Chemistry 12 Acid-Base Equilibrium IV · This allows us to formulate the following relationship for...
Transcript of Chemistry 12 Acid-Base Equilibrium IV · This allows us to formulate the following relationship for...
![Page 1: Chemistry 12 Acid-Base Equilibrium IV · This allows us to formulate the following relationship for conjugate acid-base pairs: Ka (conjugate acid) x Kb (conjugate base) = Kw = 1.0](https://reader034.fdocuments.us/reader034/viewer/2022042103/5e80df937b08ef4aa94afd9e/html5/thumbnails/1.jpg)
Chemistry12Acid-BaseEquilibriumIV
Name:Date:Block:
1. Ka/KbCalculations
Completethefollowingstatements:
1. Asasolutionbecomesmoreacidic…• [H3O+]increasesordecreases?• pHincreasesordecreases?• [OH-]increasesordecreases?• pOHincreasesordecreases?
2. IfthepHofasolutionequals5.00,the[OH-]equals_____________________M.
3. IfthepOHofasolutiondecreasesby5,thenthe[H3O+]has____________________(increasedordecreased)byafactorof____________________.
CalculationsInvolvingKaandKbRecall…
Acid=HA and Base=B
HA+H2O⇋H3O++A-
Ka =[A- ][H3O+ ]
[HA]
B+H2O⇋HB++OH-
Kb =[HB][OH- ][B]
WritethechemicalequationandKaexpressionthatrepresentsthereactionofHNO2inwater.WritethechemicalequationandKbexpressionthatrepresentsthereactionofNH3inwater.
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CALCULATIONSFORWEAKACIDS
ProblemType1:CalculatingpHExample:CalculatethepHandpOHof0.50Msolutionofhydrofluoricacid.
• IsHFastrongorweakacid?
• Whatisthechemicalequation?(Whatkindofarrowwillyouuse?)
• WhatistheKavalue(fromthetable!)andexpression?
• Sincethisisaweakacid,equilibriumisestablished.ICETABLE!!
HF(aq) + H2O(l) ⇋ F-(aq) + H3O+(aq)Initial
Change
Equilibrium
Wedon’tincludetheconcentrationofwaterbecauseweassumethe[H2O]remainsconstant.• FilloutvaluesfortheKaexpression.
• Theexpressioninthedenominator:0.50M-xcanbeassumedtobe≅0.50M.o ThevalueofKaisverysmallcomparedtotheinitialconcentrationoftheacid.Thismeans
thatthepercentoftheacidthatactuallyionizeswillnotsignificantlychangetheoriginalconcentration.
o Ifinitial[HA]isatleast103timeslargerthantheKavalue,theassumptionisvalid.
• Solveforunknown.
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Practice1:Hydrogensulphideisapoisonousflammablegaswhose“rottenegg”smellisperceptibleatconcentrationsaslowas0.00047ppm.Itisalsoaweakacidwhendissolvedinwater.CalculatethepHandpOHof0.0500MH2S.
ProblemType2:Calculatinginitial[HA]Example:WhatconcentrationofbenzoicacidisrequiredtoproduceasolutionwithapOHof10.70?
• UsinggivenpOH,calculatethe[H3O+]
• ConstructanICEtable.
• WhatistheKaforbenzoicacid?
• Solvefortheunknown.
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ProblemType3:CalculatingKaExample:A0.100Msolutionofacetylsalicylicacid,C8H7O2COOH,isfoundtohaveapHof2.27.CalculatetheKaforthisacid.
• UsinggivenpH,calculatethe[H3O+]atequilibrium.
• ConstructanICEtable.
• WhatistheKaexpression?SolveforKa.HebdenWorkbook:Pg.152#74-80,82,83
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CALCULATIONSFORWEAKBASESAswithacids,mostbasesareweak.Usingthesymbol“B”foraweakbase,wecanrepresenttheequilibriaofweakbasesinwater:
B(aq)+H2O(l)⇋HB+(aq)+OH-(aq) Kb=
WemustcalculatetheKbforthatbasebyusingtheKavalueofitsconjugateacid
Considertheconjugateacid/basepairofNH4+andNH3andtheirrespectiveKaandKbexpressions:
NH4+
Reaction:Ka=
NH3
Reaction:Kb=
Twocommontermsappearineachequation.
• Multiplythetwoexpressionstogetherandcancelthecommonterms…KaxKb=Thisallowsustoformulatethefollowingrelationshipforconjugateacid-basepairs:
Ka(conjugateacid)xKb(conjugatebase)=Kw=1.0x10-14
Forthefollowingweakbases,writeouttheequationwithwaterandcalculatetheKb.CN-
NO2-
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Thefollowingspeciesareamphiprotic.CompareKaandKb.HC2O4-
H2PO4-HPO42-
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ProblemType1:CalculatingpOHPractice:CalculatethepHandpOHofa0.50Msolutionofmethylamine,CH3NH2(Kb=4.4x10-4)Practice:CalculatethepHofasolutioncontaining0.20MCN-.
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ProblemType2:Calculatinginitial[B]Practice:WhatconcentrationofNH3wouldberequiredtoproduceasolutionwithpH=10.50?
ProblemType3:CalculatingKbPractice:Asolutionispreparedbydissolving9.90goftheweakbasehydroxylamine,NH2OHinenoughwatertoproduce500.0mLofsolution.ThepHofthesolutionisfoundtobe9.904.CalculatetheKbforhydroxylamine.
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Practice:A0.400Msolutionoftheweakbasemethylamine,CH3NH2,isfoundtohaveapHof13.30.CalculatetheKbandpKbofmethylamine.HebdenWorkbook:Pg.153-4#84-93