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Chapter 08 Activity and Systematic Treatment of Equilibrium

Chapter 08 Activity and Systematic Treatment of Equilibrium. Contents in Chapter08. PART 1: Activity Radii of Naked and Hydrated Ion Electrolyte Effect on Chemical Equilibrium Ionic Strength of a Solution Activity and Activity Coefficients Thermodynamic Equilibrium Constant

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Chapter 08 Activity and Systematic Treatment of Equilibrium

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  1. Chapter 08 Activity and Systematic Treatment of Equilibrium QCA7e Chapter08

  2. Contents in Chapter08 • PART 1: Activity • Radii of Naked and Hydrated Ion • Electrolyte Effect on Chemical Equilibrium • Ionic Strength of a Solution • Activity and Activity Coefficients • Thermodynamic Equilibrium Constant • Activity Applications in Chemical Equilibria • 1) For Solubility Equilibrium • 2) For pH Calculation QCA7e Chapter08

  3. 1. Radii of Naked and Hydrated Ion The smaller naked ion, attract more solvent molecules QCA7e Chapter08

  4. (e.g., Ca2+(aq)) (e.g., SO42–(aq)) • Inert Electrolyte Effect on Chemical Equilibrium • e.g., CaSO4(s) Ca2+(aq) + SO42–(aq) • Ionic atmosphere: The region of solution around an ion or a charged particle. It contains an excess of oppositely charged ions. • Ionic strength effect = inert electrolyte effect QCA7e Chapter08

  5. Ca2+ SO42– Example1: What is happened if CaSO4(s) dissolved in inert electrolyte KNO3? Solution: δ- δ+ 水合離子周圍吸引反價離子(counter ion) ，在此 Ca2+和SO42–分別受到 NO3–和 K+的影響，阻礙 Ca2+和 SO42–之間的作用力，因此沈澱的比例較去離子水的環境差。 換句話說，CaSO4在 KNO3(aq)溶液中的溶解度會比在去離子水中高。 QCA7e Chapter08

  6. Example 2: For Fe3+ + SCN– Fe(SCN)2+ What is happened if KNO3 was added? Solution: Equilibrium quotient (K’) value decreases if more [KNO3] increased. QCA7e Chapter08

  7. Example 3: Explanation the solubility of potassium hydrogen tartrate in following solution. HOOCC(OH)HC(OH)HCOOK(s) HOOCC(OH)HC(OH)HCOO–(aq) + K+(aq) • Solution: • Adding inert electrolyte, e.g., MgSO4 and NaCl, solubility increase • Adding neutral molecule glucose, no effect • For adding common ion contained KCl, solubility decrease QCA7e Chapter08

  8. Ionic Strength of a Solution • Definition: • Ionic strength (): A quantitative estimation for reporting the ionic composition of a solution: Ci: Molarity (M) of the ith ion Zi: Charge of the ith ion QCA7e Chapter08

  9. Example1: Calculate the ionic strength of 0.10 M NaCl(aq). Solution: Example2: Calculate the ionic strength of 0.10 M Na2SO4(aq). Solution: QCA7e Chapter08

  10. 4. Activity and Activity Coefficients • Define activity: • The effective concentration for obtaining the (true) thermodynamic constants. • Activity vs. concentration: • aA = A[A] • aA: activity of species A • A: activity coefficient of species A • [A]: concentration of species A A is affected by ionic strength (μ) of the system, the hydrated ion size and charge of the interested ionic species. QCA7e Chapter08

  11. Activity Coefficient (γ) • Definition: The coefficient that multiplied by a species’ concentration gives that species’ activity. • Extended Debye-Huckel (for estimating γ if  < 0.1 M) γ: activity coefficient Z: charge of the ion α: effective diameter (pm) of the hydrated ion μ: ionic strength of the solution 0.51 and 305 are constants for aqueous solutions at 25oC QCA7e Chapter08

  12. Example: Calculate the activity coefficient of H+ when  = 0.025 M by Extended Debye-Huckel equation. (Given:  for H+ = 900 pm) Solution: Ans QCA7e Chapter08

  13. iii) Tabulated coefficients (γ) for  < 0.1 M QCA7e Chapter08

  14. QCA7e Chapter08

  15. iv) Interpolate from tabulated values Example: Calculate the activity coefficient of H+ when  = 0.025 M by interpolate from tabulated values. Solution: 0.914 =? H+=0.860 =0.01 0.025 0.05 Ans: H+=0.894 QCA7e Chapter08

  16. 5. Thermodynamic Equilibrium Constant For a chemical reaction: aA(aq) + bB(aq) cC(aq) + dD(aq) Concentration (apparent) equilibrium constant: Thermodynamic (real) equilibrium constant: K is the true constant, K' values depend on the solution’s environment (ionic strength). When μ→0, γ→1，K=K' QCA7e Chapter08

  17. 6. Activity Applications in Chemical Equilibrium 1) For Solubility Equilibrium • A simple solubility problem Example: Calculate the [Ca2+] in a 0.050 M NaClO4 saturated with CaF2(s). (Ksp of CaF2(s) = 3.9x10–11) Solution: Ionic strength is almost controlled by NaClO4, thus =0.050. CaF2(s) Ca2+(aq) + 2F–(aq) initial - 0 0 change - s 2s final - s 2s Ans QCA7e Chapter08

  18. ii) Consider common ion effect Example: Calculate the [Ca2+] in a 0.050 M NaF saturated with CaF2(s). (Ksp of CaF2(s) = 3.9x10–11) Solution: Ionic strength is almost controlled by NaF, thus =0.050. CaF2(s) Ca2+(aq) + 2F–(aq) initial - 0 0.05 change - s 2s final - s 2s+0.05 Assume 2s << 0.05, thus, 2s+0.05=0.05 Yes!! 2s <<0.05!! Ans QCA7e Chapter08

  19. For pH Calculation • i) Pure water at 25 oC Example: Calculate the pH of pure water at 25oC. Solution: *** QCA7e Chapter08

  20. ii) Water containing salt at 25oC Example: Calculate the pH of water containing 0.10 M KCl at 25oC. Solution: QCA7e Chapter08

  21. Contents in Chapter08 • PART 2: Systematic Treatment of Equilibrium • Charge Balance • Mass Balance • Steps for Systematic Approach • Examples of Systematic Treatment of Equilibrium Definition of systematic treatment of equilibrium: A method that uses the charge balance, mass balance(s), and equilibrium constant(s) to completely specify the system’s composition QCA7e Chapter08

  22. 1. Charge Balance 1) Define charge balance: Total concentration of positive charge in a solution equal the total concentration of negative charge. 2) Charge balance equation: n1[C1+n1] + n2[C2+n2] + ..... = m1[A1-m1] + m2[A2-m2] + ..... [Ci+ni]：+ni價Ci離子之濃度(M) [Aj-mj]：-mj價Aj離子之濃度(M) • 3) Remark: • 每一個溶液系統只有一個charge balance equation • 電中性物種不可書寫於charge balance equation • Charge balance equation 左右兩邊應等值，可作為平衡運算結果是否正確的判斷依據。 QCA7e Chapter08

  23. Example: A solution contains H+, OH–, K+, H2PO4–, HPO42–，, and PO43–，what is the charge balance equation? Solution: [H+] + [K+] = [H2PO4–] + 2[HPO42–] + 3[PO43–] + [OH–] • Remark: • 離子所帶之價數，即為該離子之倍數係數 • Charge balance 使用 concentration，不是 activity QCA7e Chapter08

  24. 2. Mass Balance • 1) Define Mass balance: An equation stating that matter is conserved, and that the total amount of a species added to a solution must equal the sum of the amount of each of its possible forms present in solution. • Remark: • 每一個溶液系統可以有許多個 mass balance equations • Mass balance(s) 使用 concentration，不是 activity QCA7e Chapter08

  25. Examples 1 (known concentrations) Q1: What is the mass balance of 0.05M aqueous CHOAc solution A1: ∵ HOAc and OAc– both existed, CHOAc= [HOAc] + [OAc–] the mass balance equation is 0.05M = [HOAc] + [OAc–] Q2: What is the mass balance of 0.025 F H3PO4? A2: CH3PO4= [H3PO4] + [H2PO4–] + [HPO42–] + [PO43–] Mass balance is: 0.025 M= [H3PO4] + [H2PO4–] + [HPO42–] + [PO43–] QCA7e Chapter08

  26. Q3: What is the mass balance of0.025 mole KH2PO4 and 0.030 mole KOH dissolved in 1 L water? A3: a) [K+] = (0.025 + 0.030) M = 0.055 M b) [H3PO4] + [H2PO4–] + [HPO42–] + [PO43–] = 0.025 M QCA7e Chapter08

  27. Examples 2 (unknown concentrations) • Q1: What is the mass balance of La(IO3)3(s) in water? • A1: La(IO3)3(s) La3+(aq) + 3IO3-(aq) • x 3x • Mass balance equation: • [IO3-] = 3[La3+] • Q2: What is the mass balance of Ag3(PO4)(s) in water • Ag3PO4(s) 3Ag+(aq) + PO43-(aq) • 3x x • [Ag+] = 3CPO43- • Mass balance equation: • [Ag+]= 3{ [H3PO4] + [H2PO4–] + [HPO42–] + [PO43–]} QCA7e Chapter08

  28. 3. Steps for Systematic Approach Step 1: Write all relevant equilibrium reactions. Step 2: Write charge balance equation Step 3: Write mass balance equations Step 4: Write equilibrium constants with their expressions. (唯一需使用使用 activity 的項目，但本課程後續仍以 concentration 代替 activity，以避免運算過於繁瑣。) Step 5: Count the number of equations and number of unknown (n 個 unknowns 需至少 n 個 不同之 equations，才能解題) Step 6: 帶入法或聯立方程式或連續運算法，進行解題。 QCA7e Chapter08

  29. 4. Examples of Systematic Treatment of Equilibrium Example 1: Q: Calculate the pH of the pure water A: H2O  H+ + OH– Charge balance: [H3O+] = [OH–] (1) Mass balance: [H3O+] = [OH–] (same as charge balance)(2) Equilibrium constant [H3O+][OH–] = 1.0×10–14(3) (2)代入(3)得 [H3O+] [H3O+] = 1.0×10–14 [H3O+] = 1.0×10–7 pH=7.00 QCA7e Chapter08

  30. Example 2: Q: Calculate the pH of 1.0x10–8 M KOH solution. A: H2O  H+ + OH– KOH  K+ + OH– Charge balance: [K+] + [H3O+] = [OH–] (1) Mass balance: [K+] = 1.0×10–8(2) Equilibrium constant [H3O+][OH–] = 1.0×10–14(3) (2)代入(1)得 [OH–] = 1.0×10–8 + [H3O+] (4) (4)代入(3)得 [H3O+](1.0×10–8 + [H3O+]) = 1.0×10–14 (5) 解一元二次方程式 [H3O+] = 9.6×10–8 pH=7.02 QCA7e Chapter08

  31. Examples: All Exercise： A-G Problems: 1-7, 9-13, 15-20, 22-23 End of Chapter08 QCA7e Chapter08

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