Chemistry 12 Acid-Base Equilibrium IV · This allows us to formulate the following relationship for...

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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

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.

Page 2: 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

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.

Page 3: 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

Practice1:Hydrogensulphideisapoisonousflammablegaswhose“rottenegg”smellisperceptibleatconcentrationsaslowas0.00047ppm.Itisalsoaweakacidwhendissolvedinwater.CalculatethepHandpOHof0.0500MH2S.

ProblemType2:Calculatinginitial[HA]Example:WhatconcentrationofbenzoicacidisrequiredtoproduceasolutionwithapOHof10.70?

• UsinggivenpOH,calculatethe[H3O+]

• ConstructanICEtable.

• WhatistheKaforbenzoicacid?

• Solvefortheunknown.

Page 4: 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

ProblemType3:CalculatingKaExample:A0.100Msolutionofacetylsalicylicacid,C8H7O2COOH,isfoundtohaveapHof2.27.CalculatetheKaforthisacid.

• UsinggivenpH,calculatethe[H3O+]atequilibrium.

• ConstructanICEtable.

• WhatistheKaexpression?SolveforKa.HebdenWorkbook:Pg.152#74-80,82,83

Page 5: 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

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-

Page 6: 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

Thefollowingspeciesareamphiprotic.CompareKaandKb.HC2O4-

H2PO4-HPO42-

Page 7: 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

ProblemType1:CalculatingpOHPractice:CalculatethepHandpOHofa0.50Msolutionofmethylamine,CH3NH2(Kb=4.4x10-4)Practice:CalculatethepHofasolutioncontaining0.20MCN-.

Page 8: 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

ProblemType2:Calculatinginitial[B]Practice:WhatconcentrationofNH3wouldberequiredtoproduceasolutionwithpH=10.50?

ProblemType3:CalculatingKbPractice:Asolutionispreparedbydissolving9.90goftheweakbasehydroxylamine,NH2OHinenoughwatertoproduce500.0mLofsolution.ThepHofthesolutionisfoundtobe9.904.CalculatetheKbforhydroxylamine.

Page 9: 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

Practice:A0.400Msolutionoftheweakbasemethylamine,CH3NH2,isfoundtohaveapHof13.30.CalculatetheKbandpKbofmethylamine.HebdenWorkbook:Pg.153-4#84-93