1. Acids & Bases CHAPTER 16 (& part of CHAPTER 17) Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop

2. CHAPTER 16: Acids & Bases Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 2 Learning Objectives:  Define Brønsted-Lowry Acid/Base  Define Lewis Acid/Base  Evaluate the strength of acids/bases  Strong vs weak acids/bases  Periodic trends  Conjugate acids/bases  Identify likely compounds that will form acids and bases from the periodic table  Acidic metal ions  Acid/Base equilibrium:  pH, pOH  Ka, Kb, pKa, pKb  Kw of water

3. CHAPTER 16: Acids & Bases Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 3 Lecture Road Map: ① Brønsted-Lowry Acids/Bases ② Trends in acid strength ③ Lewis Acids & Bases ④ Acidity of hydrated metal ions ⑤ Acid/Base equilibrium

4. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 4 Acid/Base Equilibrium CHAPTER 16 Acids & Bases

5. Acid/Base Equilibrium Weak Acids & Bases Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 5 Incompletely ionized Molecules and ions exist in equilibrium HA = any weak acid; B = any weak base HA(aq) + H 2 O A – (aq) + H 3 O + (aq) B (aq) + H 2 O B H + (aq) + OH – (aq) CH 3 COOH(aq) + H 2 O CH 3 COO – (aq) + H 3 O + (aq) HSO 3 – (aq) + H 2 O SO 3 2 – (aq) + H 3 O + (aq) NH 4 + (aq) + H 2 O NH 3 (aq) + H 3 O + (aq)

6. Acid/Base Equilibrium Weak Acids & Bases Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 6 Often simplify as HA (aq) A – (aq) + H + (aq)

7. Acid/Base Equilibrium Weak Acids & Bases Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 7 Or generally HA(aq) + H 2 O A – (aq) + H 3 O + (aq) But [H 2 O] = constant (55.6 M ) so rewrite as Where K a = acid ionization constant Acid + Water Conjugate Base + Hydronium Ion

8. Acid/Base Equilibrium Weak Acids & Bases Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 8

9. Acid/Base Equilibrium Weak Acids & Bases Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 9 But [H 2 O] = constant so rewrite as Where K b = base ionization constant CH 3 COO – (aq) + H 2 O CH 3 COOH(aq) + OH – (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH – (aq) Or generally B (aq) + H 2 O B H + (aq) + OH – (aq)

10. Acid/Base Equilibrium Weak Acids & Bases Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 10

11. Acid/Base Equilibrium pH Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 11 Lots of weak acids and bases –How can we quantify their relative strengths? Need reference –Choose H 2 O Water under right voltage –Slight conductivity –Where does conductivity come from?

12. Acid/Base Equilibrium pH Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 12 Trace ionization  self-ionization of water H 2 O + H 2 O H 3 O + (aq) + OH – (aq) acid base acid base Equilibrium law is: But [H 2 O] pure = = 55.6 M [H 2 O] = constant even for dilute solutions

13. Acid/Base Equilibrium pH Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 13 H 2 O + H 2 O H 3 O + (aq) + OH – (aq) Since [H 2 O] = constant, equilibrium law is K w = is called the ion product of water Often omit second H 2 O molecule and write H 2 O H + (aq) + OH – (aq)

14. Acid/Base Equilibrium Kw Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 14 H 2 O H + (aq) + OH – (aq) For pure H 2 O at 25 °C –[H + ] = [OH – ] = 1.0 × 10 –7 M –K w = (1.0 × 10 –7 )(1.0 × 10 –7 ) = 1.0 × 10 –14 –See Table 17.1 for K w at various temperatures H 2 O auto-ionization occurs in all solutions –When other ions present [H + ] is usually NOT equal to [OH – ] But K w = [H + ][OH – ] = 1.0 × 10 –14

15. Acid/Base Equilibrium Definition of Acidic & Basic Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 15 In aqueous solution, –Product of [H 3 O + ] and [OH – ] equals K w –[H 3 O + ] and [OH – ] may not actually equal each other –Solutions are classified on the relative concentrations of [H 3 O + ] and [OH – ] Solution Classification Neutral[H 3 O + ] = [OH – ] Acidic[H 3 O + ] > [OH – ] Basic[H 3 O + ] < [OH – ]

16. Acid/Base Equilibrium Weak Acids & Bases: Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 16 Ex. 1 In a sample of blood at 25 °C, [H + ] = 4.6  10 –8 M. Find [OH – ] and determine if the solution is acidic, basic or neutral. So2.2 × 10 –7 M > 4.6 × 10 –8 M [OH – ] > [H 3 O + ] so the solution is basic

17. Acid/Base Equilibrium pH Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 17 Arrhenius (of kinetics fame) –Sought an easy way to write the very small numbers associated with [H + ] and [OH – ] –Developed the “p” notation where p stands for the –log mathematical operation –Result is a simple number pH is defined as: –Define pOH as: –Define pK w as: Take anti-log to obtain [H + ], [OH – ] or K w

18. Acid/Base Equilibrium General Properties of Logarithms Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 18 Using Logarithms Start with Taking –log of both sides of eqn. gives So at 25 °C:

19. Acid/Base Equilibrium Definition of Acidic and Basic Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 19 As pH increases, [H + ] decreases; pOH decreases, and [OH – ] increases As pH decreases, [H + ] increases; pOH increases, and [OH – ] decreases NeutralpH = 7.00 AcidicpH < 7.00 BasicpH > 7.00

20. Acid/Base Equilibrium Measuring pH Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 20 1.pH meter –Most accurate –Calibrate with solutions of known pH before use –Electrode sensitive to [H + ] –Accurate to  0.01 pH unit 2.Acid-base indicator –Dyes, change color depending on [H + ] in solution –Used in pH paper and titrations –Give pH to  1 pH unit 3.Litmus paper –RedpH  4.7 acidic –BluepH  4.7 basic –Strictly acidic vs. basic

21. Acid/Base Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 21

22. Acid/Base Equilibrium Example pH Calculations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 22 Calculate pH and pOH of blood in Ex. 1. We found [H + ] = 4.6 × 10 –8 M [OH – ] = 2.2 × 10 –7 M pH = –log(4.6 × 10 –8 ) = 7.34 pOH = –log(2.2 × 10 –7 ) = 6.66 14.00 = pK w Or pOH = 14.00 – pH = 14.00 – 7.34 = 6.66

23. Acid/Base Equilibrium Example pH Calculations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 23 What is the pH of NaOH solution at 25 °C in which the OH – concentration is 0.0026 M? [OH – ] = 0.0026 M pOH = –log(0.0026) = 2.59 pH = 14.00 – pOH = 14.00 – 2.59 = 11.41

24. Acid/Base Equilibrium Strong Acids Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 24 Assume 100% dissociated in solution –Good ~ if dilute Makes calculating [H + ] and [OH  ] easier 1 mole H + for every 1 mole HX –So [H + ] = C HX for strong acids Thus, if 0.040 M HClO 4 [H + ] = 0.040 M And pH = –log (0.040) = 1.40 Strong Acids HCl HBr HI HNO 3 H 2 SO 4 HClO 3 HClO 4 HX (general term for a strong acid)

25. Acid/Base Equilibrium Strong Bases Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 25 1 mole OH – for every 1 mole M OH [OH – ] = C MOH for strong bases 2 mole OH – for 1 mole M(OH) 2 [OH – ] = for strong bases Strong Bases NaOH KOH LiOH Ca(OH) 2 Ba(OH) 2 Sr(OH) 2

26. Acid/Base Equilibrium Effect of Auto ionization of Water with Strong Bases Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 26 The auto-ionization of H 2 O will always add to [H + ] and [OH – ] of an acid or base. Does this have an effect on the last answer? –The previous problem had 0.00022 M [OH – ] from the Ca(OH) 2 but the [H + ] must have come from water. If it came from water an equal amount of [OH – ] comes from water and the total [OH – ] is –[OH – ] total = [OH – ] from Ca(OH) 2 + [OH – ] from H 2 O –[OH – ] total = 0.00022 M + 4.6 × 10 –11 M = 0.00022 M (when properly rounded)

27. Acid/Base Equilibrium Example pH Calculations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 27 –So [H + ] from H 2 O must also be 5.0  10 –13 M [H + ] total = 0.020 M + (5.0  10 –13 M) = 0.020 M (when properly rounded) So we see that [H + ] from H 2 O will be negligible except in very dilute solutions of acids and bases

28. Acid/Base Equilibrium Looking at Weak Acids Again Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 28 What is the pK a of HF if K a = 3.5 × 10 –4 ? HF(aq) + H 2 O F – (aq) + H 3 O + (aq) or HF(aq) F – (aq) + H + (aq) = 3.5 × 10 –4 pK a = –log K a = –log(3.5 × 10 –4 ) = 3.46

29. Acid/Base Equilibrium Conjugate Acid-Base Pairs Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 29 1. Consider ionization reaction of generic acid and water HA(aq) + H 2 O A – (aq) + H 3 O + (aq) 2. Consider reaction of a salt containing anion of this acid (its conjugate base) with water A – (aq) + H 2 O HA(aq) + OH – (aq)

30. HA (aq) + H 2 O A – (aq) + H 3 O + (aq) A – (aq) + H 2 O HA (aq) + OH – (aq) 2H 2 O H 3 O + (aq) + OH – (aq) For any conjugate acid base pair: (at 25 °C) Acid/Base Equilibrium Conjugate Acid-Base Pairs Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 30

31. Acid/Base Equilibrium More About Logarithms Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 31 Then taking –log of both sides of equation gives: Earlier we learned the inverse relationship of conjugate acid- base strengths, now we have numbers to illustrate this. The stronger the conjugate acid, the weaker the conjugate base. So (at 25 °C)