acids ii using ka

7/30/2019 Acids II Using Ka

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ACIDS WITH LARGE VALUES OF KaARE STRONGER THAN ACIDS

WITH SMALLER VALUES OF Ka

Ka = 1.74x10-5M

EXAMPLE 5: (a) Find pH of 0.0500M CH3COOH; (b) the

concentrations of the species; and (c) % dissociation of acetic acid.

(a)Init. 0.05 0 0

Change -x +x +x

Eqm. 0.05-x x x

251.74 10

0.05

xx

x

=

( )2 51.74 10 0.05x x x=

Solve quadratic equation or use method of successive approximations


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METHOD OF SUCCESSIVE APPROXIMATION

( )2 51.74 10 0.05x x x=

x is small with respect to 0.05.

First approximation, x = 0.

( )2 51.74 10 0.05x x =

x = 9.33 x 10-4

MFor small K

athis is sufficient.

2nd approximation,(2 51.74 10 0.05 9.33 10x x x = 4

x = 9.24 x 10-4 M

3rd approximation,(2 51.74 10 0.05 9.24 10x x x = 4

x = 9.24 x 10-4

M


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SHORTCUT METHOD OF SUCCESSIVE APPROXIMATION

( )51.74 10 0.05x x x = ( )2 51.74 10 0.05x x x=

0 ENTER

( )( )E-1.74 5 .05 ANS

ENTER

ENTER

ENTER

0

9.33E-4

9.24E-4

9.24E-4

TI-83

[H3O+] = 9.24 x 10-4M

Aside, log(9.33E-4)

= 3.03pH = -log [H3O

+] = -log(9.24E-4) = 3.03

[CH3COO-] = 9.24 x 10-4M

[CH3COOH] = 0.05 -x = 0.05 - 9.24 x 10-4 = 0.0491M

(b)

(c) % CH3COOH dissociated = (x/[CH3COOH]0)(100%)= (9.24 x 10-4 /0.05)(100%) = 1.8%


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% Dissociation of Acetic Acid as a Function of Its Initial Concentration

Technique of example 5 works for [CH3COOH]0 > 2x10-5M

Why not for smaller concentrations?

McQuarrie

(5 5 51 51.74 10 2.0 10 1.9 1 ; 1.18 10 0x x Mx x x x x x = = =

% CH3COOH dissociated = (x/[CH3COOH]0)(100%)= (1.18 x 10-5 /2x10-5)(100%) = 59 %


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pKa

Defined

pKa = -log Ka = -log(1.74E-5) = 4.76 foracetic acid

pKa

= -log Ka

McQuarrie


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BASES WITH LARGE VALUES OF KbARE STRONGER THAN

BASES WITH SMALLER VALUES OF Kb

McQuarrie


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BASES WITH LARGE VALUES OF KbARE STRONGER THAN

BASES WITH SMALLER VALUES OF Kb

All have the same reaction as NH3 (aq):

McQuarrie


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McQuarrie


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EXAMPLE 6: Calculate the pH and the concentrations

of all the species in a 0.755 M hydroxylamine,

HONH2(aq), solution.

Init. 0.755M 0 0

Change -x +x +x

Eqm. 0.755-x x x

28

1.07 100.755

x

xx

= ( )

2 8

1.07 10 0.755x x x

=

( )8 -51.07 10 0.755 0 8.99x10x x = =


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pOH = -log [OH-]= -log(8.99E-5) = 4.05

pH = 14 - pOH =14 4.05 = 9.95

[HONH3+] = x = 8.99 x 10-5M

[HONH2] = 0.755 x = 0.755 - 8.99 x 10-5 = 0.755M


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THE PAIR OF SPECIES HB,B- IS CALLED A CONJUGATE ACID-BASE PAIR

( ) ( ) ( )2 3H aq H O l H aB BO q+

+ +

Acid Base Conjugate

Acid

Conjugate

Base

McQuarrie


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EXAMPLE 7: Determine the conjugate base for NH4+(aq).

Conjugate

Base

Conjugate

Acid

Acid Base


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SALTS OF THE CONJUGATE BASE OF STRONG ACIDS:

ACIDIC, BASIC, OR NEUTRAL? NEUTRAL

Dissociates completelyHCl(g) +H2O(l) H3O+(aq) +Cl-(aq)

Cl-(aq) +H2O(l) No Reaction Cl-(aq) has no tendency to grab H+

from H2O

CATIONS OF STRONG BASES: ACIDIC, BASIC, OR NEUTRAL?

NEUTRAL

Dissociates completelyNaOH(aq) Na+(aq) + OH-(aq)

Na+(aq) +H2O(l) No Reaction Na+(aq) has no tendency to grap

OH- from H2O

EXAMPLES NaCl(aq) NEUTRAL

CaBr2(aq) NEUTRAL


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SALTS OF THE CONJUGATE BASE OF WEAK ACIDS:

ACIDIC, BASIC, OR NEUTRAL? BASIC

HCN(aq) +H2O(l) H3O+(aq) +CN-(aq) Mostly undissociated

CN-(aq) +H2O(l) HCN(aq) +OH-(aq) CN-(aq) can grab H+ from H2O

THE WEAKER THE ACID THE BETTER IT CAN GRAB H+ FROM H2OAND THE STRONGER THE BASE

SIMILARLY, SALTS OF THE CONJUGATE ACID OF WEAK BASES:

ACIDIC

EXAMPLES NaCN(aq) BASIC

Ca(CH3COO)2(aq) BASIC

NH4Cl(aq) ACIDIC

KI(aq) NEUTRAL

If this slide and the last one are confusing, see analogy for salts of weak acids using dogs

and bones found in the Supplemental Material.


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SALTS OF THE CONJUGATE BASE OF WEAK ACIDS ARE BASIC

EXAMPLE 8: Determine the pH of 0.400M NaCH3COO (aq).

Kb = ?

( ) ( ) ( )2 32H O l H O aq OH aq+

+ Kw

Kb

=Init. 0.400M 0 0 14 2

5

10

1.74 10

M

x M

=Change -x +x +x

Eqm. 0.4-x x x 105.75 10x M=

2105.75 10

0.4

xx

x

=

( )2 105.75 10 0.4x x x=


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( )10 5

5.75 10 0.4 0 1.52 10x x x M

= =

pOH = -log [OH-]= -log(1.52E-5) = 4.82

pH = 14 - pOH =14 4.05 = 9.18

SIMILARLY SALTS OF THE CONJUGATE ACID OF WEAK BASES ARE ACIDIC


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SIMILARLY, SALTS OF THE CONJUGATE ACID OF WEAK BASES ARE ACIDIC

EXAMPLE 9: Determine the pH of 0.400M NH4Cl (aq).

Ka = ?



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Ka = ?

( ) ( ) ( ) ( )4 2 3NH aq OH aq H O aq NH aq+

+ +1

1

b

K

K

=

( ) ( ) ( )2 32H O l H O aq OH aq+

+ Kw

wa

b

KKK

=



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Ka = ?

( ) ( ) ( ) ( )4 2 3NH aq OH aq H O aq NH aq+

+ +1

1

b

K

K

=

( ) ( ) ( )2 32H O l H O aq OH aq+

+ Kw

wa

b

KKK

=

Init. 0.400M 0 0 14 2

5

10

1.75 10

M

x M

=Change -x +x +x

Eqm. 0.4-x x x 105.71 10x M=

2105.71 10

0.4

xx

x

=

51.51 10x x =

pH = -log [H3O+] = -log(1.51E-5) = 4.82

F j b id


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For conjugate bases or acids:a b w

K K K=

Kb

=For the conjugate base:

wa

b

KK

K=For the conjugate acid:

AQUEOUS SOLUTIONS OF MANY SALTS ARE EITHER ACIDIC OR BASIC


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K>>1( ) ( ) ( ) ( )2 3HCl aq H O l H O aq Cl aq+

+ +(1)

( ) ( ) ( )NaOH aq Na aq OH aq+ + K>>1

Na+ and Cl- are neutral ions (pH does not change in their presence).

Neutral cations are cations of strong bases.

Neutral anions are anions of strong acids (except sulfuric acid).

NaCl, KClO4, CsNO

3

(2) The basic anions are the conjugate bases of weak acids.

Because most acids are weak, most anions are basic.

CH3COO-, F-, CO32-

(3) The acidic cations include the conjugate acids of weak bases.

NH4+, CH3NH3

+



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(4)The acidic cations include the cations of some solvated metal ions.

McQuarrie



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(5)The hydrogen sulfate ion and the dihydrogen phosphate ion are acidicanions because each undergoes a second acid-dissociation reaction:

EXAMPLE 10: Predict whether the following salts produce acidic, neutral, or

basic solutions when dissolved in water: NaCl, NaCH3COO, NH4Cl, and NH4F.

McQuarrie

SUMMARY


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SUMMARY

McQuarrie

HSO -More practice problems: Acidic


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HSO4More practice problems: Acidic

Ca(ClO4)2 Neutral

KCN Basic

Cs2SO4 Basic

AlCl3 Acidic

KNO3 Neutral

NaI Neutral

BasicKCCl3COO

NH4HCOO

NH4Br14

10

4 5

105.71 10

1.75 10

wa

b

KNH K x

K x

+

= = =

1411

4

10

5.61 101.78 10

wb

a

K

HCOO K xK x

= = =

Acidic

Acidic

acids ii using ka

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