Chem 120 Exam 4 Study Guide: Precipitation and Solubility Equilibrium

AcidsandBasesProtondonors:maybemolecularcompounds,cationsoranionsHNOs(aa)+H20(e)->No;Lag)+Hg0"lag)Protonacceptors:maybemolecularcompounds,cationsoranionsNHy(aq)+H20(e)->NHicaa)+OHcoalAcids:HABases:NaOHGH2SO4BaCOH)2EECHONHzAcidinwater:acidacidHA(aa)+He0(l)EAsOt(as)+ACan)base(Acid)(Base)(Conjugate(conjugatebaseacid)base)Conjugatebase:everythingthatHC)(aq)+H20(l)+2-caq)+H30+(aq)remainsoftheacidmoleculeH,POplaal+Hy0(el->H2POCaas+130 caa)afteraproteinislostNHS+H20(el+NHi caa)+OHiau)Conjugateacid:formedWhentheC5HjN+H20(e)CH5NH+(au)+OHiaalDrotonistranferredtothebaseMonoproticGuids:Bronted-LowryAcidsCapableofdonatingoneprotonPolyproticacids:Bronsted-LobryAcidsCapableofdonatingmorethanOneProtonDiprotizacid:PolyproticacidthatcandonatetwoprotonsAmphotericSpecies:FormedWhenanypolyproticacidlosesaproton.Canactasanacidorbase.Polyproticacidscontainbothacidicandnonacidichydrogenatoms.Itransfer#CO;(aq)+120(a)CoS(as)+Hg0"(ag)PH=-logCH30+]acidbaseConjugatebaseConjugateacidH"transferpOH=-log[OH]#Cojtal+H20(ag)HyCos(a)+OH -LualbaseacidConjugateacidConjugatebasePH+DOH=pXw=14kr=CH>O+](ON]*The#ofSigfigsisequalto#ofdecimalKw=1.0x10at25%placesinanswer IsautoionizationConstantforwater"pH=-log[H>0+]==log(1.5x10-2)=1.82DOH=-log[0H]PH+poH=14==10921.5x10-1)14-11.82=2.18=11.82 AcidicSolution:BasicSolutionNeutralSolutionPH=-log[H+]pOH=-logCON]CH+]=CON's[H+]<[OH]COH-]<CH+)pH<Lplt)LConcentrationofanacid,[Hjot]:[H>0+]=10-BHCOH]canbefoundfrom:LOH-]=10-POHOnce(HjO+)isknown,COH]canbefoundfrom:20H-]=[Hz0+]ViceVersa,[Izo+]=ErStrongbase[OH-Balultal-Ba+-0.1850.370HNO,LNOj+Hy0t-log(0.370)PH=-log[Hg0+]POH=0.431=-log(0.185)=0.333PH=14-0.431=13.568Leazbasebicitsabase,you-canuseEnNH,+H20-1NHi+OH-WeatdurduI0.185-SHN82+H20->NO=+Hyot2-X+X+X=0.185-E⊥ -E0.185-X-XXC-X-↓X+XEU.185X#21.8460-5=5 -4.5 x0=5-x3.33x10-"=hn8.325x10-5=xX=0.00182[OH'SX=0.00912pOH=-logL=-log(0.00152)PH=-log(Hj0+)=2.74=-log(0.09/2)=2.04pH=14-2.34=11.24*strong/weatbasewillalwayshaveaputabove7Strong/WeataddwillalwayshaveapHbelowL -reactions-⑳forbaseor-QuidHA+H20->A+Hy0+&(na)+H20ce)->HB anOHan10.36 -&EI2.4x10-g&&-X-+X+YE0.36-x-XXC-Y-+X*X-PHMLaz(1.51x0-5)2X=(Hj05=10E2.4x10-x-Xx=0.36-1.51x0-5=iEx=(3.55x10-5)PH=14-9.55=6.36 x10-104.45(2.4 x 10-3-3.55x020H -=-4.4)10P7a=-log]a=5.32x10-=3.55x10-5=-log(6.36x10-1)=9.20p7a=-loytb=-log(5.32x102)=6.27-HC2HgO2+H20->CHgO2+HgOtI0.25-OOH=A+H20+A-11+-30C---Y+XIU.28-&SE0.25-X-X*2-X-+X->X1.8x10-5=5=x)E0.20-X-X*4.5x10-"=x2.9x10-5=-x)w2-X=0.002/1.58x10-5=xX=0.00397PH=-log[Hj0+)-=-log20.002)PH=-log[H>0+]=2.67==log(0.00397)=2.40HA+120->A2-+OH-I0.004-O2-X+X-E0.004-x-X