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Chapter 18: Solubility and Simultaneous Equilibria

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Chapter 18: Solubility and Simultaneous Equilibria

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  1. Chapter 18: Solubility andSimultaneous Equilibria Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop

  2. Solubility of Salts • Precipitation reactions (Chapter 5) • Aqueous metathesis reactions in which one product is insoluble in water • CaCl2(aq) + Na2CO3(aq) CaCO3(s) + 2NaCl (aq) • Insoluble compound • Compound having water solubility of less than 0.01 mole of dissolved material per liter of solution • Solubility < 0.01 M

  3. Solubility of Salts • Chapter 5 Solubility Rules • Guidelines for what is insoluble • Doesn’t mean compound won’t dissolve at all • Just not very much • Now want to • Quantitate solubilities • Explore conditions under which some compounds precipitate and others don’t • Applications in separation of ions • Especially toxic metal ions such as Hg2+, Tl3+, U3+, etc.

  4. Why Study Solubility? • Tooth decay • Acids from foods dissolve enamel, [Ca5(PO4)3OH], hydroxyapatite • Reduced by fluoride which replaces OH– to form fluorapatite, [Ca5(PO4)3F], and CaF • Lower solubility means it doesn’t dissolve as readily in acid • Upper and lower gastrointestinal series • X ray of upper and lower gastrointestinal tract • Clarified by barium sulfate—very insoluble • BaSO4 toxic, but safe, as it doesn’t dissolve

  5. Solubility Equilibria • Solids in equilibrium with ions in solution • When ionic salt dissolves in water • Dissociates into separate hydrated ions • Initially, no ions in solution • CaF2(s)  Ca2+(aq)+ 2F–(aq) • As dissolution occurs, ions build up and collide • Ca2+(aq)+ 2F–(aq) CaF2(s) • At equilibrium • CaF2(s) Ca2+(aq)+ 2F–(aq) • Now have saturated solution

  6. Solubility and Solubility Product Solubility • Amount of salt that dissolves in given amount of solvent to give saturated solution • Concentration, M, g/100 mL, etc. Solubility product • Name for the equilibrium law for solubility reactions • Equilibrium constant, Ksp • Only one value for given solid at given temperature Temperature dependence • Solubility and thus Ksp changes with T

  7. Solubility of Salts • Consider AgCl in water • Only a very small amount dissolves • Equilibrium exists when solution is saturated • AgCl(s) Ag+(aq) + Cl–(aq) • Equilibrium law: • Ksp = [Ag+][Cl–] Ksp = solubility product constant • Solubility equilibrium • Reflects solubility of compound • Product of ion concentrations

  8. Ion Product vs. Solubility Product • For: AxBy(s)xAy+(aq) + yBx–(aq) • Solubility Product Ksp = [Ay+]x[Bx–]y • Ion product value for saturated solution • System is at equilibrium • Ion ProductQsp = [Ay+]x[Bx–]y • Same form as solubility product, except system is need not be at equilibrium. • Any dilution of salt that results in an unsaturated solution where the Qsp is less than the Ksp • Mixtures may, initially, have Qsp greater than Ksp

  9. Writing Ksp Equilibrium Laws • For: AxBy(s)x Ay+(aq) + y Bx–(aq) • Ksp = [Ay+]x[Bx–]y More examples BaSO4(s)Ba2+(aq) + SO42–(aq)Ksp = [Ba2+][SO42–] CaF2(s)Ca2+(aq) + 2F–(aq)Ksp = [Ca2+][F–]2 Ag2CrO4(s)2Ag+(aq) + CrO42–(aq)Ksp = [Ag+]2[CrO42–] AuCl3(s)Au3+(aq) + 3Cl–(aq)Ksp = [Au3+][Cl–]3

  10. Calculation using Ksp and Molar Solubilities Molar solubility • Moles of salt dissolved in one liter of saturated solution • The salt that dissolves dissociates 100% • There must be there is some solid present • Quantity is not important • Assume it is there if solution is saturated • Solid is not included in mass action expression Two Examples: A. Given solubilities, Calculate Ksp B. Given Ksp, Calculate Solubility

  11. Molar Solubility and Ksp Problems Strategy for solving: • Write balanced equation for dissociation of salt • Write equilibrium law • Get Ksp for salt (from table), if needed • Construct the concentration table 5a.Given solubility, x, calculate Ksp Or 5b. Given Ksp and saturated solution, solve for x = solubility

  12. A.Given Solubilities, Calculate Ksp Ex. At 25 °C, the solubility of AgCl is 1.34 × 10–5 M. Calculate the solubility product for AgCl. AgCl(s) Ag+(aq) + Cl–(aq) Ksp = [Ag+][Cl–] 1.34 × 10–5 M 1.34 × 10–5 M 1.34 × 10–5 M 1.34 × 10–5 M Ksp = (1.34 × 10–5)(1.34 × 10–5) Ksp = 1.80 × 10–10

  13. Your Turn! The solubility of a salt, A2B, is found to be 3.0 × 10–5 M. What is the value ofKsp? A. 1.8 × 10–14 B. 5.4 × 10–9 C. 2.4 × 10–23 D. 1.7 × 10-21 A2B 2A+ + B2– [A] = 2(3.0 × 10–5) [B] = 3.0 × 10–5 Ksp = [A]2[B] = (6.0 × 10–5)2(3.0 × 10–5) Ksp = 1.8 × 10–14

  14. B. Given Ksp Calculate Solubility Ex. What is the molar solubility of CuI in water? Determine the equilibrium concentrations of Cu+ and I–? Step 1. Write balanced equation for dissociation of salt CuI(s) Cu+(aq) + I–(aq) Step 2. Write equilibrium law Ksp = [Cu+][I–] Step 3. Ksp for salt Ksp =1.3  10–12

  15. Ex. Molar Solubilities from Ksp Step 4. Construct concentration table Step 5. Solve Ksp expression • Ksp = 1.3 × 10–12 = (x)(x) • x2 = 1.3 × 10–12 • x= 1.1 × 10–6 M= calculated molar solubility of CuI = [Cu+] = [I–] No entriesNo entries No entries +x +x x x

  16. Your Turn! Given Ksp = 1.2 × 10–7 for BaC2O4, calculate the molar solubility of this salt. • 1.4 × 10–14M • 3.5 × 10–4M • 6.0 × 10–8M • 1.2 × 10–7M BaC2O4(s) Ba2+(aq) + C2O42–(aq) Ksp = 1.2 × 10–7 = [Ba2+][C2O42–] 1.2 × 10–7 = (x )(x) x = 3.5 × 10–4

  17. B. Given Ksp, Calculate Solubilities Ex. Calculate the solubility of CaF2 in water at 25 °C, if Ksp = 3.4 × 10–11. CaF2(s) Ca2+(aq) + 2F– (aq) Step 1: Write equilibrium law Ksp = [Ca2+][F–]2 Step 2: Construct concentration table +x +2x x 2x

  18. Ex. Molar Solubilities from Ksp Step 3. Solve the Ksp expression • Ksp = [Ca2+][F–]2= (x) (2x)2 • 3.4 × 10–11 = 4x3 x = 2.0 × 10–4M= molar solubility of CaF2 [Ca2+] = x = 2.0 × 10–4M [F–] = 2x = 2(2.0 × 10–4M) = 4.0 × 10–4M

  19. Your Turn! What is the solubility of PbCl2 in grams per 100.0 mL at 25°C? Ksp = 1.7 × 10–5, PbCl2 MM = 278.11 g/mol A. 0.56 g B. 0.72 g C. 0.45 g D. 0.39 g PbCl2(s) Pb2+(aq) + 2Cl–(aq) Ksp = 1.7 × 10–5 = [Pb2+][Cl–]2 = x(2x)2 x = 1.62 × 10–2 M = 0.45 g PbCl2 /100 mL

  20. Relative Solubilities • Ksp gives information about solubility of salts • Must be careful when comparing relative solubilities • Two possible cases when comparing: • Compare salts that have the same number of ions AgI(s) Ksp = 8.5 × 10–17 CuI(s) Ksp = 1.3 × 10–12 CaSO4(s) Ksp = 4.9 × 10–5 • Each salt dissolves to produce two ions • Salt cation + anion • Ksp = [cation][anion]

  21. Relative Solubilities • Compare salts that contain the same number of ions • If solubility = x • Then [cation] = [anion] = x • Ksp = x2 • We conclude that Ksp is directly related to solubility • As Ksp increases so will solubility • CaSO4 > CuI > AgI most soluble least soluble largest Ksp smallest Ksp

  22. Relative Solubilities 2. Compare salts with different number of ions • Each produces different number of ions • Each uses different Ksp expression • Solubility related to Ksp is not easy to judge without calculating solubilities • e.g. Bi2S3 > Ag2S > CuS Most soluble least soluble

  23. Common Ion Effect • Common ion • Ion in solution that is supplied by more than one solute • Common ion effect • Lowering of solubility of ionic compound by addition of common ion

  24. Common Ion Effect • Up until now all calculations in pure water • What happens if another salt, containing one of the ions in our insoluble salt, is added to a solution? Consider PbI2(s) Pb2+(aq) + 2I–(aq) • Saturated solution of PbI2 in water • Filter • Add KI • PbI2 (yellow solid) precipitates out • Why? • Le Chatelier’s Principle • Add product I– • Equilibrium moves to left and solid PbI2 forms

  25. Ex. Common Ion Effect Consider three cases • What is the molar solubility of Ag2CrO4 in pure water? • What is the molar solubility of Ag2CrO4 in 0.10 M AgNO3? • What is the molar solubility of Ag2CrO4 in 0.10 M Na2CrO4? • Ag2CrO4(s)2Ag+(aq) + CrO42–(aq) • Ksp = [Ag+]2[CrO42–] = 1.1 × 10–12

  26. Ex. Common Ion Effect A.What is the solubility of Ag2CrO4 in pure water? +2x +x 2x x Ksp = [Ag+]2[CrO42–] = (2x)2(x) = 1.1  10–12 = 4x3 x = Solubility of Ag2CrO4 = 3.0 × 10–4M [CrO42–] = x = 3.0 × 10–4 M [Ag+] = 2x = 6.0 × 10–4M

  27. Ex. Common Ion Effect B.What is the molar solubility of Ag2CrO4 in 0.10 M AgNO3 solution? Ksp = 1.1 × 10–12 Ksp = 1.1 × 10–12= (0.10 M)2[x] • x = Solubility of Ag2CrO4 = 1.1 × 10–10M • [Ag+] = 0.10 M • [CrO42–] = 1.1 × 10–10M +2x +x x ≈0.10

  28. Your Turn! What effect would adding copper(II) nitrate have on the solubility of CuS? A. The solubility would increase B. The solubility would decrease C. The solubility would not change

  29. Ex. Common Ion Effect C. What is the solubility of Ag2CrO4 in 0.100 M Na2CrO4? +2x +x 2x ≈0.10 Ksp = (2x)2(0.10) = 1.1  10–12 = 4x2(0.10) x = Solubility of Ag2CrO4 = 1.6 × 10–5M [CrO42–] = 0.10 + x = 0.10 + 1.6 × 10–5 M = 0.10 M [Ag+] = 2x = 3.2 × 10–5M

  30. Ex. Common Ion Effect What have we learned about the solubility of silver chromate? • Dissolving it in pure water the solubility was 3.0 × 10–4M • Dissolving it in AgNO3 solution solubility was1.1 × 10–10M • Dissolving it in Na2CrO4 solution solubility was3.2 × 10–5M • Common ion appearing the most in the formula of the precipitate decreases the solubility the most

  31. Your Turn! The molar solubility of PbF2 in 0.10 M Pb(NO3)2 solution is 2.85 × 10–4M. What is Ksp for PbF2? A. 1.2 × 10–6 B. 3.1 × 10–7 C. 9.6 × 10–13 D. 3.8 × 10–12 PbF2(s) Pb2+(aq)+ 2F–(aq) [Pb2+] = 0.10 M [F–] = 2(2.85 × 10–4) M Ksp= [Pb2+][F–]2 = (0.1)(5.7 × 10–4)2 Ksp = 3.2 × 10–8

  32. Your Turn! What is the molar solubility of PbI2 in pure water? Ksp = 9.8 × 10–9 A. 2.1 × 10–3 B. 1.7 × 10–3 C. 4.9 × 10–5 • 1.4 × 10–3 PbI2(s ) Pb2+ + 2I– Ksp = 9.8 × 10–9 = [Pb2+][I–] = x (2x)2 x = 1.3 × 10–3 M

  33. Your Turn! What is the molar solubility of PbI2 in 0.20 M NaI solution? Ksp = 9.8  10–9 • 3.9 × 10–10 • 2.4 × 10–7 • 6.1 × 10–9 • 2.4 × 10–7 PbI2(s ) Pb2+ + 2I– [I–] = 0.20 M Ksp = 9.8 × 10–9 = [Pb2+][I–]2 = [Pb2+](0.20)2 [Pb2+] = 2.45 × 10–7 M Molar solubility = [Pb2+] = 2.4 × 10–7 M

  34. Your Turn! What is the molar solubility of PbI2 in 0.20 MPb(NO3) solution? Ksp = 9.8  10–9 • 4.9 × 10–8 • 3.7 × 10–3 • 1.1 × 10–4 • 2.2 × 10–4 PbI2(s ) Pb2+ + 2I– [Pb2+] = 0.20 M Ksp = 9.8 × 10–9 = [Pb2+][I–]2 = (0.20) [I–]2 [I–] = 2.2 × 10–4 M Molar solubility of PbI2 = ½ [I–] = 1.1 × 10–4 M

  35. Your Turn! From the previous two Your Turn problems, what can you conclude about the relative effect of added Pb2+ vs. added I–? A. Adding Pb2+ increases the solubility of PbI2 more than I–. B. Adding I– decreases the solubility of PbI2 more than Pb2+. C. Adding Pb2+ decreases the solubility of PbI2 more than I–. D. They both have the same effect on the solubility of PbI2.

  36. Predicting if Precipitate will Form • In making a solution containing various ions, will the salt precipitate at the given concentrations? • For precipitate of salt to form, solution must be supersaturated. • Same as asking if given mixture of reactants and products is an equilibrium mixture • Can judge by calculating ion product, Q • Comparing with solubility product, Ksp

  37. Predicting if Precipitate will Form • For: AxBy(s)xAy+(aq) + yBx–(aq) • Ion Product Qsp = [Ay+]x[Bx–]y • Solubility Product Ksp = [Ay+]x[Bx–]y

  38. Predicting Precipitation Ex. Will a precipitate of PbI2form if 100.0 mL of 0.0500 M Pb(NO3)2 are mixed with 200.0 mL of 0.100 M NaI? PbI2(s) Pb2+(aq) + 2I–(aq) Ksp = [Pb2+][I–]2 = 9.8 × 10–9 Strategy for solving • Calculate concentrations in mixture prepared • Calculate Qsp = [Pb2+][I–]2 • Compare Qsp to Ksp

  39. Ex. Predicting Precipitation Step 1. Calculate concentrations • Here diluting 100.0 mL to 300.0 mL • Vtotal = 100.0 mL + 200.0 mL = 300.0 mL • [Pb2+] = 1.67 × 10–2M • [I–] = 6.67 × 10–2M

  40. Ex. Predicting Precipitation Step 2. Calculate Qsp • Qsp = [Pb2+][I–]2 = (1.67 × 10–2M)(6.67 × 10–2M)2 • Qsp =7.43 × 10–5 Step 3. Compare Qsp and Ksp • Qsp = 7.43 × 10–5 • Ksp = 9.8 × 109 • Qsp > Ksp so precipitation will occur • How much precipitate will form and what will ion concentrations be at the end? • Ksp small • So most ions precipitate out as PbI2 • Reaction essentially goes to completion

  41. Ex. Predicting Precipitation • If reaction goes essentially to completion, what will final equilibrium concentrations be? • Do limiting reactant calculations for precipitate formation • Then do equilibrium calculations to determine ion concentrations in solution. Step 1. Stoichiometric Calculation 5.00 – 5.00 = 0.00 mmol 20.00 – 2(5.00) = 10 mmol

  42. Ex. Predicting Precipitation Step 2. System at equilibrium • Some small amount of PbI2 redissolves to form equilibrium [Pb2+] • Basically a common ion problem 3.33 × 10–2M+ 2x ≈ 3.33 × 10–2M +x Ksp= 9.8 × 10–9 = [Pb2+][I–]2 = (x)(3.33 × 10–2)2

  43. Ex. Predicting Precipitation • 3.33 × 10–2M >> 2x, so approximation valid • Final equilibrium concentrations of Pb2+ and I– in 100.0 mL of 0.0500 M Pb(NO3)2 mixed with 200.0 mL of 0. 100 M NaI • [I–] = 3.33 × 10–2M • [Pb2+] = 1.3 × 10–5M

  44. Ex. Predicting Precipitation Suppose you mix 100.0 mL of 0.200 M BaCl2 with 50.0 mL of 0.0300 M Na2SO4. Will BaSO4precipitate? BaSO4(s) Ba2+(aq) + SO42–(aq)Ksp = 1.1 × 10–10 • Ksp = [Ba2+][SO42–] Step 1. Calculate concentrations • [Ba2+] = 0.133 M

  45. Ex. Predicting Precipitation [SO42–] = 0.0100 M Step 2. Calculate Qsp • Qsp = [Ba2+][SO42–] = (0.133)(0.0100) • Qsp = 1.33 × 10–3 Step 3. Compare Qsp and Ksp • Ksp = 1.1 × 10–10 • 1.33 × 10–3 >> 1.1 × 10–10 • Qsp >> Ksp • So BaSO4 will precipitate

  46. Solubility of Acid and Basic Salts • Acid Salts: • Many acid salts have NH4+ as the conjugate acid and are very soluble. Some metal cations are slightly acidic but that is beyond our scope. • Basic Salts: • Anions from weak acids are conjugate bases (e.g. F–) and we have worked equilibrium problems with them. Solubility of these salts involves the solubility process and the conjugate base ionization constant.

  47. Solubility of Acid and Basic Salts • Examples of Basic Salts: (all in aq sol’n) • PbF2 Pb2+ + 2F– Ksp • 2F– + 2H2O 2HF + 2OH– Kb2 • PbF2 +2H2O 2HF + 2OH– +Pb2+Koverall = KspKb2 • Koverall = [HF]2[OH–]2[Pb2+] = (2s)2(2s)2(s) =16s5 • BaSO4 Ba2+ + SO42– Ksp • SO42– + H2O HSO4– + OH– Kb1 • BaSO4+ H2O Ba2++ HSO4– + OH– Koverall = KspKb1 • Koverall = [Ba2+][HSO4–][OH–] = (s)(s)(s) = s3 • (Note: we’ve assumed that [OH–] is much less than that contributed by water)

  48. pH and Solubility • Mg(OH)2(s)Mg2+(aq) + 2OH–(aq) • Increase OH– shift equilibrium to left • Add H+shift equilibrium to right • Le Châtelier’s Principle • Ag3PO4(s)3Ag+(aq) + PO43–(aq) • Add H+ increase solubility • H+(aq) + PO43–(aq)HPO42–(aq) • AgCl(s)Ag+(aq) + Cl–(aq) • Adding H+ has no effect on solubility Why? • Cl– is very, very weak base, so neutral anion • So adding H+ doesn’t effect Cl– concentration

  49. Metal Oxides Undergo Reaction with Water • Usually ignore reaction of ionic solid with water • If the anion of a salt is very basic, a subsequent reaction of the anion with water occurs • Such is the case of many metal oxides • Kb for O2– = 1 × 1022 • So Ag2O actually dissociates to form Ag+ and OH– • Kspvalue listed takes this subsequent reaction into account Ag2O(s) 2Ag+(aq) + O2–(aq)Ksp O2–(aq) + H2O 2OH–(aq)Kb Ag2O(s) + H2O 2Ag+(aq) + 2OH–(aq)Knet

  50. Metal Sulfides Also Undergo Reaction With Water • Sulfide ion (S2–) is also very basic • Doesn’t exist in aqueous solution • Metal sulfides also undergo a subsequent reaction with water Ag2S(s) 2Ag+(aq) + S2–(aq)Ksp S2–(aq) + H2O OH–(aq) + HS–(aq)Kb1 Ag2S(s) + H2O 2Ag+(aq) + OH–(aq) + HS–(aq)Ksp Actual Ksp =[Ag+]2[OH–][HS–]