1. Acids & Bases Ch.15

2. (15-1) Acids Inc. the H 3 O + conc. (Arrhenius) Properties: –Sour taste –Conduct electricity –React w/ metals to produce H 2 (g) 2H 3 O + (aq) + Zn(s)  2H 2 O(l) + Zn 2+ (aq) + H 2 (g)

3. Bases Inc. the OH - conc. (Arrhenius) Properties: –Slippery –Bitter –Electrical conductors

4. Acid Nomenclature

5. Acid Naming Practice HBr  hydrobromic acid H 2 CO 3  carbonic acid H 2 SO 3  sulfurous acid

6. Acid Strength Strong Acids: ionize completely –HNO 3 (l) + H 2 O(l)  H 3 O + (aq) + NO 3 - (aq) Weak Acids: only small # of molecules ionize –CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO - (aq)

7. Base Strength Strong bases: fully dissociate in soln –NaOH(s)  Na + (aq) + OH - (aq) Weak bases: few molecules dissociate –NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) Strength = conc.

8. Neutralization Rxn H 3 O + from an acid & OH - from a base react to produce H 2 O H 3 O + (aq) + OH - (aq)  2H 2 O(l)

10. Self-ionization of Water 2H 2 O(l) H 3 O + (aq) + OH - (aq) K eq = K w = autoionization constant [OH - ] = [H 3 O + ] @ 25°C = 1 x 10 -7 M K w = [H 3 O + ][OH - ] = (1 x 10 -7 )(1 x 10 -7 ) = 1 x 10 -14

11. (15-2) Arrhenius Drawbacks: –Only applies to aq solns –Not all bases donate OH - Ex: NH 3 Bronsted-Lowry: more modern

12. B-L Acid Proton donor (H + ) –H 2 SO 4 + H 2 O  H 3 O + + HSO 4 - Monoprotic: donates 1 proton –HNO 3 Diprotic: donates 2 protons –H 2 SO 4 Triprotic: donates 3 protons –H 3 PO 4

13. B-L Base Proton acceptor –NH 3 + H 2 O NH 4 + + OH - Amphoteric: acts as either an acid or a base –Ex: H 2 O –H 2 O + H 2 O OH - + H 3 O +

14. Conjugates Conjugate acid: acid formed when a base accepts a proton Conjugate base: base formed when an acid donates a proton Stronger A/B  weaker conj. B/A

15. B A CA CB

16. Lewis Definitions Lewis acid: accepts an e- pair Lewis base: donates an e- pair

17. Acid-ionization Constant K a : equil. constant for a rxn in which an acid donates a proton to H 2 O CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO - (aq) K a = [aq products] = [H 3 O + ][CH 3 COO - ] [aq reactants] [CH 3 COOH] = 1.75 x 10 -5

18. (15-3) pH pH = -log[H 3 O + ] –As [H 3 O + ] inc., pH dec. pH scale: 1-14 –Acidic = 7 Neutral: [H 3 O + ] = [OH - ]

19. Calculating pH What’s the pH of a soln w/ a [H 3 O + ] of 1.8 x 10 -5 M? pH = -log [H 3 O + ] = -log (1.8 x 10 -5 ) = 4.74

20. Calculating [H 3 O + ] If the pH of Diet Coke is 3.12, what’s the [H 3 O + ]? [H 3 O + ] = 10 -pH = 10 -3.12 = 7.6 x 10 -4 M

21. pOH Since acids & bases are opposites, pH & pOH are opposites pH + pOH = 14 pOH = -log [OH - ]

22. pOH Practice What’s the pH of a 1.2 x 10 -4 M NaOH soln? pOH = -log[OH - ] = -log (1.2 x 10 -4 ) = 3.9 pH = 14 – pOH = 14 – 3.9 = 10.1

23. pOH Practice Another method would be to use K w rather than pOH. [H 3 O + ] = K w = 1 x 10 -14 = 8.3 x 10 -11 M [OH - ]1.2 x 10 -4 pH = -log[H 3 O + ] = -log (8.3 x 10 -11 ) = 10.1

24. pH of Weak Acids Use K a to solve for [H 3 O + ] C 6 H 5 COOH(aq) + H 2 O(l) C 6 H 5 COO - (aq) + H 3 O + (aq) K a = [C 6 H 5 COO - ][H 3 O + ] [C 6 H 5 COOH]

25. pH of Weak Bases Use equil. expression to solve for [OH - ] Use K w to solve for [H 3 O + ] or pOH

27. Indicator Dye that changes colors in solns of different pH –Ex: phenolphthalein bases = pink, acids = clear pH meter

29. Buffer Soln that resists changes in pH –Blood Mixture of weak acid & its conj. base in approx. = amts. Follows Le Chatelier’s principle to counteract effect of acid or base addition

30. (15-4) Titration Gradually adding 1 soln to another to reach an equivalence pt. Titrant: soln being added Standard soln: known conc.

31. Equivalence Point Amt. of added base or acid = the amt. of acid or base originally in soln 2 ways to detect: –pH meter –Indicator End point: indicator changes color

32. Titration Curve SA/SB Not all equiv.pts at pH = 7

34. Titration Calculations M 1 V 1 = M 2 V 2 Where, –M = molarity = mol/L –V = L or mL

35. Titration Practice 40 mL of HCl of unknown conc. is titrated w/ 0.55 M NaOH. The V of base needed to reach the equiv. pt. is 24.64 mL. What’s the M of the HCl? 1.List known V A = 40 mLV B = 24.64 mL [HCl] = ? M[NaOH] = 0.55 M

36. Titration Practice 2.Write titration rxn HCl + NaOH  NaCl + H 2 O 3.Write eq. M A V A = M B V B  M A = M B V B V A 4.Substitute & solve M A = (0.55 M)(24.64 mL) = 0.3388 M 40. mL

37. More Titration If the titration rxn is not a 1:1 ratio, you must compensate w/ the eq. Ex: Ba(OH) 2 + 2HCl  BaCl 2 + 2H 2 O M A V A = 2M B V B If coef. is in front of acid, multiply base side of eq. by it