Class 31 Wednesday April 7 Mastering Maintenance Fri 4 16 11 pm Sat 4 17 7 am Today s Coverage 17 8 17 10 Salts Structure and Strength of Acids Helpful items for today Equations sheet located on Canvas Pencil pen and paper or tablet Reef open and ready to go Calculator Activity 4 available today at noon due next Wed at 10 pm MT great practice for LME 2 Friday Review during lecture lecture slides available today on Canvas at noon Posted on the Unit 2 LME Logistics and Resources page and on the Unit 3 page LME 2 breakdown Mastering Problem Set 6 Chp 17A Due Wed 4 7 at 10 pm MT Mastering Problem Set 7 Chp 17B 19B Due Wed 4 21 at 10 pm MT Additional Optional Resources Video Helps by Professors J2 for Chps 15 16 and 17A Practice Problem Sets for Chps 15 16 and 17A Mastering for Chps 15 16 17A Lecture Notes Pace your studying LME 2 April 12 M 7 30 9 pm MT Chapters 15 16 17A pace your preparation ACS LME 2 Help Session Wed April 7th from 5 7pm Fri April 9th 1 3 pm Weekend sessions with Dr Beren Sat and Sun at 8 pm Coverage Chp 15 16 17 1 17 6 Final LME May 7 Friday 8 10 am Updated grade calculator posted use only the updated No quadratic or polynomial calculations on LME perfect squares or x is small In class review on Fri 4 9 LME 2 Resources and Logistics page is available located next Unit 2 link on the homepage HA aq H2O l H3O aq A aq A aq H2O l OH aq HA aq Strong Acid H3O HA initial pH log H3O Strong Base OH OH initial pOH log OH pH 14 pOH Weak Acid Ka H3O eq A eq HA eq Weak Base OH eq HA eq Kb A eq Kb x 2 A i x Ka x 2 HA i x H3O x pH log H3O OH x pOH log OH pH 14 pOH What is the pH of an aqueous solution containing 0 050 M KOH and 0 015 M Sr OH 2 A 13 00 B 12 96 C 12 69 D 12 90 E 13 29 Strategy Weak or Strong OH 0 050 2 0 015 0 080 M pOH log OH 1 10 pH 14 00 pOH pH 14 00 1 10 12 90 Calculate the pH of a 15 0 M solution of NH3 Ka 5 56 x 10 10 B aq H2O l BH aq OH aq A 13 0 B 12 2 C 1 8 D 0 0164 E 10 4 Strategy Weak or Strong B 15 0 x 15 0 x I C E OH 1 BH 1 0 x x 0 x x Kb 1 8 x 10 5 Kb BH OH B 1 8 x 10 5 x x 15 0 x OH 0 0164 pOH log 0 0164 pOH 1 8 pH 12 2 Relationship between Ka of an Acid and Kb of Its Conjugate Base Add equations multiply K s Kw 1 0 x 10 14 Given Ka solve for Kb Given Kb solve for Ka Ka pKa Ka pKa Kb pKb Kb pKb HA aq H2O l aq H2O l A H3O OH aq A aq aq HA aq Stronger the acid the weaker the CB Stronger the base the weaker the CA Conjugate of WA is basic Conjugate of WB is acidic Conjugate of SA is neutral Conjugate of SB is neutral Some acid base reactions make salts HBr aq NaOH aq H aq Br aq Na aq OH aq NaBr aq H2O H aq OH aq H2O But what about the salt NaBr Does it impact the solution pH Na aq H2O l reaction Cation Br aq H2O l reaction Anion pH of Salt Solutions Cation Acidic Conjugate acid of a weak base NH4 H2O NH3 H3O Small highly charged metal ions Example Al 3 Fe 3 Anion Basic Conjugate base of a weak acid A aq H2O HA aq OH aq Examples F CH3CO2 CN Not Basic negligible Conjugate base of a strong acid Examples Cl 1 NO3 1 ClO4 1 Not Acidic neutral Counterions of strong bases Examples alkali alkaline earth Li Na K Ca2 Sr2 Ba2 Acidic Basic or Neutral HC7H5O2 WA C7H5O2 I HI SA Sr2 SB cation Mn3 Small highly charged NO2 HNO2 WA C5H5NH C5H5N WB Li SB cation basic neutral neutral acidic basic acidic neutral Acidic Basic or Neutral HC7H5O2 WA C7H5O2 I HI SA Sr2 SB cation Mn3 Small highly charged NO2 HNO2 WA C5H5NH C5H5N WB Li SB cation Salt Cation neutral LiC7H5O SrI2 neutral acidic C5H5NHI Mn NO2 3 acidic Anion basic neutral neutral basic basic neutral neutral acidic basic acidic neutral pH basic neutral acidic depends Which of the following is a basic salt K cation of SB Cl anion of SA neutral NH4 Na cation of SB F conj base of WA HF basic CA of WB NH3 Cl anion of SA acidic a KCl b NH4Cl c NaF d a b and c are all acidic e a b and c are all basic Answer c Which one of the following salts is properly paired A NaCl B LiCN C KClO4 D LiBrO E All of the above will form basic solutions acidic neutral basic basic Answer D NH4 2CO3 Is it acidic or basic NH4 aq H2O l CO3 aq H2O l 2 aq NH3 aq H HCO3 aq OH aq 5 6 x 10 10 Ka 1 8 x 10 4 Kb Basic Ka Kb Three possibilities Ka Kb solution will contain excess of H ions Ka Kb solution will contain excess of OH ions Ka Kb solution will contain approximately equal concentrations of H and OH ions pH 7 acidic pH 7 basic pH 7 neutral