Applications of Aqueous - General Chemistry and Qualitative Analysis - Lecture Slides, Slides of Chemistry

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Chapter 15, Applications of Aqueous

EquilibriaWe will focus on 3 areas:

1) titrations2) buffers (incl. the Henderson-

Hasselbalch Transformation),

3) solubility equilibria.

I. Neutralization Reactions

A. Strong acid-strong base

1. Let’s start by looking at an example:

HCl(

aq

)^

NaOH(

aq

)^

W

H

O( 2

l ) +

NaCl(

aq

2. Can you write a net ionic equation for the above?

+^

W

3. Will the above rxn have mostly products or reactants

present at equilibrium?

Logic?

B. Weak acid-strong base

1. Again, let’s start by looking at an example:

Try finding 2 equations that sum to the one above:

CH

COOH + H 3

O 2

W

H

O 3

+^

+ CH

COO 3

!^

K

= 1.8 x 10a^

!^5

H

O 3

+^ + OH

!^

W

2 H

O 2

1/ K

w^

= 1.0 x 10

14

CH

COOH + OH 3

!^

W

H

O + CH 2

COO 3

!^

K

n^

= K

(1/Ka

)w

K

n^ = 1.8 x 10

9

3. Therefore, the rxn. goes essentially to completion.

C. Strong acid-weak base. Same as B. above.D. Weak acid-weak base

1. In B, we could ignore Na

+^

because it has essentially

no acid-base properties. In this of problem, neither

component is weak enough to ignore.

2. Look at the rxn. of acetic acid with ammonia: CH

COOH( 3

aq

) + NH

aq

)^

W

NH

• 4 ( aq

) + CH

COO 3

aq

What rxns. will sum to give us that rxn.?

W

K

=a^

W

K

b^

W

CH

COOH 3

( aq

+ NH)

3( aq

)^

W

NH

• 4 ( aq

)^

+ CH

COO 3

!^ ( aq

)

K

n^

= K

x Ka^

b^

x

1. Consider mixing acetic acid and sodium acetate: CH

COOH 3

( aq

+ CH)

COO 3

! ( aq

W)

CH

COOH 3

( aq

+ CH)

COO 3

! ( aq

)

Can we calculate [CH

COOH 3

],^

[H

O 3

+ ], & [CH

COO 3

!]

at equilibrium?

([Na

• ] usually not of interest.)

2. Use the same approach we developed in Chapter 14

(Fig. 14.7). Main difference: [CH

COO 3

!]

initial

a) Step #1, Identify reactive (interesting) species: CH

COOH 3

Na

+^

H

O 2

CH

COO 3

!

acid

inert

acid/base

base

b) Step #2-3, Identify principle reaction:

CH

COOH 3

H

O 2

W

H

O 3

+^

+ CH

COO 3

!^

K

a^ = 1.8 x 10

!^5

c) Step #4, Set up the table.d) Step #5, Substitute values into the

K

expression a^

e) Steps remaining, do the algebra. (See Fig. 15.2) Do Prob. 15.

Try Key Concept Prob. 15.5, p. 594.

HCN + OH

!^

ÿ

H

O + 2

CN

!

b) What happens if you add a H

O 3

+^ to the buffer?

CN

!^

+ H

O 3

+^

ÿ

H

O + HCN 2

2. Buffer capacity. There is a limit to how much acid or

base the buffer can absorb.a) The amount of acid that can be absorbed is related to how

much basic component (CN

!^ above) of the buffer is

present. b) The amount of base that can be absorbed is related to how

much acidic component (HCN above) of the buffer ispresent.

E. Where (on the pH scale) does a buffer work?

1. Recall the

K

a^ expression:

[H

O 3

+ ] [A

!]

K

a^

[HA]

2. This can be rearranged to obtain:

K

a^

[HA]

[H

O 3

+ ] =

[A

!]

3. This tells us:

a) The [H

O 3

+^ ] (and therefore pH) is determined by the ratio

of acid and conjugate base. b) The pH of effective buffering depends on K

.a

Do Key Concept Prob. 15.6, p. 598.

A. Derivation.

Let’s start with the

K

a^

expression:

[H

O 3

+] [A

!]

K

a^

[HA]

Distributive law to get:

[A

!]

K

a^

= [H

O 3

+] ×

[HA]

Take log of both sides:

log

K

a^

= log [H

O 3

+] + log([A

!]/[HA])

Rearrange:

!

log [H

O 3

+] =

!

log

K

a^

+ log ([A

!]/[HA])

Finally, substitute pH and pK

a^

definitions:

pH = p

K

a^

log ([A

!]/[HA])

This is the H-H equation.

(Note: p

K

a^

B. What use is this, anyway? Let’s see what

happens when we mix equimolar quantities ofbuffer components (HA and A

pH = p

K

a^

log (

x /

x )

Because log 1 = 0,

pH = p

K

a

pH

pH = p

K

a^

+ log([A

!]

'

[HA])

P

COa^

2

P

Oa^

2

SaO

2

HCO

! (^3) b) Is HCO

! 3 acting like an acid or a base?

See pK

valuesa

above, think of CO

leaving the body. 2

V. pH Titration Curves

A. Titration: quantitative analysis method in

chemistry. Based on chemical rxns. To do one,you need to know:1. The stoichiometry for the reaction.2. The concentration of the “known” component.3. The volume of known component added.

B. If you know these things, you can calculate the

quantity of

unknown

present in a sample.

C. You can also get p

K

a^

information from a

titration.

VII. Weak Acid-Strong Base Titrations

A. You get quantitative & K

a^

information w/ these.

B. Again, a weak acid reaction with a strong base

goes to completion:

100%

HA

( aq

+ OH)

!^ (

aq

)^

ÿ

H

O 2

( l )

+ A

!

1. See Fig. 15.8 for titration of CH

COOH with NaOH. 3

2. Note equivalence point at pH

…

7.0.

(Indicator?)

Equivalence point

: point in titration where

stoichiometrically equal amounts of acid and base havebeen added.

Seen as steep inflection point in graph.

3. Comment on the shape (vs. pH location) of the Fig. 15.

curve for different weak acids. If you understand this figure,you are in good shape re. acid-base chemistry and buffers.

To reinforce see Key Concept Prob 15.15, p. 607VIII. Weak Base-Strong Acid Titrations (VII)IX. Polyprotic Acid-Strong Base Titrations

A. Analogous to VII, above. Fig. 15.11, p. 610.B. Try Prob. 15.19, p. 612 on your own.