Chemistry12Acid-BaseEquilibriumIII

Name:Date:Block:

1. pHandpOH

pHandpOHCompletethefollowingtable:Solution [H3O+] [OH-]1.0MNaOH

1.0MHCl

• Concentrationofacidsandbasescanrangefromextremelyhightoextremelylow.• Itiseasiertoexpresstheseconcentrationsaslogarithms.

Logarithms• “Powerof10”waytospecifytheconcentrationofhydroniumorhydroxideionsinasolution.• Thelogarithmofanumberisthepowertowhich10mustberaisedtoobtainthatnumber.

10y=x⇔ Log10x=y

Practice.Takethelogofthefollowingnumbers.

log(1.0x10-9)=

log(1.0x10-7)=

log(1.0x10-5)=

log(1.0x10-3)=

log(5.0x10-9)=

log(2.4x10-7)=

log(1.6x10-5)=

log(7.9x10-3)=

Wewanttoavoidnegativenumberssowemultiplyby–1.Thisiscalledtakingthe“negativelog”.

–log(7.9x10-3)=________

Practice.TaketheNEGATIVElogofthefollowingnumbers.

–log(1.0x10-8)=

–log(1.0x10-6)=

–log(1.0x10-4)=

–log(1.0x10-2)=

–log(5.0x10-8)=

–log(2.4x10-6)=

–log(1.6x10-4)=

–log(7.9x10-2)=

• Thereverseof“takingthelog”isto“taketheantilog”àEXPONENTIALFORM

• Itjustsimplymeanstowritethenumberasapowerof10.

Antilog(2.0)=102.0=100 Antilog(-2.0)=10-2.0=0.01

Practice.Calculatethefollowing.

10–4.23=

10–0.34=

10–6.89=

10–5.790=

10–2.1=

10–6.71=

10–5.33=

10–1.1=

SignificantFiguresforLogs:o Onlythedigitsafterthedecimalplaceofalogvaluearesignificant:

Ex:–log(5.28x10-5)=[–log(5.28)]+[–log(10-5)]=–0.723+(5)=4.277

Molarity:5.28x10-5M(3SF) pH=4.277(3SF)

Practice.Whichsolutionshavethecorrectnumberofsignificantfigures?Fortheincorrectsolutions,writethecorrectanswerbelow.–log(5.61x10-8)=7.25

–log(8.9x10-5)=4.051

–log(3.0912x10-2)=1.509872895

–log(1.0x10-10)=10.00

10–4.52=3.02x10-5

10–3.1=8x10-4

10–1.11=0.078

10–0.96=0.1096

Kw=[H3O+][OH-]

Takethenegativelog…

-logKw=-log[H3O+]+-log[OH-]

Resultsin…

pKw=pH+pOH

Whichmeans:

pH=-log[H3O+]andpOH=-log[OH-]

and

[H3O+]=10–pHand[OH-]=10–pOH

Practice:

1. WhatisthepHofa0.010Mnitricacidsolution?

2. WhatisthepHofasolutionwith[H3O+]=3.2x10-4M?

3. Whatis[H3O+]ofasolutionwithpH=2.31?

4. WhatisthepOHofa0.05MNaOHsolution?

5. WhatisthepOHofasolutionwith[OH-]=2.08x10-12?

6. WhatisthepHofasolutionwithapOH=11.022?

Kw=[H3O+][OH-]=1.00x10-14

Takethenegativelog…

-logKw==-log[H3O+]+-log[OH-]=-log(1.00)+-log(10-14)

Resultsin…

pKw=pH+pOH=0+14

Whichmeans:

pH+pOH=14Practice:

1. IfpH=0.355,whatispOH?

2. IfpH=6.330,whatis[OH-]?

[H3O+]

[OH-]

pH

pOH

HebdenWorkbook:Pg.139#49-53,Pg.141#55-56:

DeterminethepHofthesolutionthatresultswhen50.0mLof0.200MH2SO4ismixedwith100.0mLof0.400MNaOH.Astudentadds35.0mLofanHClsolutionwithapHof2.00to15.0mLofNaOHsolutionwithapHof12.00.CalculatethepHofthefinalsolution.WhatmassofNaOHmustbeaddedto500.0mLofasolutionof0.020MHItoobtainasolutionwithapHof2.50?

HebdenWorkbookPg.143#58-68