CM4106 Review of Lesson 4

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CM4106 Review of Lesson 4

  1. 1. CM4106 Chemical Equilibria & ThermodynamicsLesson 4Solubility EquilibriaA Chemistry Education Blog by Mr Tanhttp://chemistry-mr-tan-yong-yao.blogspot.sg/
  2. 2. Solubility of Common Salts Courtesy of Chemistry, 10th Edition by Raymond Chang
  3. 3. Fundamentals: AxBy (s) ⇌ xAy+(aq) + yBx(aq)• The equilibrium constant for this heterogeneous equilibrium is called the solubility product, Ksp, is written as: Ksp = [Ay+(aq)]xeqm [Bx-(aq)]yeqm• The solubility product, Ksp, of a sparingly soluble salt is defined as the product of the concentration of the ions (in M) in a saturated solution at a given temperature raised to the power of its coefficient in the equilibrium equation. Page 69
  4. 4. Calculating Ksp value from solubility• The solubility of AgCl at 18°C is 1.46 x 10-3 g/L, what is the solubility product of AgCl at 18°C? Since the solubility of AgCl = 1.46 x 10-3 g / L; [Ag+] = (1.46 x 10-3 g/L) / (143.5 g/mol) = 1.017 x 10-5 mol/L AgCl (s) ⇌ Ag+ (aq) + Cl (aq) Initial / M - 0 0 Change / M -s +s +s Eqm / M - s s Ksp = [Ag+] [Cl] = (1.017 x 10-5)2 = 1.03 x 10-10 The solubility product of AgCl at 18°C is 1.03 x 10-10
  5. 5. Calculating Solubility from Ksp• The Ksp for Ag2CO3 is 8.0 x 10-12 M3, calculate its solubility at this temperature. Let s be the solubility of Ag2CO3 in mol / L Ag2CO3(s) ⇌ 2Ag+(aq) + CO32(aq) Initial / M - 0 0 Change / M -s +2s +s Eqm / M - 2s s Ksp = [Ag+]2[CO32] 8.0 x 10-12 = [2s]2[s] 8.0 x 10-12 = 4 s3 s = 1.26 x 10-4 M = 1.3 x 10-4 M (2 s.f.) Solubility of Ag2CO3 = 1.3 x 10-4 M
  6. 6. Quantitative ProblemsCalculate Ksp given solubilityCalculate solubility given Ksp
  7. 7. Relationship between Ksp and Solubility (M)
  8. 8. Predicting Precipitation - Will a PPTform?Knowing the value of Ksp allows us to predict if a ppt will beformed when two solutions containing ions to form an insolublesalt are mixed.Step 1: Calculate reaction quotient QspStep 2: Compare Qsp with Ksp Inference Q>K Precipitation occurs till Q = Ksp No precipitation is seen because the solution is not saturated Q<K with the ions hence they remain dissolved in the solution Q=K A saturated solution is obtained Page 75
  9. 9. Predicting Precipitation (Ksp vs Q) Will a precipitate form when 0.10 L of 8.0 x 10-3 M Pb(NO3)2 is added to 0.40 L of 5.0 x 10-3 M Na2SO4? (Ksp = 6.3 x 10-7) REMEMBER: TAKE INTO ACCOUNT OF CONCENTRATION OF IONS IN MIXTURE (DILUTION UPON MIXING) PbSO4 (s) ⇌ Pb2+ (aq) + SO42- (aq)[Pb2+] = (0.10/0.50) x 8.0 x 10-3 [SO42-] = (0.40/0.50) x 5.0 x 10-3 = 0.0016 M = 0.0040 M Q = [Pb2+][SO42-] = (0.0016) (0.0040) = 6.4 x 10-6 > Ksp = 6.3 x 10-7 Q > Ksp  ppt will form
  10. 10. Factors affecting solubility1. Common Ion Effect2. pH of solution3. Formation of Complexes
  11. 11. 1. Common Ion EffectSolubility of a substance is affected by the presence of othersolutes, especially if there is a common ion present CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)What happens to the solubility of CaF2 if the solution alreadycontains Ca2+ (aq) or F- (aq)?1. Equilibrium position shifts to the left2. Solubility of CaF2 decreases
  12. 12. 1. Common Ion Effect Calculate the molar solubility of CaF2 at 25 oC in 0.010 M Ca(NO3)2 solution CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq) Initial / M - 0.010 0 Change / M - +s + 2s Eqm / M - 0.010 + s 2sKsp = [Ca2+][F-]2 = (0.010 + s)(2s)2Since s is small, assume 0.010 + s ≈ 0.010(CaF2 is a sparingly soluble salt and the solubility is further suppressed by thepresence of common ion effect)3.9 x 10-11 = (0.010)(2s)2 Assumption is valid, s << 0.010s = solubility = 3.1 x 10-5 mol/L
  13. 13. 2. pH of Solution If a substance has a basic anion, it will be more soluble in an acidic solution. If a substance has an acidic cation, it will be more soluble in an basic solution.
  14. 14. 2. pH of Solution Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH- (aq) Ksp = 1.8 x 10-11 (25 oC)Calculate the molar solubility of Mg(OH)2 in pure water.What is the pH of the resulting solution? Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH- (aq) Initial / M - 0 0Ksp = [Mg2+][OH-]2 = (x)(2x)2 Change / M - +x +2x1.8 x 10-11 = 4x3 Eqm / M - x 2xx = Solubility = 1.651 x 10-4 mol/L = 1.7 x10-4 M (2 s.f.)[OH-] = 2x = 3.302 x 10-4 MpOH = - lg(3.302 10-4) = 3.48pH = 10.5 (1 d.p.) Page 72
  15. 15. 2. pH of SolutionWhat is the solubility of Mg(OH)2 in a less alkaline solution bufferedat pH 9?pOH = 5 Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH- (aq)[OH-] = 1.0 x 10-5 M Initial / M - 0 1.0 x 10-5 Change / M - +x - Eqm / M - x 1.0 x 10-5 Ksp = [Mg2+] [OH-]2 Ksp = 1.8 x 10-11 (25 oC) = [Mg2+](1.0 x 10-5)2 [Mg2+] = 0.18 M (2 s.f.) Page 73
  16. 16. With theaddition of acid
  17. 17. 3. ComplexationWhen aqueous ammonia is added dropwise to a Cu2+ solution, a light blue precipitate isobserved. As aqueous ammonia is added in excess, the precipitate dissolves to give a deepblue solution. Page 74
  18. 18. 3. Complexation1. When aqueous ammonia is added dropwise to a Cu2+ solution, a light blue precipitate is observed. As aqueous ammonia is added in excess, the precipitate dissolves to give a deep blue solution. NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH (aq)When added to Cu2+(aq) solution, Cu2+(aq) + 2OH (aq) ⇌ Cu(OH)2(s) – (1)Hence a blue precipitate is observed.As more NH3(aq) is added, Cu2+(aq) + 4NH3(aq) ⇌ Cu(NH3)42+(aq)[Cu2+] drops as Cu(NH3)42+(aq) is formed, thus position of equilibrium in equation1 shifts left to produce more Cu2+(aq), hence Cu(OH)2(s) dissolves. [Cu(H2O)6]2+ + 2 OH- (aq) → [Cu(H2O)4(OH)2] (s) [Cu(H2O)6]2+ + 4 NH3 (aq) → [Cu(NH3)4(H2O)2]2+ (aq) Page 74
  19. 19. Ksp values

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