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Chapter 18 Principles of Reactivity: Other Aqueous Equilibria

Chapter 18 Principles of Reactivity: Other Aqueous Equilibria. The Common-ion Effect Buffer Solutions Composition and action of buffers Buffer capacity and pH Buffer calculations Acid-Base Titrations Strong acid - strong base titrations Weak acid titrated with strong base

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Chapter 18 Principles of Reactivity: Other Aqueous Equilibria

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  1. Chapter 18 Principles of Reactivity: Other Aqueous Equilibria

  2. The Common-ion Effect • Buffer Solutions • Composition and action of buffers • Buffer capacity and pH • Buffer calculations • Acid-Base Titrations • Strong acid - strong base titrations • Weak acid titrated with strong base • Weak base titrated with strong acid • Titration of polyprotic acids • Acid-base indicators • Solubility Equilibria • Ksp and Molar Solubility • Factors Affecting Solubility • Selective Precipitation of Ions

  3. adding A– pushes equilibrium to the left adding BH+ pushes equilibrium to the left I. The Common-Ion Effect HA H3O+ + A– “common ion” BH+ + OH– B e.g. What is the pH of a) 1.0 M HF? (Ka = 6.8 x 10–4) b) 1.0 M HF & 1.0 M NaF?

  4. A. Composition and action of buffers acid buffer: solution of a weak acid HA and its conjugate base A– base buffer: solution of a weak base B and its conjugate acid BH+ usually high, similarconcentrations of each ([HA]/[A–], [B]/[BH+]  0.1 to 10) II. Buffer Solutions = solutions that resist changes in pH when small amounts of strong acid or strong base are added to them

  5. high [A–] both equilibria lie very far to the left high [HA]  Ka = or [H3O+] = Henderson-Hasselbalch equation for acid buffers  pH = pKa + log [H3O+][A–] [HA] [A–] [HA] Ka·[HA] [A–] pH = II. Buffer Solutions A. Composition and action of buffers 1. acid buffers HA H3O+ + A– A– OH– + HA  assume [HA]eq = [HA]i [A–]eq = [A–]i e.g., Solution of 1.0 M HC2H3O2 and 1.0 M NaC2H3O2 (pKa = 4.74)

  6. B OH– + BH+ high [BH+] both equilibria lie very far to the left high [B] BH+ H3O+ + B  Kb = or [OH–] = Henderson-Hasselbalch equation for base buffers  pOH = pKb + log [BH+][OH–] [B] [BH+] [B] Kb·[B] [BH+] pOH = II. Buffer Solutions A. Composition and action of buffers 2. base buffers  assume [B]eq = [B]i [BH+]eq = [BH+]i e.g., Solution of 1.0 M NH3 and 0.5 M NH4Cl (pKb = 4.74)

  7. either way, log doesn’t change much [A–] [HA] II. Buffer Solutions A. Composition and action of buffers 3. buffering action acid buffer: HA & A– add strong acid (H3O+): H3O+ + A– HA + H2O add strong base (OH–): OH– + HA  A– + H2O

  8. Buffer Water II. Buffer Solutions A. Composition and action of buffers 3. buffering action e.g., Buffer: 1.0 L of 1.0 M HC2H3O2 and 1.0 M NaC2H3O2 (pKa = 4.74) initial: pH = 4.74 pH = 7.00 add 0.1 mol HCl: now [HA] = 1.1 M [A–] = 0.9 M pH = 4.74 + log(0.9/1.1) = 4.65 DpH = 0.09 now [H3O+] = 0.10 M pH = 1.00 DpH = 6.00 !

  9. Buffer Water II. Buffer Solutions A. Composition and action of buffers 3. buffering action e.g., Buffer: 1.0 L of 1.0 M HC2H3O2 and 1.0 M NaC2H3O2 (pKa = 4.74) initial: pH = 4.74 pH = 7.00 add 0.1 mol NaOH: now [HA] = 0.9 M [A–] = 1.1 M pH = 4.74 + log(1.1/0.9) = 4.83 DpH = 0.09 now [OH–] = 0.10 M pOH = 1.00 pH = 13.00 DpH = 6.00

  10. useful pH range for an acid buffer = pKa 1 Useful pH range for a base buffer: [BH+] [B] [BH+] [B] [A–] [HA] [A–] [HA] = 0.10 pOH = pKb – 1 useful pH range for a base buffer = (14 – pKb)  1 = 10.0 pOH = pKb + 1 II. Buffer Solutions B. Buffer capacity and pH Useful pH range for an acid buffer: = 0.10 pH = pKa – 1 = 10.0 pH = pKa + 1 e.g., HCHO2/ NaCHO2: pH range = 3.74  1 (2.74–4.74) e.g., NH3/NH4Cl: pH range = (14 – 4.74)  1 (8.26–10.26)

  11. II. Buffer Solutions C. Buffer calculations e.g., How many grams of sodium acetate (fw 82.03) must be added to 1.00 L of 1.00 M acetic acid (pKa = 4.74) to give a buffer with a pH = 5.00?

  12. II. Buffer Solutions C. Buffer calculations • e.g., A buffer solution has a propanoic acid concentration of 0.52 M and a sodium propanoate concentration of 0.88 M. (pKa = 4.89) • a) What is the pH of the buffer? • b) What will be the pH when 0.11 mol of NaOH is added to 1.0 L of the buffer solution?

  13. II. Buffer Solutions C. Buffer calculations e.g., What will be the pH change when 20.0 mL of 0.100 M NaOH is added to 100.0 mL of a buffer solution consisting of 0.169 M NH3 and 0.183 M NH4Cl? (pKb = 4.74)

  14. Equivalence point: • 50.0 mL titrant • pH = 7.0 Initial pH = 1.0 III. Acid-Base Titrations A. Strong acid - strong base titrations • e.g., 50.0 mL of 0.10 M HCl titrated with 0.10 M NaOH • HCl + NaOH  NaCl + H2O (H3O+ + OH–  H2O)

  15. 1. No titrant added •  solution of weak HA • Ka pH  14 • 2. Some titrant added •  solution of HA & A– • = buffer • pH = pKa + log pHeq  pH • 3. Equivalence point (Veq, pHeq) • - complete neutralization •  solution of A– • Kb =  pOH  pHeq  pH½ • 4. Half-equivalence point (V½, pH½) • - exactly half neutralized • V½ = ½Veq • [HA] = [A–]  pH½ = pKa   2 V½ Veq [A–] [HA] Kw Ka • 5. Excess titrant added •  solution of OH–  pOH  pH Volume of titrant added III. Acid-Base Titrations B. Weak acid titrated with strong base • HA + OH– A– + H2O • Plot pH vs volume of titrant:

  16. HF H3O+ + F– x2 (1.0) x2 (1.0 - x) Ka = = 7.2 x 10–4 =  [H3O+][F–] [HF] x = [H3O+] = __________ pH = III. Acid-Base Titrations B. Weak acid titrated with strong base e.g., 25.0 mL of 1.0 M HF titrated with 1.25 M NaOH HF + OH– F– + H2O • 1. No titrant added •  1.0 M solution of HF

  17. 25 mL x 1.0 M = 25 mmol 8 mL x 1.25 M = 10 mmol 15 mmol HF 33 mL 10 mmol F– 33 mL = 0.30 M F– = 0.45 M HF pH = III. Acid-Base Titrations B. Weak acid titrated with strong base e.g., 25.0 mL of 1.0 M HF titrated with 1.25 M NaOH (cont.) 2. 8.0 mL of titrant added • HF + OH– F– + H2O • init: 25 10 0 mmol • final: 15 0 10 mmol = buffer

  18. pHeq? in solution: = 0.56 M F– F– HF + OH– 25 mmol F– 45 mL Kb = = = = Kw Ka x = [OH–] = ___________ pOH =____, pHeq = III. Acid-Base Titrations B. Weak acid titrated with strong base e.g., 25.0 mL of 1.0 M HF titrated with 1.25 M NaOH (cont.) 3. Equivalence point Veq? M1V1 = M2V2 (1.0 M)(25.0 mL) = (1.25 M) ·Veq Veq = 20.0 mL

  19. V½ = 10.0 mL [HF] = [F–] pH½ = pKa = 3.14 = _________ pOH = ____ pH = III. Acid-Base Titrations B. Weak acid titrated with strong base e.g., 25.0 mL of 1.0 M HF titrated with 1.25 M NaOH (cont.) 4. Half-equivalence point 5. 30.0 mL titrant added 10.0 mL excess OH– x 1.25 M = 12.5 mmol OH–(OH– from F– is negligible)

  20. 1. No titrant added •  solution of weak B • Kb pOH  pH • 2. Some titrant added •  solution of B & BH+ • = buffer • pOH = pKb + log  pH 14 [BH+] [B]   • 3. Equivalence point (Veq, pHeq) • - complete neutralization •  solution of BH+ • Ka =  pHeq pH½  pH Kw Kb  pHeq  • 4. Half-equivalence point (V½, pH½) • - exactly half neutralized • V½ = ½Veq • [B] = [BH+]  pOH½ = pKb pH½ 2 V½ Veq Volume of titrant added • 5. Excess titrant added •  solution of H3O+  pH III. Acid-Base Titrations C. Weak base titrated with strong acid • B + H3O+ BH+ + H2O

  21. III. Acid-Base Titrations C. Weak base titrated with strong acid • e.g., 50.0 mL of 0.20 M NH3 titrated with 0.25 M HCl • (Kb = 1.8 x 10–5, pKb = 4.74) • What is the pH of the solution • a) at the start of the titration? • b) after 15.0 mL of titrant have been added? • c) at the equivalence point? • d) at the half-equivalence point? • e) after 50.0 mL of titrant have been added? • Sketch the titration curve.

  22. III. Acid-Base Titrations D. Titration of polyprotic acids • e.g., H2C2O4 titrated with NaOH • H2C2O4 + OH– HC2O4– + H2O • HC2O4– + OH– C2O42– + H2O 14 10 pH 6 2 Volume of Titrant

  23. different colors predominant species at higher pH predominant species at lower pH pHeq of titration should fall within range of indicator III. Acid-Base Titrations E. Acid-base indicators HIn H3O+ + In– e.g., Congo red blue  red 3.0 - 5.0 Bromthymol blue yellow  blue 6.0 - 7.6 Phenolphthalein colorless  pink 8.2 - 10.0

  24. III. Acid-Base Titrations E. Acid-base indicators Weak acid titrated with strong base: 14 10 Phenolphthalein pH Bromothymol blue 6 Congo red 2 Volume of Titrant

  25. III. Acid-Base Titrations E. Acid-base indicators Weak base titrated with strong acid: 14 Phenolphthalein 10 pH Bromothymol blue 6 Congo red 2 Volume of Titrant

  26. < stoichiometric  buffer (HA, A–) = stoichiometric  solution of A– > stoichiometric  solution of OH– Summary: General Strategy for Acid-Base Problems 1. Recognize: strong acids/bases (H3O+, OH–) weak acids/bases (HA, B) weak conjugate acids/bases (BH+, A–) • 2. Carry out stoichiometric reactions, if any, and determine what species result: • HA + OH– A– + H2O B + H3O+ BH+ + H2O A– + H3O+ HA + H2O BH+ + OH– B + H2O • Only 4 likely reactions! • (strong/strong trivial, • weak/weak too difficult)

  27. pOH = pKb + log Kb = = [HA][OH–] [A–] pH = pKa + log Kw Ka Kw Kb Ka = = [B][H3O+] [BH+] [H3O+][A–] [HA] Ka = [BH+] [B] [A–] [HA] [BH+][OH–] [B] Kb = Summary: General Strategy • 3. Write equilibrium, if any: • HA H3O+ + A– • B BH+ + OH– • A– HA + OH– • BH+B + H3O+ 4. Solve equilibrium expression: solution of HA: B: A–: BH+: HA & A–: B & BH+:

  28. MA(s) MA2(s) IV. Solubility Equilibria saturated solution = ionic solid ions in solution M+(aq) + A–(aq) M2+(aq) + 2A–(aq) Kc small: “insoluble” Kc large: “soluble”

  29. for MA(s) M+(aq) + A–(aq) [M+][A–] [MA(s)] Kc´ = constant, not included Kc´[MA(s)] = Ksp = [M+][A–] for MA2(s) M2+(aq) + 2A–(aq) Ksp = [M2+][A–]2 V. Ksp and Molar Solubility A. Ksp

  30. V. Ksp and Molar Solubility B. Molar solubility 1. The solubility of BaC2O4 is 0.390 g in 500 mL of water. What is Ksp? (fw 225.3)

  31. V. Ksp and Molar Solubility B. Molar solubility 2. AgBr has Ksp = 5.0 x 10–13. What is the molar solubility of AgBr?

  32. V. Ksp and Molar Solubility B. Molar solubility 3. Mg(OH)2 has Ksp = 1.2 x 10–11. What is the pH of a saturated solution of Mg(OH)2?

  33. For MA(s) M+(aq) + A–(aq) Q = [M+][A–], not necessarily at equilibrium V. Ksp and Molar Solubility C. Ksp and Q if Q < Ksp: unsaturated solution Q = Ksp: saturated solution Q > Ksp: supersaturated solution  precipitate will form • e.g., 1.00 mL of 0.025 M CaCl2 and 1.00 mL of 0.0050 M Na2CO3 are mixed. Will a precipitate form? • Ksp(CaCO3) = 9 x 10–9

  34. For MA(s) M+(aq) + A–(aq), the solubility of MA decreases as [M+] or [A–] increases (Le Châtelier). • e.g., What is the molar solubility of CaCO3 in • 1) water? • 2) 0.50 M Na2CO3? Ksp(CaCO3) = 9 x 10–9 VI. Factors Affecting Solubility A. Common-ion effect

  35. VI. Factors Affecting Solubility B. pH e.g., What is the molar solubility of Mg(OH)2 at pH = 12.00? (Ksp = 1.2 x 10–11)

  36. M+ + X– MX(s) precipitate MX(s) + X– MX2– MX(s) + 2Y– MY2– + X– MX(s) + 4Z MZ4+ + X– complex ions e.g., AgCN(s) + CN– Ag(CN)2– Zn(OH)2(s) + 2OH– Zn(OH)4 2– Al(OH)3(s) + 6F– AlF63– + 3OH– AgCl(s) + 2NH3 Ag(NH3)2+ + Cl– VI. Factors Affecting Solubility C. Formation of complex ions

  37. e.g., Fe2+ + 6CN– Fe(CN)64– [Fe(CN)64–] [Fe2+][CN–]6 Kform = = 1.0 x 1024 VI. Factors Affecting Solubility C. Formation of complex ions • 1. formation constant (stability constant) • Kform = equilibrium constant for formation of a complex ion • from its individual ions

  38. Fe(OH)2(s) + 6CN– Fe(CN)64– + 2OH– Fe(OH)2(s) Fe2+ + 2OH–Ksp = 7.9 x 10–16 Fe2+ + 6CN– Fe(CN)64–Kform = 1.0 x 1024 Fe(OH)2(s) + 6CN– Fe(CN)64– + 2OH–Kc = Ksp · Kform = 7.9 x 108 Kc = = 7.9 x 108 [Fe(CN)64–][OH–]2 [CN–]6 VI. Factors Affecting Solubility C. Formation of complex ions 2. dissolving insoluble salts by formation of complex ions two simultaneous equilibria:

  39. VI. Factors Affecting Solubility C. Formation of complex ions 2. dissolving insoluble salts by formation of complex ions e.g., What is the molar solubility of AgCl in 5.0 M NH3? (Ignore the dissociation of NH3.) Ksp(AgCl) = 1.8 x 10–10, Kform(Ag(NH3)2+) = 1.6 x 107

  40. VI. Factors Affecting Solubility C. Formation of complex ions 2. dissolving insoluble salts by formation of complex ions e.g., What pH is necessary to dissolve 0.10 mol of Zn(OH)2(s) in 1.0 L of solution? Ksp(Zn(OH)2) = 3.0 x 10–16, Kform(Zn(OH)42–) = 2.8 x 1015

  41. increase [X–] selective precipitation VII. Selective Precipitation of Ions - separation of one metal ion in the presence of others Solution of M1+, M2+, X–Ksp1 < Ksp2 (i.e., M1X less soluble) while [M1+][X–] < Ksp1: no precipitate when [M1+][X–] Ksp1: M1X begins to precipitate while [M1+][X–] >Ksp1 and [M2+][X–] <Ksp2: M1X continues to precipitate when [M2+][X–] Ksp2: M2X starts to precipitate

  42. VII. Selective Precipitation of Ions • e.g., In a solution containing 0.010 M Ba2+ ions and 0.10 M Ca2+ ions, • Ksp(BaSO4) = 1.5 x 10–9 • Ksp(CaSO4) = 2.0 x 10–4 a. At what [SO42–] does BaSO4 start to precipitate? b. At what [SO42–] does CaSO4 start to precipitate? c. What [Ba2+] remains in solution at this point? d. What is the percent recovery of Ba2+?

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