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Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry , 2007 (John Wiley)      ISBN: 9 78047081 0866 . CHEM1002 [Part 2]. Dr Michela Simone Weeks 8 – 13 Office Hours: Monday 3-5, Friday 4-5

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  1. Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille,Chemistry, 2007 (John Wiley)     ISBN: 9 78047081 0866

  2. CHEM1002 [Part 2] Dr Michela Simone Weeks 8 – 13 Office Hours: Monday 3-5, Friday 4-5 Room: 412A (or 416) Phone: 93512830 e-mail:michela.simone@sydney.edu.au

  3. Completely ionise in water: e.g. HCl(aq) H+(aq) + Cl–(aq) Strong Acids and Bases • Equilibrium lies completely to the right, Ka  . Strong acids H2SO4, HCl, HBr, HI, HNO3, HClO4 Strong basesAll hydroxides of Groups 1 & 2 (except Be): NaOH, Ca(OH)2, …

  4. Examples What is the pH of a 0.1 M HCl solution? HCl(aq H+(aq) + Cl-(aq) [H+] = 0.1 M • Thus [H+] = 0.1 M and pH = – log10 [H+] = 1.0 • What is the pH of a 0.002 M NaOH solution? • Completely ionised, so [OH– ] = 0.002 M, • pOH = – log10 [OH– ] = – log10 (0.002) = 2.7 • pH = 14 – 2.7 = 11.3

  5. Acid dissociation constant: HA(aq) H+(aq) + A–(aq) pKa = – log10Ka Weak Acids • Most acids or bases areweak • they do not completely ionise in water

  6. Relationship between Ka and pKa • The larger the value of Ka, the stronger the acid and the lowerthe value of pKa. • Ka = 1.02 x 10-2 then pKa = -log10(1.02 x 10-2) = 1.991 • pKa = 1.991 then Ka = 10-1.991 = 1.02 x 10-2

  7. Find the pH of 0.1 M acetic acid (CH3COOH (HAc)) Example • DATA: pKa = 4.7, Ka = 10-4.7 HAc(aq) H+(aq) + Ac–(aq)

  8. Find % ionisation of 0.50 M HF (pKa = 3.1) Example HF(aq) H+(aq) + F–(aq) • % ionisation = x / 0.50  100 =

  9. Ionisation of a weak base: Weak Bases NH3(aq) + H2O(l) NH4+(aq) + OH–(aq) • Equilibrium constant is called base ionisation constant, Kb : • Tactic: calculate pOH and then pH, given Kb.

  10. Acid (HA):HA(aq) H+(aq) + A–(aq) Relationship Between pKa and pKb • For conjugate systems (Brønsted-Lowry acid-base pairs) • Conjugate base (A– ):A–(aq) + H2O(l) HA(aq) + OH–(aq) pKa + pKb = 14 • We only need values of pKa , since pKb = 14 – pKa

  11. Find the pH of 1.0 x 10–2 M NaHCO2 (pKa of formic acid (HCO2H) is 4.1) Example • pKb = 14 – pKa = 14 – 4.1 = 9.9 HCO2–(aq) + H2O(l) OH–(aq) + HCO2H(aq) +x -x +x x x 0.01 - x

  12. Example (Continued) • x = [OH–] so pOH = -log10(10-5.95) = 5.95 • pH = 14 – 5.95 = 8.05 = 8.1 (one significant figure)

  13. removing more protons is harder: increasing pKa = decreasing Ka :Ka1 > Ka2 >Ka3 reason: harder to remove +ve charge against increasing -ve charge large difference in pKa values only need to consider one equilibrium at a time (simplifies maths!) H3PO4(aq) H+(aq) + H2PO4–(aq)pKa1 = 2.2 H2PO4–(aq) H+(aq) + HPO42–(aq)pKa2 = 7.2 HPO42-(aq) H+(aq) + PO43–(aq)pKa3 = 12.4 Polyprotic Acids

  14. Practice Examples • 1. Which one of the following is NOT a conjugate acid-base pair? • (a) HCO3– and CO32– • (b) H3O+ and H2O • (c) OH– and O2– • (d) SO3 and HSO3– • (e) NH2OH2+ and NH2OH • What is the pH of a 0.20 M solution of boric acid? The pKa of boric acid is 9.24. • (a) 0.70 (b) 2.73 (c) 4.97 (d) 5.12 (e) 5.87 • 3. Rank the following solutions (all 1.0 M) in DECREASING order of pH. • NaCl, NaCN, KOH, HCl, CH3COOH • (a) HCl > CH3COOH > NaCN > NaCl > KOH • (b) KOH > NaCl > NaCN > CH3COOH > HCl • (c) KOH > NaCN > NaCl > HCl > CH3COOH • (d) KOH > NaCl > NaCN > HCl > CH3COOH • (e) KOH > NaCN > NaCl > CH3COOH > HCl

  15. Summary: Acids & Bases 2 • Learning Outcomes - you should now be able to: • Complete the worksheet • Complete acid/base calculations • Use pH, pKw, pKa and pKb • Define strong and weak acids & bases • Answer Review Problems 11.12-11.35 in Blackman • Next lecture: • Buffer systems

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