C. Y. Yeung (CHW, 2009) p.01 Acid-Base Eqm (2): K a and K b Review Questions Calculate the pH :...

C. Y. Yeung (CHW, 2009)

p.01

Acid-Base Eqm (2): Acid-Base Eqm (2): KK

aa and K and Kbb

Acid-Base Eqm (2): Acid-Base Eqm (2): KK

aa and K and Kbb

Review QuestionsReview Questions

Calculate the pH :Calculate the pH : Q.1Q.1 1010-8-8 HCl(aq) HCl(aq) [100% ionized][100% ionized]

Q.2Q.2 0.01M NaOH(aq) 0.01M NaOH(aq) [100% ionized][100% ionized]

[H[H33OO++] = (] = (1010-7-7 + 10 + 10-8-8) M = 1.1) M = 1.11010-7 -7 M,M,

pH = -log(1.1pH = -log(1.11010-7-7) = 6.96) = 6.96

[OH[OH--] = 0.01M, pOH = -log(0.01) = -2] = 0.01M, pOH = -log(0.01) = -2

pH = 14 – 2 = 12pH = 14 – 2 = 12

weak acidweak acid

p.02

Acid Dissociation Constant Acid Dissociation Constant (K(Kaa))[temperature dependent!!][temperature dependent!!]

HAHA + + HH22OO AA-- + + HH33OO++

KKcc = =[H[H33OO++(aq)](aq)] [A[A--(aq)](aq)]

[HA(aq)][HA(aq)] [H[H22O(l)]O(l)]

KKcc [H[H22O(l)]O(l)] = =[H[H33OO++(aq)](aq)] [A[A--(aq)](aq)]

[HA(aq)][HA(aq)]

KKaa = =[H[H33OO++(aq)](aq)] [A[A--(aq)](aq)]

[HA(aq)][HA(aq)]

p.03


[HA(aq)][HA(aq)]

Q:Q: KKaa of methanoic acid (HCOOH) at 298K is 1.78 of methanoic acid (HCOOH) at 298K is 1.781010-4-4 mol dm mol dm-3-3. . Calculate the pH value of 1.00M HCOOH.Calculate the pH value of 1.00M HCOOH.

HCOOHHCOOH + + HH22OO HCOOHCOO-- + + HH33OO++

at startat start

at eqmat eqm

1.001.00 0000

1.00 – x1.00 – x xx xx

KKaa = =xx22

1 – x1 – x= 1.78= 1.781010-4-4

xx22 = 1.78 = 1.781010-4-4 (x << 1, (x << 1, KKaa is is

small)small)x = 0.0133x = 0.0133

pH = -log(0.0133) = 1.88pH = -log(0.0133) = 1.88solved eqn without approximation, solved eqn without approximation,

x = 0.01325, pH = 1.88x = 0.01325, pH = 1.88

p.04

KKaa = =xx22

0.01 – x0.01 – x= 1.78= 1.781010-4-4

xx22 = 1.78 = 1.781010-6-6 (x << 1, (x << 1, KKaa is is

small)small)x = 1.33 x = 1.33 1010-3-3

pH = -log(1.33 pH = -log(1.33 1010-3-3) = 2.87) = 2.87

Note: Note: Different conc. of HA has different Different conc. of HA has different “% of dissociation”“% of dissociation”

For 1.00M HCOOH, % of dissociation = (0.01325/1)For 1.00M HCOOH, % of dissociation = (0.01325/1)100% 100% = 1.33 = 1.33

%%

Q:Q: KKaa of methanoic acid (HCOOH) at 298K is 1.78 of methanoic acid (HCOOH) at 298K is 1.781010-4-4 mol dm mol dm-3-3. . Calculate the pH value of 0.01 M HCOOH.Calculate the pH value of 0.01 M HCOOH.

For 0.01M HCOOH, % of dissociation = (1.242For 0.01M HCOOH, % of dissociation = (1.2421010--

33/0.01)/0.01)100% 100% = 12.4 % = 12.4 %

solved eqn without approximation, solved eqn without approximation, x = 1.242x = 1.2421010-3-3, pH = 2.91, pH = 2.91

*** i.e. [HA] decreases, % of dissociation increases.*** i.e. [HA] decreases, % of dissociation increases.

p.05

For For Polyprotic AcidPolyprotic Acid: : KKa1a1, K, Ka2a2, ….., …..[Note: K[Note: Ka1a1 > K > Ka2a2 > … ] > … ] WHY? WHY?

Note:Note: It is more difficult to remove a HIt is more difficult to remove a H++ from the negatively from the negatively charged HSOcharged HSO44

--..

HH22SOSO44 + + HH22OO HSOHSO44-- + + HH33OO++

HSOHSO44-- + + HH22OO SOSO44

2-2- + + HH33OO++

KKa1a1

KKa2a2

HH33OO++ is already produced in the first dissociation, it is already produced in the first dissociation, it

would suppress the formation of Hwould suppress the formation of H33OO++ from the second from the second

dissociation. dissociation. [COMMON ION EFFECT (p.148)][COMMON ION EFFECT (p.148)]

Overall KOverall Kaa of Polyprotic acid = K of Polyprotic acid = Ka1a1 K Ka2a2 … …

p.06

Q:Q: Assuming that the first ionization of HAssuming that the first ionization of H22SOSO44 is 100%, and the is 100%, and the

KKa2a2 of sulphuric acid is 0.01M. Find the pH of 0.10M H of sulphuric acid is 0.01M. Find the pH of 0.10M H22SOSO44..

HSOHSO44-- + + HH22OO SOSO44

2-2- + + HH33OO++ KKa2 a2 = 0.01M= 0.01M

at startat start

at eqmat eqm

0.100.10 0.10.100

0.10 – x0.10 – x xx 0.1+x0.1+x

KKa2a2 = =x(0.1+x)x(0.1+x)

0.1 – x0.1 – x= 0.01= 0.01

x = 8.44x = 8.441010-3-3

pH = -log (0.1 + 8.44pH = -log (0.1 + 8.441010-3-3) = 0.965) = 0.965

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Importance of KImportance of Kaa ?? ??

1.1. Indicates the Indicates the strengthstrength of acid: of acid:

(a)(a) larger Klarger Kaa stronger acid stronger acid

(b)(b) smaller pKsmaller pKaa stronger acid stronger acid

(ref.: p. 1(ref.: p. 141)41)

2.2. Indicates the Indicates the relative stabilityrelative stability of of conjugate basesconjugate bases: :

e.g.e.g. KKaa of ethanoic acid = 1.80 of ethanoic acid = 1.801010-5-5 mol dm mol dm-3-3,,

KKaa of butanoic acid = 1.48 of butanoic acid = 1.481010-5-5 mol dm mol dm-3-3..

i.e.i.e. CHCH33CHCH22CHCH22COOCOO-- is less stable than CH is less stable than CH33COOCOO--. .

p.08

Base Dissociation Constant Base Dissociation Constant (K(Kbb))[temperature dependent!!][temperature dependent!!]

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

KKbb = =[HB[HB++(aq)](aq)] [OH[OH--(aq)](aq)]

[B(aq)][B(aq)]

Q:Q: KKbb of NH of NH33 at 298 K is 1.78 at 298 K is 1.781010-5-5 mol dm mol dm-3-3. What is the pH value of . What is the pH value of

1M NH1M NH33 at 298K? at 298K?

KKbb = =xx22

1 – x1 – x= 1.78= 1.781010-5-5

x = 4.22 x = 4.22 1010-3-3

pOH = -log(4.22 pOH = -log(4.22 1010-3-3) = 2.37) = 2.37

pH = 14 – pOH = pH = 14 – pOH = 11.611.6

p.09

Q:Q: KKbb of NH of NH33 at 298 K is 1.78 at 298 K is 1.781010-5-5 mol dm mol dm-3-3. What is the pH value of . What is the pH value of

a solution which is both 1M NHa solution which is both 1M NH33 and 1M NH and 1M NH44Cl at 298K?Cl at 298K?

NHNH33 + + HH22OO NHNH44++ + + OHOH--

at startat start

at eqmat eqm

11 0011

1 – x1 – x 1+x1+x xx

KKbb = =x(1+x)x(1+x)

1 – x1 – x= 1.78= 1.781010-5-5

x = 1.78x = 1.781010-5-5

pOH = -log(1.78 pOH = -log(1.78 1010-5-5) = 4.75) = 4.75

pH = 14 - 4.75 = 9.25pH = 14 - 4.75 = 9.25

p.10

Importance of KImportance of Kbb ?? ??

1.1. Indicates the Indicates the strengthstrength of base: of base:

(a)(a) larger Klarger Kbb stronger base stronger base

(b)(b) smaller pKsmaller pKbb stronger base stronger base

(ref.: p. 1(ref.: p. 142)42)

2.2. Indicates the Indicates the relative stabilityrelative stability of of conjugate acidconjugate acid: :

e.g.e.g. KKbb of NH of NH33 = 1.74 = 1.741010-5-5 mol dm mol dm-3-3,,

KKbb of (CH of (CH33))22NH = 9,55NH = 9,551010-4-4 mol dm mol dm-3-3..

i.e.i.e. (CH(CH33))22NHNH22++ is more stable than NH is more stable than NH44

++. .

p.11

Relationship between KRelationship between Kww, K, Kaa and K and Kbb ?? ??



[HA(aq)][HA(aq)]

KKbb = =[HA(l)][HA(l)] [OH[OH--(aq)](aq)]

[A[A--(aq)](aq)]

AA-- + + HH22OO HAHA + + OHOH--

KKaa K Kbb = =[H[H33OO++(aq)](aq)] [A[A--(aq)](aq)]

[HA(aq)][HA(aq)]

[HA(l)][HA(l)] [OH[OH--(aq)](aq)]

[A[A--(aq)](aq)][H[H33OO++(aq)](aq)] [A[A--(aq)](aq)]

[HA(aq)][HA(aq)]

[HA(l)][HA(l)] [OH[OH--(aq)](aq)]

[A[A--(aq)](aq)]= K= Kww

AssignmentAssignment

Study all examples in p.142 - 149Study all examples in p.142 - 149

p.152 Q.3 –12 p.152 Q.3 –12 [due date: 26/3(Thur)] [due date: 26/3(Thur)]

p.12

Next ….Next ….Explain the Strength of Organic Explain the Strength of Organic Acids & Bases (Book 3A p. 113 – Acids & Bases (Book 3A p. 113 – 133)133)

Pre-Lab: Expt. 12 Determination of KPre-Lab: Expt. 12 Determination of Kaa

p.13

Expt. 12Expt. 12 Dissociation Constant of Weak AcidsDissociation Constant of Weak Acids



[HA(aq)][HA(aq)]

log Klog Kaa = =[H[H33OO++(aq)](aq)] [A[A--(aq)](aq)]

[HA(aq)][HA(aq)]loglog

pKpKaa = =[A[A--(a(a

q)]q)][HA(aq)][HA(aq)]+ log+ logloglog [H [H33OO++(aq)](aq)]


q)]q)][HA(aq)][HA(aq)]+ log+ logpHpH

p.14



q)]q)][HA(aq)][HA(aq)]+ log+ logpHpH

Step 6 Step 6 (10c(10cmm33 NaOH add NaOH added)ed)

111010-3-3 mol mol0 mol0 mol

Step 5Step 5 111010-3-3 mol mol 0 mol0 mol

Step 7 Step 7 (another 10cm(another 10cm33 HA added)HA added)

111010-3-3 mol mol 111010-3-3 mol mol

at eqmat eqm 111010-3-3 mol – x mol – x 111010-3-3 mol + x mol + x

in the same in the same volume, i.e. same volume, i.e. same

conc.!conc.!

= pH + log (1) = pH= pH + log (1) = pH

measured by pH metermeasured by pH meter

p.15

Simple explanation on the Simple explanation on the Relative Acidity of Organic AcidsRelative Acidity of Organic Acids

The more stable is the conjugate base, The more stable is the conjugate base, the more acidic is the acid.the more acidic is the acid.

1. 1. If the If the –ve charge–ve charge of conjugate base is of conjugate base is localizedlocalized, it i, it is more likely to recombine with the Hs more likely to recombine with the H33OO++ to form th to form th

e acid again. e acid again. the acid is less acidic! the acid is less acidic!

2. 2. If the If the –ve charge–ve charge of conjugate base is of conjugate base is delocalizeddelocalized, , it is less likely to recombine with the Hit is less likely to recombine with the H33OO++ to form t to form t

he acid again. he acid again. the acid is more acidic! the acid is more acidic!

p.16

Alkyl group (e.g. CHAlkyl group (e.g. CH33) is ) is ee-- releasing releasing, ha, ha

lo group (e.g. Cl) is lo group (e.g. Cl) is ee-- withdrawing withdrawing..

HH

CC

HH

CHCH33 CCOO

OO--

HH

CC

HH

ClCl CCOO

OO--

e- releasing e- releasing groupgroup

e- withdrawing e- withdrawing groupgroup

more stable conjugate base, more stable conjugate base, ClCHClCH22COOH is more acidicCOOH is more acidic!!

(i.e. larger K(i.e. larger Kaa, smaller pK, smaller pKaa))

*** Longer alkyl group,

more e- realeasing.

*** More halogen atoms,

more e- withdrawing.

C. Y. Yeung (CHW, 2009) p.01 Acid-Base Eqm (2): K a and K b Review Questions Calculate the pH :...

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