Chapter 14/15 Structured Overview Chapter 14/15 Structured Overview follows on the next slide. You will find that the Chapter 14 material is already filled in. The blanks that will be filled in through the animation are from the content from chapter 15 because I wanted you to focus on that. All content is fair game for the test however. As you go through the presentation fill in the blanks on your own structured overview. After filling in the structured overview work through the problem on the slide that follows it (the back of your hard copy) by clicking through the presentation. After working that problem, info about the on-line test is given on the last slide.

x= Ch. 14/15 A.P. Chemistry ACIDS AND BASES Nature of Them As Equilibrium Rxtns. Practical Definitions Conceptual Definitions Conjugate Pairs Dissociation Constants ACIDS BASES SourSour Turns litmus redTurns litmus red Reacts with metals to produce H 2Reacts with metals to produce H 2 BitterBitter Turns litmus blueTurns litmus blue slipperyslippery Arrhenius B-L Lewis A: Liberates H + ions in sol’n B: Liberates OH - ions in sol’n A: proton donor B: proton acceptor A: electron pair acceptor B: electron pair donor [H + ] [OH - ] [the autoionization of water] KaKaKaKa KbKbKbKb KwKwKwKw Tells about: Strength pH Defined: -log [H 3 O + ] Calculations Strong Species Weak Species Polyprotic Acids % Dissociation Changes with dilution Easiest… concentration on bottle dictates [H 3 O + ] Harder… Use K a or K b values Volumetric analysis to determine unknown [ ]’s… Indicators… used to find end point of titration Range pKa ± 1 pH curves SA/SB WA/SB WB/SA Buffers Sol’ns which resist changes in pH with small additions of acids or bases How they work Calculations Henderson-Hasselbalch Eq. SaltsOxides neutral acidic basic Covalent (acidic) Ionic (basic) Ionic (basic) Electron Withdrawing Capacity explains this Titration

A Beaker of Water Adding a Strong Base Adding a Weak Base To make a 0.10 M sol’n of ______ 50.0 mL of 2.00 M _______ Adding a ___________ Adding a __________________ 50.0 mL of 2.0 M __________ mL of 1.5M __________66.66 mL of 1.5M __________ Adding 2.0 mL of a 0.10 M strong base To make a buffer H 2 O + H 2 O ↔ H 3 O + + OH - K w = 1.0 x pH = 7 pOH = 7 NaOH pOH = 1pH= 13 CH 3 NH 2 CH 3 NH 2 + H 2 O ↔ CH 3 NH OH - K b = 4.38 x [OH - ] = K b = [OH - ] [CH 3 NH 3 + ] / [CH 3 NH 2 ] = x 2 / x pOH = 1.53 pH = salt of the conj. acid CH 3 NH 3 NO 3 [CH 3 NH 3 + ] = 100. mmol / mL [CH 3 NH 2 ] = 100. mmol / mL = 1.00 M pH = pK a + log [CH 3 NH 2 ]/ [CH 3 NH 3 + ] pH = pH = 10.6K a = K w /K b = 2.28x CH 3 NH OH - ↔ CH 3 NH 2 + H 2 O 99.8 mmol / mL= M mmol / mL = M pH = log (0.982/0.978) pH = 10.6 Titration Strong Acid CH 3 NH 2 + H 3 O + ↔ CH 3 NH H 2 O HCl 50.0 mmol of H 3 O + added [CH 3 NH 2 ] = 50.0 mmol / mL = M= [CH 3 NH 3 + ] too! pH = 10.6 CH 3 NH H 2 O ↔ CH 3 NH 2 + H 3 O + HCl x x K a = x mmol mL [H 3 O + ] = 4.42 x Half-Neutralized !!!!Equivalence Point !!!! pH = 5.35

Chapter 14/15 On-Line Test Click on the “On-Line Tests” link on the class web page. Read the instructions and click on the Login button. User Name: last name (lower case) Password: student ID # Read directions and find the link for the answer sheet. Solve the problems and put the answers on the answer sheet. answer sheet to me as instructed. (no attachments) Available beginning 3:30 PM on prior to 8:00 am on 4/5/11 GOOD LUCK !!!!!!