1. Chapter 12 A Deeper Look at Chemical Equilibrium

2. why K a, K w, K sp ↑while adding [NaCl]↑ Effect of electrolyte concentration on concentration-based equilibrium constants

3. 12-1 The effect of ionic strength on solubility of salts

4. The explanation dissociation ↑by↑ionic strength of soln.

5. What do you mean by “ionic strength”? ionic strength = μ= ½ΣCιZι 2 [I]ionic charges example at p. 244

6. Effect of charge on μ ElectrolyteMoralityμex. 1 : 1MMNaCl 2 : 1M3 MNa 2 SO4 3 : 1M6 MAl(NO 3 ) 3 2 : 2M4 MMgSO 4

7. (1) rxn: aA + bB → cC + dD Without considering μ, k = (2) The effect of μ; the [C] activities (A c ) [C] c [D] d [A] a [B] b ∴ k = AAaABbAAaABb ACcADdACcADd γ A a γ B b [A] a [B] b γ C c γ D d [C] c [D] d = 12-2 Activity Coefficients

8. The Debye – H ü ckel eqn.  Fine γ X of the ions from Z X (ionic charge) & α X (average size) -log γ X = 1 + 3.3 α X 0.51 Z X 2 α X = effective diameter of the hydrated ion X (10 -9 m) works fairly well for   0.1 M

9.  for most single charged ions : α X ≈ 0.3 nm -log γ X ≈  the larger charged & smaller ions ;  the larger α X ( Table 12-1 )  For μ< 0.01  3.3 α X << 1  1 + 3.3 α X ≈ 1 ∴ -log γ X = 0.51 Z X 2 1 + 0.51

12. Effect of ionic strength   increase  decrease. at μ→ 0, γ → 1, A x → [x] (2) at μ= const. ion with larger charge, the bigger △ γ ex. △ γ BaSO4 > △ γ AgCl at μ = k (3) at any μ, for same charge ions: γ X ≈ γ Y their difference (minor) could be from at any effective diameter of hydrated ion was formed

13. How to interpolate   can be also found for  that is between values in Table 12-1 by using linear interpolation.

14. High ionic strengths at high μ(μ> 0.1 M), γ↑& could > 1

15. Ex. at p. 249 for a better estimate of solubility of PbI 2 PbI 2 (s)  Pb 2+ + 2I - x 2x (2) From , interpolation in Table 12-1 to get  Pb2+ &  I-, and Ksp = γ Pb 2+ γ I - 2 [Pb 2+ ][I - ] 2  find x2 &  (1) Ksp = x(2x) 2 = 7.9 x 10 -9  find x1 &  (3) From  2 & Ksp = γ Pb 2+ γ I - 2 [Pb 2+ ][I - ] 2  find x3 Compare x1, x2, and x3

16. Real definition of pH pH = - logA H+ = - log([H + ]  H+ ) Ex: at P. 251

17. Ex: What weight of Na 2 HPO 4 and KH 2 PO 4 would be required to prepare 200 mL of a buffer solution of pH 7.40 that has an ionic strength of 0.20?

18. Sol: (1)pH = pK a2 + log [HPO 4 -2 ]/[ H 2 PO 4 - ] 7.40 = 7.20 + log X/Y (2) µ = 1/2 Σ CiZi 2 0.20 = 1/2 {[Na + ](1) 2 + [X](2) 2 + [K + ](1) 2 + [Y](1) 2 } = 3X + Y 解聯立方程式 (1) and (2)

19. 12-3 Charge & Mass Balances Ex. Consider the charge balance for a solution prepared by weighing out 0.0250 mol of KH 2 PO 4 plus 0.0300 mol of KOH and diluting to 1.00 L. Solution: H 2 PO 4 -  H + + HPO 4 2- HPO 4 2-  H + + PO 4 3- [K + ] = 0.0300 + 0.0250 = 0.0550M [HPO 4 2- ] = 0.020M [PO 4 3- ] = 0.005M charge balance: [H + ] + [K + ] =[OH - ] + [H 2 PO 4 - ] + 2[HPO 4 2- ]+ 3[PO 4 3- ]

20. Mass balance K 2 HPO 4 in H 2 O Mass balance: K 2 HPO 4  2 K + + HPO 4 - HPO 4 2-  H + + PO 4 3- HPO 4 2- + H +  H 2 PO 4 - H 2 PO 4 - H +  H 3 PO 4 [K + ] = 2 {[H 3 PO 4 ] + [H 2 PO 4 - ] + [HPO 4 2- ] + [PO 4 3- ]}

21. 12-4 Systematic Treatment of Equilibrium Ex. 1 at p.254 pH of 10 -8 M KOH  KOH  K + + OH - H 2 O  H + + OH - Sol: Charge balance: [K + ] + [H + ] = [OH - ] Mass balance: [K + ] = 1.0 ×10 -8 M K balance: K w = [H + ]γ H + x [OH - ]γ OH - ∵ μ is very low in this soln., so γ≈ 1 K w = [H + ](1.0 ×10 -8 + [H + ]) [H + ] 2 + 1.0 ×10 -8 + [H + ] - 1×10 -14 = 0 [H + ] = = 9.6 ×10 -8 or -1.1 ×10 -7 M pH = -log (9.6 ×10 -8 ) = 7.02

22. Ex. 2 at P.255 Solubility of CaF 2 CaF 2(s)  Ca 2+ + 2F - Ksp = 3.9 ×10 -11 F - + H 2 O  HF + OH - Kb = 1.5 ×10 -11 H 2 O  H + + OH - Kw = 1 ×10 -14 Sol: Charge balance: 2[Ca 2+ ] + [H + ] = [F - ] + [OH - ] Mass balance: 2[Ca 2+ ] = [F - ] + [HF] K balance: Ksp = [Ca 2+ ]γ Ca 2+ x [F - ] 2 (γ F - ) 2 = 3.9 ×10 -11 Kb = = 1.5 ×10 -11 Kw = [H + ]γ H + x [OH - ]γ OH - = 1.0 ×10 -14

23. Ignore μ & allγ= 1, (check later), & use buffer pH=3  [H + ] = 1 ×10 -3 ∴ [OH - ] = 1 ×10 -11 From Kb: 1.5 ×10 -11 =  = 1.5 From mass balance: 2[Ca 2+ ] = [F - ] + 1.5[F - ] = 2.5[F - ] [Ca 2+ ] = 1.25[F - ] From Ksp: 1.25[F - ] 3 = 3.9 ×10 -11 [F - ] = = 3.15 ×10 -4 ∴ [Ca 2+ ] = 3.9 ×10 -4

24. 12-5 Fraction Composition Equilibrium HA  H + + A - F F = formal conc. = [HA] + [A - ] Ka = = [HA]Ka = [H + ]F – H + [HA] [HA] (Ka + H + ) = [H + ]F