C. Y. Yeung (CHW, 2009)

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Acid-Base Eqm (3): Acid-Base Eqm (3): Explain the Strength of Explain the Strength of Organic Acids & Bases Organic Acids & Bases

Acid-Base Eqm (3):Acid-Base Eqm (3): Explain the Strength of Explain the Strength of Organic Acids & Bases Organic Acids & Bases

Review QuestionsReview Questions

Q.1:Q.1: pKpKbb of CH of CH33NHNH22 (an organic weak base) at 298K is 4.75. (an organic weak base) at 298K is 4.75.

Find the pH of 0.05M CHFind the pH of 0.05M CH33NHNH22 at 298K. at 298K.

KKbb = =xx22

0.05 – x0.05 – x= 10= 10-4.75-4.75

xx22

0.050.05= 1.778= 1.7781010-5-5

pOH = -log(9.43 pOH = -log(9.43 1010-4-4) = 3.03) = 3.03

pH = 14 – pOH = pH = 14 – pOH = 11.011.0

x = 9.43x = 9.431010-4-4

p.02Q.2:Q.2: 25.0 cm25.0 cm33 of a monobasic organic weak base B(aq) requires of a monobasic organic weak base B(aq) requires

20.25 cm20.25 cm33 of 0.50M HCl(aq) for complete neutralization. Give of 0.50M HCl(aq) for complete neutralization. Given that the pH of the B(aq) at 298K is 11.25, what is Kn that the pH of the B(aq) at 298K is 11.25, what is Kbb of B at of B at

298K?298K?

As pH = 11.25, i.e. pOH = 14 – 11.25 = 2.75As pH = 11.25, i.e. pOH = 14 – 11.25 = 2.75

conc. of B(aq) =conc. of B(aq) =(20.25/1000)(20.25/1000)0.500.50

= 0.405 M= 0.405 M(25.0/1000)(25.0/1000)

BB + + HH22OO HBHB++ + + OHOH--

at startat start

at eqmat eqm

0.4050.405 0000

0.405 – x0.405 – x xx xx

x = [OHx = [OH--] = 10] = 10-2.75-2.75 = 1.778 = 1.7781010-3-3

KKbb = =(1.778(1.7781010-3-3))22

0.405 – 1.7780.405 – 1.7781010-3-3= 7.84= 7.841010-6 -6 MM

p.03

Stability of Stability of Conjugate BaseConjugate Base … …

e.g. CCle.g. CCl33COOCOO-- is the conjugate base of CCl is the conjugate base of CCl33COOHCOOH

i.e. i.e. CClCCl33COOCOO-- is stable, thus it is is stable, thus it is less likelyless likely to react to react

with Hwith H33OO++ to regenerate CCl to regenerate CCl33COOH.COOH.

CClCCl33COOCOO-- is stabilized by is stabilized by

““negative inductive negative inductive effecteffect”.”.

electronegative electronegative atom !!atom !!

The -ve charge is The -ve charge is dispedispersedrsed /delocalized over t /delocalized over the anion.he anion.

The -ve charge is The -ve charge is less concless concentratedentrated on the O atom. on the O atom.

CClCCl33COOHCOOH + + HH22OO CClCCl33COOCOO-- + + HH33OO++

ClCl

CC

ClCl

ClCl CCOO

OO--

p.04

Conjugate BaseConjugate Base may be stabilized may be stabilized by “by “resonance effectresonance effect””

OH O-

+ H+ H22OO + H+ H33OO++

phenolphenol phenoxidephenoxide

OO-- OO

--

OO

--

OO

--

The -ve charge is delocalized over the aromatic ring.The -ve charge is delocalized over the aromatic ring. (resonance effect) (resonance effect)

i.e. phenoxide ion is stabilized by resonance effect.i.e. phenoxide ion is stabilized by resonance effect.

(resonance structures)(resonance structures)

p.05pKpKaa of Phenol = 9.95 of Phenol = 9.95 pp

KaKa of Ethanoic acid = 4.76 of Ethanoic acid = 4.76stronger acid!stronger acid!

How to explain …?How to explain …?

Although CHAlthough CH33COOCOO-- is destabilized by is destabilized by positive inductive effectpositive inductive effect, ,

it is stabilized by resonance effect!it is stabilized by resonance effect!

CHCH33 CCOO

OO-- CHCH33 CC

OO

OO

-- both are both are electronegative! electronegative!

i.e. i.e. –ve charge is s–ve charge is shared equally.hared equally.

In the case of phenoxide, In the case of phenoxide, delocalization of –ve chargedelocalization of –ve charge by re by resonance is sonance is less effectiveless effective, because the , because the O atom is more electO atom is more electronegative than the carbon atomsronegative than the carbon atoms in aromatic ring. in aromatic ring.

The –ve charge is more delocalized in CHThe –ve charge is more delocalized in CH33COOCOO--, and th, and th

us it is more stable. us it is more stable. CH CH33COOH is the stronger acid.COOH is the stronger acid.

(resonance structures)(resonance structures)

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Strength of Strength of BaseBase … … (depends on the (depends on the availabilityavailability of lone pair e of lone pair e--.).)

CHCH33 NN HH

HH

push epush e--

availabilityavailability of lone pair e of lone pair e-- is is increased by +I effectincreased by +I effect

stronger base (than NHstronger base (than NH33)!)!

NN HH

HH

NN HH

HH

availabilityavailability of lone pair e of lone pair e-- is is decreased by resonance.decreased by resonance.

weaker base (than NHweaker base (than NH33)!)!

NN HH

HH--

++HHNN

HH

--++

HHNN

HH

--++

Methylamine Methylamine (pK(pKbb = 3.38) = 3.38)

phenylamine / aniline phenylamine / aniline (pK(pKbb = 9.40) = 9.40) Ref: pKRef: pKbb of NH of NH33 = 4.80 = 4.80

a.1Advanced Understanding on the Strength of Organic Acid Advanced Understanding on the Strength of Organic Acid

(1) --- (1) --- 4-nitrophenol 4-nitrophenol (p-nitrophenol)

OOOO

OO

NN --++

--How can How can 4-nitrophenoxide ion4-nitrophenoxide ion be stabilized by resonance effect? be stabilized by resonance effect?

OOOO

OO

NN --++

--

OOOO

OO

NN --++

--

OOOO

OO

NN --++

--

OOOO

OO

NN++--

--

OOOO

OO

NN++--

--

The -ve charge on the The -ve charge on the pp-nitro phe-nitro phenoxide ion is noxide ion is more dispersedmore dispersed tha than that on the phenoxide ion.n that on the phenoxide ion.

Therefore, it is stronger than pheTherefore, it is stronger than phenol.nol.

pKpKaa of phenol = 9.95 of phenol = 9.95

pKpKaa of of pp-nitrophenol = 7.14-nitrophenol = 7.14

a.2Advanced Understanding on the Strength of Organic Acid (2)Advanced Understanding on the Strength of Organic Acid (2)

--- --- 33-nitrophenol-nitrophenol (m(m-nitrophenol)-nitrophenol)Is the pKIs the pKaa of of mm-nitrophenol smaller than -nitrophenol smaller than pp-nitrophenol?-nitrophenol?

OO

OO

OO

NN

--

++

--

OO

OO

OO

NN

--

++

--

OO

OO

OO

NN

--

++--

OO

OO

OO

NN

--

++

--

The -ve charge The -ve charge cannot be dcannot be dispersedispersed over the nitro grou over the nitro group.p.

Therefore, it is weaker than Therefore, it is weaker than pp-nitrophenol, and its pK-nitrophenol, and its pKaa is is

larger.larger.

((But it is still stronger than But it is still stronger than phenol due to negative induphenol due to negative inductive effectctive effect.).)

pKpKaa of phenol = 9.95 of phenol = 9.95

pKpKaa of of pp-nitrophenol = 7.15-nitrophenol = 7.15

pKpKaa of of mm-nitrophenol = 8.35-nitrophenol = 8.35

a.3Advanced Understanding on the Strength of Organic Acid (3) Advanced Understanding on the Strength of Organic Acid (3)

--- --- pp-hydroxybenzaldehyde-hydroxybenzaldehydeHow can the conjugate base be stabilized by resonance effect?How can the conjugate base be stabilized by resonance effect?

HHOO

OO

CC

--

HHOO

OO

CC

--

HHOO

OO

CC--

HHOO

OO

CC--

HHOO

OO

CC--

HHOO

OO

CC

--

The -ve charge is The -ve charge is disperseddispersed over the anion, over the anion, pp-hydroxy b-hydroxy benzaldehyde is acidic. Therenzaldehyde is acidic. Therefore, it is stronger than phefore, it is stronger than phenol.enol.

pKpKaa of phenol = 9.95 of phenol = 9.95

pKpKaa of of pp-hydroxybenzaldhyde = 7.66-hydroxybenzaldhyde = 7.66

a.4

Look at their pKLook at their pKaa values again …… values again ……

HHOO

OHOH

CC

OHOH

OOOO

OHOH

NN --++

OO

OO

OHOH

NN

--

++

pKpKaa (at 298(at 298

K)K)

9.959.95 7.157.15 8.358.35 7.667.66

nitro group (– NOnitro group (– NO22) at the ) at the pp-position -position

stabilizes the conjugate base by delstabilizes the conjugate base by delocalizing the – ve charge through:ocalizing the – ve charge through:

negative inductive effectnegative inductive effect

resonance effectresonance effect

a.5

Compare the pKCompare the pKbb values of values of mm-nitroaniline, -nitroaniline, pp-nitroaniline -nitroaniline

and aniline.and aniline.

NHNH22

OOOO

NHNH22

NN --++

OO

OO

NHNH22

NN

--

++

pKpKbb (at 298(at 298

K)K)

9.409.40 12.912.9 11.611.6

The availability of lone pair eThe availability of lone pair e-- on the on the –NH–NH22 group is group is reducedreduced by : by :

negative inductive effectnegative inductive effect

resonance effectresonance effect

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In Conclusion ….In Conclusion ….

Stronger Acid:Stronger Acid: conjugate base stabilized by - I effect / resconjugate base stabilized by - I effect / resonance effect. onance effect. (delocalizati (delocalization of –ve charge)on of –ve charge)

Stronger Base:Stronger Base: availability of lone pair e availability of lone pair e-- increased by + I increased by + I effect. effect. (localization of –ve (localization of –ve charge)charge)

Weaker Acid:Weaker Acid: conjugate base destabilized by + I effect. (lconjugate base destabilized by + I effect. (localization of –ve charge)ocalization of –ve charge)

Weaker Base:Weaker Base: availability of lone pair eavailability of lone pair e-- decreased by - I decreased by - I effect / resonance. effect / resonance. (delocalization of –ve charge) (delocalization of –ve charge)

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More practice on the calculation of KMore practice on the calculation of Kaa, K, Kbb and pH and pH

Q.1:Q.1: At body temperature, human blood pH is 7.40. Lactic acid, At body temperature, human blood pH is 7.40. Lactic acid, CHCH33CH(OH)COOH, is produced in muscle tissues during pCH(OH)COOH, is produced in muscle tissues during p

hysical exertion. Whether the lactic acid produced exists hysical exertion. Whether the lactic acid produced exists mainly in form of “undissociated molecule” or “dissociatemainly in form of “undissociated molecule” or “dissociated lactate ions”?d lactate ions”?

(Given: K(Given: Kaa of lactic acid at body temperature = 8.50 of lactic acid at body temperature = 8.501010-4-4 mol dm mol dm--

33))

KKaa = =[H[H33OO++][CH][CH33CH(OH)COOCH(OH)COO--]]

[CH[CH33CH(OH)COOH]CH(OH)COOH]

pKpKaa = =[CH[CH33CH(OH)COOCH(OH)COO--]]

[CH[CH33CH(OH)COOH]CH(OH)COOH]pH –pH – loglog

2.14 2.14 10104 4 ==[CH[CH33CH(OH)COOH]CH(OH)COOH]

[CH[CH33CH(OH)COOCH(OH)COO--]]

Lactic acid formed is mainly in form of Lactic acid formed is mainly in form of dissociated lactic ionsdissociated lactic ions. .

With a given Ka, the ratio of [salt] and [acid] could be found from the pH value.

p.09Q.2:Q.2: Calculate the pH value of the resultant solution for Calculate the pH value of the resultant solution for

10cm10cm33 of 0.20M CH of 0.20M CH33COOH are mixed with 10cmCOOH are mixed with 10cm33 of of

0.40M CH0.40M CH33COONa. (KCOONa. (Kaa = 1.75 = 1.751010-5-5 mol dm mol dm-3-3))

x(0.20+x)x(0.20+x)

0.10 – x0.10 – x1.751.751010-5-5 = =

x = 8.75 x = 8.75 1010-6-6

pH = 5.06pH = 5.06

After mixing,After mixing, new [CHnew [CH33COOH] = 0.10MCOOH] = 0.10M

new [CHnew [CH33COONa] = 0.20MCOONa] = 0.20M

An acid-salt mixture (acid and salt have An acid-salt mixture (acid and salt have comparablecomparable conc.) conc.)

ACIDIC BUFFER! ACIDIC BUFFER!

pH of pH of acidicacidic buffer buffer could be adjusted by

could be adjusted by [acid] and [salt].[acid] and [salt].

p.10Q.3:Q.3: Calculate the pH value of the resultant solution for Calculate the pH value of the resultant solution for

10cm10cm33 of 1.00M NH of 1.00M NH33 are mixed with 10cm are mixed with 10cm33 of 1.00M of 1.00M

NHNH44Cl. (KCl. (Kbb = 1.78 = 1.781010-5-5 mol dm mol dm-3-3))

x(0.50+x)x(0.50+x)

0.50 – x0.50 – x1.781.781010-5-5 = =

x = 1.78 x = 1.78 1010-5-5

pOH = 4.75pOH = 4.75

After mixing,After mixing, new [NHnew [NH33] = 0.50M] = 0.50M

new [NHnew [NH44++] = 0.50M] = 0.50M

A base-salt mixture (base and salt have A base-salt mixture (base and salt have comparablecomparable conc.) conc.)

BASIC BUFFER! BASIC BUFFER!

pH = 9.25pH = 9.25

pH of pH of basicbasic buffer

buffer could be adjusted by

could be adjusted by [base] and [salt].

[base] and [salt].

AssignmentAssignment

Lab report Lab report [due date: 6/4(Mon)] [due date: 6/4(Mon)]

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Next ….Next ….Acidic Buffers and Basic Buffers Acidic Buffers and Basic Buffers (Book 2 p. 153 – 163)(Book 2 p. 153 – 163)

Pre-Lab: Pre-Lab: Expt. 13 Expt. 13 Analysis of Two Commercial Analysis of Two Commercial Brands of Bleaching SolutionBrands of Bleaching Solution

Book 3A p.141 Q.12, 13, 14 Book 3A p.141 Q.12, 13, 14 [due date: 20/4 (Mon)] [due date: 20/4 (Mon)]