Bonus * Bonus * Bonus * Bonus * Bonus * Bonus Bonus * Bonus * Bonus - - PowerPoint PPT Presentation

SLIDE 1

Bonus * Bonus * Bonus * Bonus * Bonus * Bonus Bonus * Bonus * Bonus * Bonus * Bonus * Bonus

SLIDE 2

Suppose you have [H Suppose you have [H3

3O

O+

+] from added

] from added HCl HCl = 0.1 M = 10 = 0.1 M = 10-

• 1

1 M

M Then [10 Then [10 -

• 1

1][OH

][OH-

• ] = 10

] = 10-

• 14

14

[OH [OH-

• ] = 10

] = 10-

• 13

13

In pure H In pure H2

2O, [OH

O, [OH-

• ] = [H

] = [H3

3O

O+

+] = 10

] = 10-

• 7

7,

, but but Weak Acids and Bases Weak Acids and Bases Add acetic acid to H Add acetic acid to H2

2O:

O: CH CH3

3COOH + H

COOH + H2

2O

O → → → → → → → → H H3

3O

O+

+ + CH

+ CH3

3COO

COO-

• K = [H3O+][CH3COO− ]

[CH3COOH] = 1.85 ×10

−5

What are [CH What are [CH3

3COOH], [H

COOH], [H3

3O

O+

+], [CH

], [CH3

3COO

COO-

• ]?

]? in acid solution [OH in acid solution [OH-

• ] < than in pure H

] < than in pure H2

2O

O K= K=K Ka

a

Acetic Acid Acetic Acid Acetate ion Acetate ion

SLIDE 3

If If we ignore H

we ignore H3

3O

O+

+ from H

from H2

2O + H

O + H2

2O

O → → → → → → → → H H3

3O

O+

+ + OH

+ OH-

• →

→ → → → → → →

( (Stoichiometry Stoichiometry) ) Let C Let Co

• = initial [CH

= initial [CH3

3COOH]

COOH]≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ [ [HOAc HOAc] ] [ [HOAc HOAc]= C ]= Co

• -
• [H

[H3

3O

O+

+] (Simple mass balance)

] (Simple mass balance) Guess, [ Guess, [HOAc HOAc] ]≅

≅ ≅ ≅ ≅ ≅ ≅ ≅ C

Co

• , since K=

, since K=K Ka

a<<<1 (

<<<1 (→ → → → → → → → [H [H3

3O

O+

+]<<< C

]<<< Co

• )

) K= K=K Ka

a

→ → → → → → → → have [H have [H3

3O

O+

+] = [CH

] = [CH3

3COO

COO-

• ]

]

K = [H3O

+ ]2

CO −[H3O

+ ] = 1.85 ×10 −5

K = [H3O+]2 CO {[H {[H3

3O

O+

+] from H

] from H2

2O

O ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤ 10 10-

• 7

7}

} [H [H3

3O

O+

+] =

] = KCo CH CH3

3COOH + H

COOH + H2

2O

O → → → → → → → → H H3

3O

O+

+ + CH

+ CH3

3COO

COO-

• C

Co

• -
• [H

[H3

3O

O+

+]

] [H [H3

3O

O+

+] [H

] [H3

3O

O+

+]

] Small K means Small K means equilibrium is equilibrium is far to left! far to left!

SLIDE 4

Example : C Example : Co

• = 1 M, [H

= 1 M, [H3

3O

O+

+]

]2

2 = 1.85

= 1.85 × × × × × × × × 10 10-

• 5

5 ×

× × × × × × × 1 M = 18.5 1 M = 18.5 × × × × × × × × 10 10-

• 6

6

[H [H3

3O

O+

+] = 4.3

] = 4.3 × × × × × × × × 10 10-

• 3

3

C Co

• -
• [H

[H3

3O

O+

+] = 0.9957 M

] = 0.9957 M ≅ ≅ ≅ ≅ ≅ ≅ ≅ ≅ C Co

• = 1 M (0.43% error)

= 1 M (0.43% error) Example : C Example : Co

• = 0.01 M

= 0.01 M → → → → → → → → [H [H3

3O

O+

+]= 4.3

]= 4.3 × × × × × × × × 10 10-

• 4

4

C Co

• -
• [H

[H3

3O

O+

+] = 100

] = 100 × × × × × × × × 10 10-

• 4

4 -

• 4.3

4.3 × × × × × × × × 10 10-

• 4

4 = 95.7

= 95.7 × × × × × × × × 10 10-

• 4

4

vs vs C Co

• = 100

= 100 × × × × × × × × 10 10-

• 4

4 (4.3% error!)

(4.3% error!) Example : C Example : Co

• = .001 M

= .001 M → → → → → → → → [H [H3

3O

O+

+]

]2

2 = 1.85

= 1.85 × × × × × × × × 10 10-

• 8

8

[H [H3

3O

O+

+] = 1.36

] = 1.36 × × × × × × × × 10 10-

• 4

4

C Co

• -
• [H

[H3

3O

O+

+] = 10

] = 10 × × × × × × × × 10 10-

• 4

4 -

• 1.36

1.36 × × × × × × × × 10 10-

• 4

4 = 8.64

= 8.64 × × × × × × × × 10 10-

• 4

4

vs vs C Co

• = 10

= 10 × × × × × × × × 10 10-

• 4

4 (13.6% error!!)

(13.6% error!!) But [CH But [CH3

3COOH]= C

COOH]= Co

• -
• [H

[H3

3O

O+

+]

] [H [H3

3O

O+

+] =

] = KCo

K = [H3O+ ]2 CO −[H3O

+ ] = 1.85 ×10 −5

SLIDE 5

Note: Note: In all cases [H In all cases [H3

3O

O+

+] >>10

] >>10-

• 7

7

Thus, OK to Thus, OK to neglect H neglect H3

3O

O+

+ from H

from H2

2O + H

O + H2

2O

O →

→ → → → → → → H

H3

3O

O+

+ + OH

+ OH-

• Could obtain an exact solution by solving a quadratic:

Could obtain an exact solution by solving a quadratic: Such equations are trivial to solve with a modern Such equations are trivial to solve with a modern graphing calculator. graphing calculator. * * * The approximation that [ * * * The approximation that [HOAc HOAc] ] ≅ ≅ ≅ ≅ ≅ ≅ ≅ ≅ C Co

• is good

is good

• nly for
• nly for small

small K Ka

a,

, large C large Co

• .

. A powerful, approximate alternative method: A powerful, approximate alternative method: K = [H3O+]2 CO −[H3O+] →  [H3O+]2 + K[H3O+]− KCO = 0(y2 + by + c = 0) (K= (K=K Ka

a)

)

SLIDE 6

Take [OH Take [OH-

• ]>>10

]>>10-

• 7

7 (ignore hydrolysis of H

(ignore hydrolysis of H2

2O)

O)→ → → → → → → → [OH [OH-

• ]=[CH

]=[CH3

3NH

NH3

3 + +]

] Kb = [CH3NH3

+ ][OH−]

[CH3NH2] = 5.0 × 10

−4

Let initial concentration [CH Let initial concentration [CH3

3NH

NH2

2] = C

] = Co

• = 0.1

= 0.1 First Approximation : First Approximation : [OH [OH-

• ] = [CH

] = [CH3

3NH

NH3

3 + +] = x

] = x1

1

(0.1)(5 (0.1)(5 × × × × × × × × 10 10-

• 4

4) = 50

) = 50 × × × × × × × × 10 10-

• 6

6

Kb = x1

2

Co → x1

2 =

[CH [CH3

3NH

NH2

2] = C

] = Co

• -
• x

x1

1,

, Assume x Assume x1

1 << C

<< Co

• →

→ → → → → → → CH CH3

3NH

NH2

2 + H

+ H2

2O

O → → → → → → → → CH CH3

3NH

NH3

3 + + + OH

+ OH-

• (weak base)

(weak base) C Co

• -
• x

x i

i

x x i

i

x x i

i

(Stoichiometry) x x1

1 = 7.1

= 7.1 × × × × × × × × 10 10-

• 3

3

Method of Successive Method of Successive Approximations Approximations

SLIDE 7

x x1

1 = 7.1

= 7.1 × × × × × × × × 10 10-

• 3

3

C Co

• -
• x

x1

1 = (100

= (100 -

• 7.1)

7.1) × × × × × × × × 10 10-

• 3

3

Second Approximation : Second Approximation : [OH [OH-

• ] = [CH

] = [CH3

3NH

NH3

3 + +] = x

] = x2

2 (unknown)

(unknown) [CH [CH3

3NH

NH2

2]

] ≅ ≅ ≅ ≅ ≅ ≅ ≅ ≅ C Co

• -
• x

x1

1 =

= (100 (100 -

• 7.1)

7.1) × × × × × × × × 10 10-

• 3

3 = 92.9

= 92.9 × × × × × × × × 10 10-

• 3

3

Kb = (x 2)2 (Co − x1) = (x2)2 92.9 ×10

−3

(x (x2

2)

)2

2 = (92.9

= (92.9 × × × × × × × × 10 10-

• 3

3)(5.0

)(5.0 × × × × × × × × 10 10-

• 4

4) = 46.45

) = 46.45 × × × × × × × × 10 10-

• 6

6

x x2

2 = 6.8

= 6.8 × × × × × × × × 10 10-

• 3

3 (Note x

(Note x1

1 = 7.1

= 7.1 × × × × × × × × 10 10-

• 3

3 see above)

see above) % error in taking C % error in taking Co

• -
• x

x1

1 = C

= Co

• is thus

is thus × × × × × × × × 100 = 7.1 % 100 = 7.1 % 7.1 100     CH CH3

3NH

NH2

2 + H

+ H2

2O

O → → → → → → → → CH CH3

3NH

NH3

3 + + + OH

+ OH-

• (weak base)

(weak base) C Co

• -
• x

x 1

1 ≅

≅ ≅ ≅ ≅ ≅ ≅ ≅ C Co

• x

x 1

1

x x 1

1

CH CH3

3NH

NH2

2 + H

+ H2

2O

O → → → → → → → → CH CH3

3NH

NH3

3 + + + OH

+ OH-

• (weak base)

(weak base) C Co

• -
• x

x 1

1

x x 2

2

x x 2

2

SLIDE 8

Thus [CH Thus [CH3

3NH

NH2

2] must equal

] must equal ≅ ≅ ≅ ≅ ≅ ≅ ≅ ≅ C Co

• -
• x

x2

2 = 93.2x10

= 93.2x10-

• 3

3

We approximated [CH We approximated [CH3

3NH

NH2

2] as C

] as Co

• -
• x

x1

1 = 92.9x10

= 92.9x10-

• 3

3

Value we chose was only off by 0.3 out of 93.2 (about 0.3%) Value we chose was only off by 0.3 out of 93.2 (about 0.3%) Try further iterations (x3) and will find no change in values to three significant figures.

SLIDE 9

HOAc HOAc is an acid, therefore is an acid, therefore OAc OAc-

• is a (conjugate) base:

is a (conjugate) base:

HOAc HOAc=CH =CH3

3COOH,

COOH, OAc OAc -

• =CH

=CH3

3COO

COO -

• Hydrolysis

Hydrolysis HOAc HOAc + H + H2

2O

O → → → → → → → → H H3

3O

O+

+ +

+ OAc OAc -

• K

Ka

a = 1.85

= 1.85 × × × × × × × × 10 10-

• 5

5

Add Add NaOAc NaOAc to H to H2

2O, which dissociates completely to

O, which dissociates completely to Na Na+

+,

, OAc OAc -

• OAc

OAc -

• + H

+ H2

2O

O → → → → → → → → HOAc HOAc + OH + OH -

• (

(“ “hydrolysis hydrolysis” ” constant) constant) Kh = [HOAc][OH− ] [OAc

−]

OAc- is a proton acceptor.

SLIDE 10

Note: Note: K Kh

h is very small, indicating that little

is very small, indicating that little OAc OAc -

• combines

combines With H With H2

2O to form

O to form HOAc HOAc

K Kw

w=10

=10-

• 14

14,

, K Ka

a=1.85x10

=1.85x10-

• 5

5 →

→ → → → → → → K Kh

h=5.4x10

=5.4x10-

• 10

10

K Ka

a=[

=[OAc OAc -

• ][H

][H3

3O

O+

+]/[

]/[HOAc HOAc] is the ionization constant for ] is the ionization constant for the reaction: the reaction: Multiply Multiply K Kh

h by 1 =

by 1 = [ H 3 O + ]

[ H 3 O

+ ]

K Kh

h =

= , but , but K Kw

w = [OH

= [OH-

• ][H

][H3

3O

O+

+]

] [HOAc][OH

−][H3O +]

[OAc

−][H3O +]

K Kh

h =

= K Kw

w

, where , where

[HOAc] [OAc

− ][H3O +] = Kw

Ka

HOAc HOAc + H + H2

2O = H

O = H3

3O

O+

+ +

+ OAc OAc -

• Since

Since K Kh

h=

=K Kw

w/

/K Ka

a , the smaller

, the smaller K Ka

a (weaker the acid) the

(weaker the acid) the Larger Larger K Kh

h (or

(or OAc OAc

• is more extensively hydrolyzed)

is more extensively hydrolyzed)

SLIDE 11

QuickTime™ and a Video decompressor are needed to see this picture.

pK pKa

a=

=

• log

log10

10(

(K Ka

a)

)

SLIDE 12

K= K=K Ka

a is very small. For reasonable concentrations of

is very small. For reasonable concentrations of HOAc HOAc initially added to solution, have: initially added to solution, have: Now suppose add 0.60 moles Now suppose add 0.60 moles NaOAc NaOAc, which dissociates , which dissociates completely to completely to Na Na+

+,

, OAc OAc -

• Consider acetic acid:

Consider acetic acid: HOAc HOAc + H + H2

2O

O → → → → → → → → H H3

3O

O+

+ +

+ OAc OAc -

• HOAc

HOAc=CH =CH3

3COOH,

COOH, OAc OAc -

• = CH

= CH3

3COO

COO -

• e.g., add 0.70 mole of

e.g., add 0.70 mole of HOAc HOAc to make 1 liter of solution: to make 1 liter of solution: Buffer Solutions Buffer Solutions K = [OAc−][H3O+] [HOAc] = 1.85 ×10

−5

[ [HOAc HOAc] = [ ] = [HOAc HOAc] ]o

• (initial concentration

(initial concentration HOAc HOAc) ) [ [HOAc HOAc] ]o

• = 0.70 M

= 0.70 M ≅ ≅ ≅ ≅ ≅ ≅ ≅ ≅ [ [HOAc HOAc] ]

SLIDE 13

First, look at First, look at HOAc HOAc + H + H2

2O

O → → → → → → → → H H3

3O

O+

+ +

+ OAc OAc -

• How keep

How keep K Ka

a=1.85x10

=1.85x10-

• 5

5 when

when throw in all of this throw in all of this OAc OAc -

• ?

?

This just uses up H This just uses up H3

3O

O+

+ that came from ionization of

that came from ionization of HOAc HOAc, , thereby forcing [ thereby forcing [HOAc HOAc] closer to [ ] closer to [HOAc HOAc] ]o

• !!

!! Inverse teeter Inverse teeter-

• totter effect:

totter effect: Must decrease [H Must decrease [H3

3O

O+

+] and increase [

] and increase [HOAc HOAc] ] Do this via Do this via OAc OAc -

• + H

+ H3

3O

O+

+ →

→ → → → → → → HOAc HOAc + H + H2

2O

O Addition of large amount of Addition of large amount of common common ion ion OAc OAc -

• suppresses the ionization of

suppresses the ionization of HOAc HOAc (Common Ion Effect) (Common Ion Effect) K Ka

a = [

= [OAc OAc -

• ][H

][H3

3O

O+

+] / [

] / [HOAc HOAc] ]

SLIDE 14

Now, look at: Now, look at: OAc OAc -

• + H

+ H2

2O

O → → → → → → → → HOAc HOAc + OH + OH -

• Very little OH

Very little OH -

• made this way (lose little

made this way (lose little OAc OAc -

• and make

and make little little HOAc HOAc) since: ) since: So when put 0.6 moles So when put 0.6 moles OAc OAc -

• in a liter of pure water

in a liter of pure water have [ have [OAc OAc -

• ]=[

]=[OAc OAc -

• ]

]o

• = 0.6 M

= 0.6 M Must decrease OH Must decrease OH -

• , increase

, increase OAc OAc -

• via: OH

via: OH-

• +

+HOAc HOAc→ → → → → → → →H H2

2O+

O+OAc OAc -

• Here, however, we add 0.6 moles

Here, however, we add 0.6 moles OAc OAc -

• to a

to a solution with a large [ solution with a large [HOAc HOAc] ] Result is to force Result is to force OAc OAc -

• even closer to its

even closer to its initial value [ initial value [OAc OAc -

• ]

]o

• = 0.60 M

= 0.60 M K Kh

h = [

= [HOAc HOAc][OH ][OH-

• ] / [

] / [OAc OAc -

• ] = 5.4

] = 5.4 × × × × × × × × 10 10 -

• 10

10 << 1

<< 1 How to keep How to keep K Kh

h = [

= [HOAc HOAc] [OH ] [OH-

• ] / [

] / [OAc OAc -

• ] =

] = 5.4 5.4 × × × × × × × × 10 10 -

• 10

10 when [

when [HOAc HOAc] is large? ] is large?