PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Suppose you have [HSuppose you have [H33OO++] from added HCl = 0.1 M = 10] from added HCl = 0.1 M = 10-1 -1 MMThen [10 Then [10 -1-1][OH][OH--] = 10] = 10-14-14In pure HIn pure H22O, [OHO, [OH--] = [H] = [H33OO++] = 10] = 10-7-7, , butbutWeak Acids and BasesWeak Acids and BasesAdd acetic acid to HAdd acetic acid to H22O:O:What are [CHWhat are [CH33COOH], [HCOOH], [H33OO++], [CH], [CH33COOCOO--]?]?K=KK=KaaIfIf we ignore Hwe ignore H33OO++ from H from H22O + HO + H22O O  H H33OO+ + + OH+ OH-- (Stoichiometry)(Stoichiometry)Let CLet Coo = initial [CH = initial [CH33COOH]COOH] [HOAc] [HOAc]Guess, [HOAc]Guess, [HOAc] C Co o , since K=K, since K=Kaa<<<1 (<<<1 ( [H [H33OO++]<<< C]<<< Coo))K=KK=Kaa have [Hhave [H33OO++] = [CH] = [CH33COOCOO--]]K =[H3O+]2CO−[H3O+]=1.85×10−5{[H{[H33OO++] from H] from H22O O  10 10-7-7}}CHCH33COOH + HCOOH + H22O O  H H33OO++ + CH + CH33COOCOO--Small K meansSmall K meansequilibrium is equilibrium is far to left!far to left!CCoo - [H - [H33OO++] = 0.9957 M ] = 0.9957 M  C Coo = 1 M (0.43% error) = 1 M (0.43% error)Example : CExample : Coo = 0.01 M = 0.01 M  [H [H33OO++]= 4.3 ]= 4.3  10 10-4-4CCoo - [H - [H33OO++] = 10 ] = 10  10 10-4-4 - 1.36 - 1.36  10 10-4-4 = 8.64 = 8.64  10 10-4-4 vs Cvs Coo = 10 = 10  10 10-4 -4 (13.6% error!!)(13.6% error!!)[H[H33OO++] = ] = KCoK =[H3O+]2CO−[H3O+]=1.85×10−5Note: Note: In all cases [HIn all cases [H33OO++] >>10] >>10-7-7 Thus, OK to Thus, OK to neglect Hneglect H33OO++ from H from H22O + HO + H22O O  H H33OO+ + + OH+ OH--Could obtain an exact solution by solving a quadratic:Could obtain an exact solution by solving a quadratic:Such equations are trivial to solve with a modern Such equations are trivial to solve with a modern graphing calculator.graphing calculator.* * * The approximation that [HOAc] * * * The approximation that [HOAc]  C Coo is good is good only for only for small Ksmall Kaa, , large Clarge Coo..A powerful, approximate alternative method:A powerful, approximate alternative method:(K=K(K=Kaa))Take [OHTake [OH--]>>10]>>10-7-7 (ignore hydrolysis of H (ignore hydrolysis of H22O)O)[OH[OH--]=[CH]=[CH33NHNH33++]]Kb=[CH3NH3+][OH−][CH3NH2]=5.0×10−4Let initial concentration [CHLet initial concentration [CH33NHNH22] = C] = Coo = 0.1 = 0.1First Approximation :First Approximation :[OH[OH--] = [CH] = [CH33NHNH33++] = x] = x11[CH[CH33NHNH22] = C] = Coo - x - x11, , Assume xAssume x11 << C << Coo (Stoichiometry)Method of Successive Method of Successive ApproximationsApproximationsxx11 = 7.1 = 7.1  10 10-3-3CCoo - x - x11 = (100 - 7.1) = (100 - 7.1)  10 10-3-3Second Approximation :Second Approximation :[OH[OH--] = [CH] = [CH33NHNH33++] = x] = x22 (unknown) (unknown)Kb=(x2)2(Co−x1)=(x2)292.9×10−3(x(x22))22 = (92.9 = (92.9  10 10-3-3)(5.0 )(5.0  10 10-4-4) = 46.45 ) = 46.45  10 10-6-6% error in taking C% error in taking Coo - x - x11 = C = Coo is thus is thus  100 = 7.1 % 100 = 7.1 %7.1100⎛ ⎝ ⎞ ⎠ CHCH33NHNH22 + H + H22O O  CH CH33NHNH33++ + OH + OH-- (weak base) (weak base)CCoo - x - x 1 1  C Coo xx 1 1 x x 1 1Value we chose was only off by 0.3 out of 93.2 (about 0.3%)Value we chose was only off by 0.3 out of 93.2 (about 0.3%)Try further iterations (x3) and will find no change in valuesto three significant figures.HOAc is an acid, therefore OAcHOAc is an acid, therefore OAc-- is a (conjugate) base: is a (conjugate) base:HOAc=CHHOAc=CH33COOH, OAc COOH, OAc -- =CH=CH33COO COO --HydrolysisHydrolysisHOAc + HHOAc + H22O O  H H33OO++ + OAc + OAc --KKaa = 1.85 = 1.85  10 10-5-5Add NaOAc to HAdd NaOAc to H22O, which dissociates completely to NaO, which dissociates completely to Na++, OAc , OAc --Note: KNote: Khh is very small, indicating that little OAc is very small, indicating that little OAc -- combines combinesWith HWith H22O to form HOAcO to form HOAcKKaa=[OAc =[OAc --][H][H33OO++]/[HOAc] is the ionization constant for]/[HOAc] is the ionization constant forthe reaction:the reaction:Multiply KMultiply Khh by 1 = by 1 = [H3O+][H3O+]HOAc + HHOAc + H22O = HO = H33OO++ + OAc + OAc --Since KSince Khh=K=Kww/K/Kaa , the smaller K , the smaller Kaa (weaker the acid) the (weaker the acid) the Larger KLarger Khh (or OAc (or OAc -- is more extensively hydrolyzed) is more extensively hydrolyzed)pKpKaa= = -log-log1010(K(Kaa))QuickTime™ and aVideo decompressorare needed to see this picture.Now suppose add 0.60 moles NaOAc, which dissociatesNow suppose add 0.60 moles NaOAc, which dissociatescompletely to Nacompletely to Na++, OAc , OAc --Consider acetic acid: HOAc + HConsider acetic acid: HOAc + H22O O  H H33OO+ + + OAc+ OAc --HOAc=CHHOAc=CH33COOH, OAc COOH, OAc -- = CH= CH33COO COO --e.g., add 0.70 mole of HOAc to make 1 liter of solution:e.g., add 0.70 mole of HOAc to make 1 liter of solution:Buffer SolutionsBuffer Solutions[HOAc] = [HOAc][HOAc] = [HOAc]oo (initial concentration (initial concentration HOAc)HOAc)First, look at HOAc + HFirst, look at HOAc + H22O O  H H33OO++ + OAc + OAc --How keep KHow keep Kaa=1.85x10=1.85x10-5-5 when whenthrow in all of this OAc throw in all of this OAc -- ??This just uses up HThis just uses up H33OO++ that came from ionization of HOAc, that came from ionization of HOAc,thereby forcing [HOAc] closer to [HOAc]thereby forcing [HOAc] closer to [HOAc]o o !!!!Inverse teeter-totter effect:Inverse teeter-totter effect:Now, look at: OAc Now, look at: OAc -- + H+ H22O O  HOAc + OH HOAc + OH --Very little OH Very little OH -- made this way (lose little OAc made this way (lose little OAc -- and make and makelittle HOAc) since: little HOAc) since: So when put 0.6 moles OAc So when put 0.6 moles OAc -- in a liter of pure water in a liter of pure waterhave [OAc have [OAc --]=[OAc ]=[OAc --]]o o = 0.6 M= 0.6 MResult is to force OAc Result is to force OAc -- even closer to its even closer to its initial value [OAc initial value [OAc --]]oo = 0.60 M = 0.60 MHow to keep KHow to keep Khh = [HOAc] [OH = [HOAc] [OH--] / [OAc ] / [OAc --] =] = 5.4 5.4  10 10 -10 -10 when [HOAc] is large? when [HOAc] is large?Addition of large amounts of a weak acid and its conjugate