = 6.4x10-8 mol/L

  12. pH and Solubility • In the dissociation Mg(OH)2(s) Mg2+ + 2OH- • The addition of OH- ions (rise in pH) forces equilibrium to the left • The addition of H+ ions (drop in pH) forces equilibrium to the right • And has the added effect of making Mg(OH)2 (s)more soluble!

  13. Ag3PO4(s) 3Ag+ + PO43-  Let’s see just how that works…. Ag3PO4(s) is more soluble in an acidic solution PO43- reacts with H+ to form HPO42- …thus shifting the reaction…. …to the right…making Ag3PO4 more soluble. So adding H+ (lowering the pH) increases solubility of Ag3PO4. If X- is an effective base (HX is weak acid) the salt (MX) shows increased solubility in an acidic solution. Effective bases: HO-, S2-, CO32-, C2O42-, CrO42-

  14. Precipitates and Quantitative Analysis • We’ll now show how we will form solids from solutions. • We must decide if a solid will form when 2 solutions with ions are mixed. • Use ion product: Q=[Ca2+][F-]2 • Notice this is initial [ ], not equilibrium [ ] (Ksp)

  15. Unsaturated solution: • When the ion product < Ksp • No precipitate occurs • Saturated solution: • When the ion product = Ksp • No new precipitate occurs (equilibrium) • Supersaturated solution: • When the ion product > Ksp • Yes, a precipitate will occur.

  16. Ce(IO3)3(s) Ce3+ + 3IO3-  750mLCe(NO3)34.00x10-3 mmolCe(NO3)3=3 mmolCe(NO3)3 1 mL 1050mL 300mL KIO32.00x10-2 mmol KIO3 = 6 mmol KIO3 1 mL 1050mL 15.16 A solution is prepared by adding 750.0 mL of 4.00x10-3 M Ce(NO3)3 to 300.0 mL of 2.00x10-2 M KIO3. Will Ce(IO3)3 (Ksp 1.9x10-10) precipitate from this solution? Lots of solid 2.86x10-3 5.71x10-3 2.86x10-3 M Ce3+ 5.71x10-3 M IO3- Q = [Ce3+][IO3-]3 = (2.86x10-3)(5.71x10-3)3 Q = 5.32x10-10 Q > Ksp thus a precipitate will form

  17. Calculate the equilibrium concentration of Pb2+ and I- ions formed by mixing 100.0 mL of 0.0500 M Pb(NO3)2 and 200.0 mL of 0.100 M NaI. Assume PbI2 reacts to completion Pb2+(aq) + 2 I-(aq) PbI2(s) [Pb2+]o = 100.0 mL  .0500 mmol  1 = 1.67x10-2M mL 300 mL [I-]o = 200.0 mL  .100 mmol  1 = 6.67x10-2M mL 300 mL Determine if PbI2 (s) forms (Ksp=1.4x10-8) 100.0 mL  .0500 mmol = 5.00 mmol Pb2+ 1 mL 200.0 mL  .100 mmol = 20.00 mmol I- 1 mL Q = [Pb2+][I-]2 = 7.43x10-5 Q > Ksp so a ppt will form 5.00 mmol Pb2+ reacts with 10.00 mmol I- to form 5.00 mmol PbI2in 300 mL forming .0167M PbI2 After the complete reaction: 0 M Pb2+, 10.0 mmol I- in 300 mL (.0033 M I-) and .0167 M PbI2 formed.

  18. PbI2(s) Pb2+ + 2 I-  After the complete reaction: 0 M Pb2+, 10.0 mmol I- in 300 mL (.0033 M I-) and .0167 M PbI2 formed. .0167 0 .0033 -x +x +2x .0167-x x .0033+2x Ksp = 1.4x10-8 = x (.0033+2x)2 x = [Pb2+] = 1.3x10-5 [I-] ≈ 3.33x10-2 M

  19. Mg2+(aq) + 2F-(aq) MgF2(s)  150.0 mL  .0100 mmol  1 = 3.75x10-3 M Mg2+ 1 mL 400.0mL 250.0 mL  .0100 mmol  1 = 6.25x10-2 M F- 1 mL 400.0mL 15.17 A solution is prepared by mixing 150.0 mL of 1.00x10-2 M Mg(NO3)2 and 250.0 mL of 1.00x10-1 M NaF. Calculate the concentrations of Mg2+ and F- at equilibrium with solid MgF2 (Ksp=6.4x10-9). = 1.5 mmol Mg2+ = 25.0 mmol F- All 1.5 mmol of Mg2+ will react with 2(1.5) mmol F- leaving 22.0 mmol F- / 400.0 mL = 5.50x10-2M F- Q = [Mg2+][F-]2 = 1.46x10-5 Q > Ksp thus MgF2 will form Now MgF2 will dissociate slightly to reform Mg2+ and F-

  20. MgF2(s) Mg2+(aq) + 2F-(aq)  lots – x 0 + x 5.5x10-2 + x Ksp = 6.4x10-9 6.4x10-9 = x ( 5.5x10-2+ x)2 x = [Mg2+] = 2.1x10-6 M [F-] = 5.5x10-2 M

  21. Selective Precipitation • Mixtures of metal ions are separated with this method • Select the anion which forms a precipitate with only selected ion or ions • For example, in a mixture of Ba2+ and Ag+, select NaCl as AgCl forms a white precipitate but BaCl2 is soluble.

  22. 15.18 A solution contains 1.0x10-4 M Cu+ and 2.0x10-3 M Pb2+. If a source of I- is added gradually to this solution, will PbI2 (Ksp = 1.4x10-8) or CuI (Ksp=5.3x10-12) precipitate first? Specify the concentration of I- necessary to begin precipitation of each salt. Any [I-] > 2.6x10-3 will start the precipitation of PbI2. 1.4x10-8 = Ksp = [Pb2+][I-]2 1.4x10-8 = (2.0x10-3)[I-]2 [I-] = 2.6x10-3 M Any [I-] > 5.3x10-12 will start the precipitation of CuI. 5.3x10-12 = Ksp = [Cu+][I-] 5.3x10-12 = (1.0x10-4)[I-] [I-] = 5.3x10-8 M CuI will precipitate first!

  23. Sulfide Ions • Sulfide ions are often used to precipitate with since sulfide salts have very different solubilities. • A solution with 10-3 M Fe2+ and 10-3 M Mn2+… • FeS (Ksp=3.7x10-19) is much less solubile than… • MnS (Ksp = 2.3x10-13). • Adding S2- to the mixture will precipitate Fe2+ and leave Mn2+ • H2S  H+ + HS- Ka1=1.0x10-7 • HS-  H+ + S2- Ka2=1.0x10-19 • Since HS- is so very weak, S2- has a very high affinity for protons.

  24. HS- H+ + S2- Ka2 = 1.0x10-19  In a mixture with Cu2+ , Hg2+ , Mn2+ , Ni2+ CuS Ksp = 8.5x10-45 HgS Ksp = 1.6x10-54 MnS Ksp = 2.3x10-13 NiS Ksp = 3.0x10-21 Adding H2S to solution will make solution acidic… …attracted to Cu & Hg so they precipitate Then adding OH- to raise pH (more basic) forces more production of S2- (drives reaction to the right) …resulting in more S2- thus precipitating Mn2+ and Ni2+

  25. Qualitative Analysisfor the following cations, separate into 5 groups: K+ Hg22+ NH4+ Ag+ Cd2+ Mn2+ Pb2+ Hg2+ Sn2+ Co2+ Ba2+ Bi3+ Cu2+ Zn2+ Mg2+ Ca2+ Ni2+ Fe2+ Al3+ Na+ Cr3+

  26. Ag+ Pb2+ Hg22+ Precipitate out when dilute HCl is added Hg2+ Cd2+ Bi3+ Cu2+ Sn2+ Precipitate out when H2S is added Co2+ Zn2+ Fe2+ Ni2+ Mn2+ Cr3+ Al3+ Precipitate out as you make solution basic then add more H2S Precipitate out as insoluble hydroxides Mg2+ Ca2+ Ba2+ Precipitate by the addition of CaCO3. Na+ K+ NH4+ Are all soluble but can be distinguished by characteristic colors produced during flame test.

  27. Complex Ions and Their Equilibria • Complex Ion: a metal ion which becomes strongly attached to anions or neutral molecules—the resulting “complex ion” has an overall charge. • The number of ligands attached to the central atom is called the coordination number. • The coordination number is usually 2, 4, or more commonly, 6 • Cu(H2O)42+ • Al(H2O)63+ • Co(H2O)62+

  28. Cu(NH3)42+ Cu2+ + 4NH3  Kinst = [Cu2+][NH3]4 [Cu(NH3)42+] Kform = [Cu(NH3)42+] [Cu2+][NH3]4 Dissociation equation for Cu(NH3)42+: The equilibrium constant for this reaction is called an instability constant, Kinst. The reciprocal of Kinst is called Kform which is equal to Keq. When a complex ion is formed in a solution of an insoluble salt, it reduces the concentration of free metal ion…as a result, more solid must dissolve.

  29. [NH3]o = 100.0 mL  2.0 mmol NH3  1 = 1.0 M 1 mL 200 mL  NH3 + H2O NH4+ + OH-   Ag+ + NH3 Ag(NH3)+ K1 = 2.1x103   Ag(NH3)+ + NH3 Ag(NH3)2+ K2 = 8.2x103  [Ag+]o = 100.0 mL  1x10-3 mmol Ag+  1 = 5.0x10-4M 1 mL 200 mL Consider a solution prepared by mixing 100.0 mL of 2.0 M NH3 with 100.0 mL of 1.0x10-3 M AgNO3. All of the Ag+ will be attracted to the Lewis base, NH3 and leave 5.0x10-4M Ag(NH3)2+ and ~1.0M NH3 unreacted (1.0 - .001) Before any reaction occurs, the major species are Ag+, NO3-, NH3 and H2O. This reaction might occur but since Kb = 1.8x10-5, we should be able to ignore the NH3 used up. with K1=2.1x103 & K2=8.2x103 so big, both go essentially to completion so the net reaction is used.

  30. Ag+ + 2NH3  Ag(NH3)2+ K2 = [Ag(NH3)2+] [Ag(NH3)+][NH3] K1 = [Ag(NH3)+] [Ag+][NH3]  Ag+ + NH3 Ag(NH3)+ K1 = 2.1x103   Ag(NH3)+ + NH3 Ag(NH3)2+ K2 = 8.2x103  Stoichiometry: 0 1.0 5.0x10-4 5.0x10-4 8.2x103 1.0 6.0x10-8 2.1x103 1.0 NH3≈ 1.0 M Ag(NH3)2+ = 5.0x10-4 M Ag(NH3)+ = Ag+ = 6.1x10-8 M   2.9x10-11 M

  31. Calculate the concentrations of Ag+, Ag(S2O3)-, and Ag(S2O3)23- in a solution prepared by mixing 150.0 mL of 1.00x10-3 M AgNO3 with 200.0 mL of 500 M Na2S2O3. The stepwise formation equilibria are: Ag+ + S2O32- Ag(S2O3)- K1=7.4x108 Ag(S2O3)- + S2O32- Ag(S2O3 )23- K2=3.9x104     [Ag+]o = 150.0 mL  1x10-3 mmol Ag+  1 = 4.29x10-4M 1 mL 350 mL [S2O32-]o = 200.0 mL 5.0 mmol S2O32-  1 = 2.86 M 1 mL 350 mL [S2O32-]o >>[Ag+]o and K1 and K2 are big, both formation reactions go to completion Before any reaction occurs, find Ag+ and S2O32- Ag+ + 2S2O32- Ag(S2O3)23- Stoichiometry: 0 ≈2.86 4.29x10-4

  32. Ag+ + 2S2O32- Ag(S2O3)23-  Ag+ + S2O32- Ag(S2O3)- K1=7.4x108 Ag(S2O3)- + S2O32- Ag(S2O3 )23- K2=3.9x104   K2 = 3.9x104 = [Ag(S2O3)23-] [Ag(S2O3)-][S2O32-]  K1 = 7.4x108 = [Ag(S2O3)-] [Ag+][S2O32-] Stoichiometry: 0 ≈2.86 4.29x10-4 4.29x10-4 2.86 3.8x10-9 2.86 S2O32-≈ 2.86 M Ag(S2O3)23- = 4.29x10-4 M Ag(S2O3)- = Ag+ = 3.8x10-9 M 1.8x10-18 M

  33. The End

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