pH Ka Kb Kw Ch. 14/15 A.P. Chemistry ACIDS AND BASES x = Titration Covalent (acidic) Ionic (basic) Nature of Them As Equilibrium Rxtns. pH Salts Oxides neutral acidic Practical Definitions Conceptual Definitions Conjugate Pairs Dissociation Constants Defined: -log [H3O+] Calculations basic ACIDS Arrhenius Ka x Kb = Kw Strong Species Weak Species Polyprotic Acids Sour Turns litmus red Reacts with metals to produce H2 A: Liberates H+ ions in sol’n B: Liberates OH- ions in sol’n [H+] [OH-] [the autoionization of water] Easiest… concentration on bottle dictates [H3O+] Harder… Use Ka or Kb values B-L Tells about: Strength A: proton donor B: proton acceptor Sol’ns which resist changes in pH with small additions of acids or bases BASES Volumetric analysis to determine unknown [ ]’s… % Dissociation Changes with dilution Bitter Turns litmus blue slippery Bond Strength and Electron Withdrawing Capacity explains this Titration SA/SB Henderson-Hasselbalch Eq. Calculations How they work Indicators… used to find end point of titration pH curves WA/SB Buffers Range pKa ± 1 WB/SA

Adding a __________________ salt of the conj. acid pH = 10.642 Ka= Kw/Kb= 2.28x10-11 pOH = 1.00 pH= 13.00 Adding 2.0 mL of a 0.50 M strong base CH3NH3NO3 To make a 0.10 M sol’n of ______ NaOH 50.0 mL of 2.00 M __________ CH3NH3+ + OH- ↔ CH3NH2 + H2O Adding a Strong Base Adding a __________________ salt of the conj. acid [CH3NH3+] = 100. mmol / 100.0 mL = 1.00 M 99. mmol / 102 mL = 0.97 M [CH3NH2] = 100. mmol / 100.0 mL = 1.00 M 101 mmol / 102 mL = 0.990 M pH = pKa + log [CH3NH2]/ [CH3NH3+] pH = 10.642 + log (0.990/0.97) p.H = 10.642 + 0 pH = 10.651 pH = 7.000 pOH = 7.000 A Beaker of Water Adding a ___________ Strong Acid Titration H2O + H2O ↔ H3O+ + OH- CH3NH2 + H3O+ ↔ CH3NH3+ + H2O pH = 5.355 Kw = 1.00 x 10-14 HCl HCl 33.33 mL of 1.50 M __________ 66.66 mL of 1.50 M __________ To make a buffer Half-Neutralized! (midpoint) Equivalence Point !!!! 50.0 mmol of H3O+ added CH3NH3+ + H2O ↔ CH3NH2 + H3O+ pOH = 1.529 [CH3NH2] = 50.0 mmol / 83.3 mL x x pH = 12.471 = 0.600 M = [CH3NH3+] too! Ka = x2 Adding a Weak Base Kb= 4.38 x 10-4 pH = 10.642 100. mmol CH3NH2 [OH-] = 0.0296 116.7 mL 50.0 mL of 2.00 M _______ CH3NH2 + H2O ↔ CH3NH3+ + OH- Kb = [OH-] [CH3NH3+] / [CH3NH2] [H3O+] = 4.42 x 10-6 = x2 / 2.00 - x

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