Principles of Chemistry II © Vanden BoutTodaySolubilityThe easiest of all the equilibriaPolyprotic Acidsdetermining something about an unknownby reacting it with a known solution Principles of Chemistry II © Vanden BoutSolubility EquilibriaMg(OH)2 (s) Mg2+(aq) + 2OH-(aq)K = [Mg2+][OH-]2spsolubility product Principles of Chemistry II © Vanden Boutexample Principles of Chemistry II © Vanden BoutIn which solution with Mg(OH)2 have the highest solubility?! A.! ! pure water ! B.! ! 1 M NaOH ! C.! ! 1 M MgCl2 !! D.! ! 1 M HClThe acid will neutralize the OH-allowing more to dissolvePrinciples of Chemistry II © Vanden BoutIf I add 1 mole of Al(OH)3 to1 L of a 1 M HCl solutionwhat is the pH?Al(OH)3 (s) Al3+(aq) + 3OH-(aq)H+(aq) + OH-(aq) H2O(l)Ksp = 4.6 x 10-33First neutralize. Even though the [OH-] concentration will be small, whatever dissolves will neutralizeThen you can solve the equilibrium Principles of Chemistry II © Vanden BoutSilver Nitrate (AgNO3) and Potassium Chloride (KCl) are both soluble salts. What will happen if I mix 100 mL of 1 M AgNO3 solution with 200 ml of 1 M KCl solution given that Ksp for AgCl is 1.8 x 10-10! A.! ! I'll have a solution with Ag+, Cl-, K+, and NO3- ions ! B.! ! some solid AgCl will form ! C.! ! both B & C ! Principles of Chemistry II © Vanden BoutPrecipitationLike neutralization problemsFirst react, the solve the equilibriumKsp is generally small.First assume as much solid as possible formsThen look at what "re-dissolves" into solution Principles of Chemistry II © Vanden BoutProblemPrinciples of Chemistry II © Vanden BoutPolyprotic AcidsAcids that have more than one proton to loseNow we need to keep track of all the "forms" of the acidMonoprotic HA , A-Diprotic H2A, HA-, A2-Triprotic H3A, H2A-, HA2-, A3- Principles of Chemistry II © Vanden BoutFor exampleSulfuric AcidH2SO4(aq) H+(aq) + HSO4-(aq)H2AHSO4-(aq) H+(aq) + SO42-(aq)HA-A2-HA-Ka1 = [H+][HSO4-][H2SO4]Equilibrium for the firstproton coming "off"= 103Ka2 = [H+][SO42-][HSO4-]=1.2 x10-2 Equilibrium for the nextproton coming "off" Principles of Chemistry II © Vanden BoutKey QuestionWhat is in solution!H2A(aq) H+(aq) + HA-(aq)HA-(aq) H+(aq) + A2-(aq)Ka1 = [H+][HA-][H2A]Ka2 = [H+][A2-][HA-]we'll reduce all such problems to 1 or 2 major forms of the acid.First figure out which ones will be in solution Principles of Chemistry II © Vanden BoutCitric AcidKa1 = 7.4 x 10-4 Ka2 = 1.7 x 10-5 Ka3 = 4.0 x 10-7 What is the pH of 1M Citric Acid?Imagine that it was monoproticH3A H+H2A-ICECa-xCa-x0+x+x0+x+xKa1 = [H+][H2A-][H3A](x)(x)Ca - x(x)(x)Ca==[H+] = x = KaCaPrinciples of Chemistry II © Vanden BoutCitric AcidKa1 = 7.4 x 10-4 Ka2 = 1.7 x 10-5 Ka3 = 4.0 x 10-7 Imagine that it was monoprotic[H+] = x = KaCa= (7.4 x 10-4)(1)= 0.027 Lets look at Ka2 Ka2 = [HA2-][H2A-][H+][HA2-][H2A-]=Ka2[H+]=1.7 x 10-5 0.027 =6.3 x 10-4 This is a very small numbervery very little HA2- the second proton doesn't come offpH is dominated by the first proton equilibrium Principles of Chemistry II © Vanden BoutWhen do I care about the other protons?When I neutralize the acid.As you neutralize the first protons,the second will come off,....If I add 0.1 moles of NaOH to 0.05 moles of H3PO4 what will be the dominant species in solution? Principles of Chemistry II © Vanden BoutIf I add 0.1 moles of NaOH to 0.05 moles of H3PO4 what will be the dominant species in solution?! A.! ! H3PO4 and H2PO4-! B.! ! H2PO4- ! C.! ! H2PO4- and HPO42-!! D.! ! HPO42-! E.! ! HPO42- and PO43--.05 moles OH would neutralize all theH3PO4 making 0.5 moles of H2PO4-.05 moles would neutralize all theH2PO4- making 0.5 moles of HPO42-OH- no all neutralizedwhat is left? 0.5 moles of HPO4- Principles of Chemistry II © Vanden BoutWhat is the pH of a solution with 0.5 M HPO42-?H3PO4 Ka1 = 7.1 x 10-3Ka2 = 6.3 x 10-8Ka3 = 4.5 x 10-13to simplify we'll use the generic notation HPO42- is HA2-HA2- is found in equilibria 2 & 3Ka2 = [H+][HA2-][H2A-]Ka3 = [H+][A3-][HA2-]Principles of Chemistry II © Vanden BoutWhat is the pH of a solution with 0.5 M HPO42-?H3PO4 Ka1 = 7.1 x 10-3Ka2 = 6.3 x 10-8Ka3 = 4.5 x 10-13Ka2 = [H+][HA2-][H2A-]Ka3 = [H+][A3-][HA2-][HA2-] = [H+][A3-]Ka3 Ka2 = [H+][H2A-][H+][A3-]Ka3 [H+]Ka2 x Ka3 =assume thesmall change informing both acid and base Principles of Chemistry II © Vanden BoutTitration of a polyproticTwo equivalence pointsDiprotic H2A Principles of Chemistry II © Vanden BoutTitration of a polyproticall H2A weak acid Principles of Chemistry II © Vanden BoutTitration of a polyproticOH- neutralizes some H2A to HA-buffer around Ka1halfwayto equivalence point 1pH = pKa1Principles of Chemistry II © Vanden BoutTitration of a polyproticequivalence point 1moles OH- = moles H2AAll H2A converted to HA- Principles of Chemistry II © Vanden BoutTitration of a polyproticOH- neutralizes HA- to A2-HA- and A2-buffer around Ka2halfwayto equivalence point 1pH = pKa2 Principles of Chemistry II © Vanden BoutTitration of a polyproticequivalence point 2moles OH- = 2 x moles H2Anow all H2A is
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