Lecture #13 Acids & Bases: Polyprotics Benjamin, Chapter 4 - - PowerPoint PPT Presentation

Lecture #13 Acids & Bases: Polyprotics Benjamin, Chapter 4

(Stumm & Morgan, Chapt.3 )

David Reckhow CEE 680 #13 1

Updated: 11 February 2020

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Rapid Method for Log C vs. pH Graph

 1. Plot diagonal [H+] and [OH-] lines  2. Draw a light horizontal line corresponding to log CT  3. Locate System Point

 i.e., pH = pKa, log C = log CT  make a mark 0.3 units below system point

 4. Draw 45º lines (slope = ±1) below log CT line, and aimed

at system point

 5. Approximate curved sections of species lines ±1 pH unit

around system point

 6. Repeat steps as necessary for more complex graphs

 #3-#5 for additional pKas of polyprotic acids  #2-#5 for other acid/base pairs

David Reckhow CEE 680 #13 2

Diprotic acids: calculations

 Start with CT and Ka equations

David Reckhow CEE 680 #13 3

2 2 1 1

] [ ] [ 2 2 2 1 1 2 2 2 2 1 2 1 2

1 1 ] [ ] [ ] [ 1 ] [ ] [ ] [ ] [ ] [ ] [

+ + +

+ =         + + = + + =

+ + + + H K K H K T T T

C A H H K K H K A H C H A H K K H A H K A H C ] [ ] [ ] [

2 2 − − +

+ = A HA A H CT

] [ ] ][ [

2 1

A H HA H K

− +

= ] [ ] ][ [

2 2 − − +

= HA A H K ] [ ] [ ] [

2 1 + − =

H A H K HA

2 2 2 1 2 2

] [ ] [ ] [ ] [ ] [

+ + − −

= = H A H K K H HA K A

Diprotic acids: calculations (cont.)

 Use [H2A]/CT and Ka equations to get other α’s

David Reckhow CEE 680 #13 4

] [ ] ][ [

2 1

A H HA H K

− +

=

α0

For distribution diagrams

Note: α0 + α1 + α2 = 1

] [ ] ][ [

2 2 − − +

= HA A H K ] [ ] [ ] [

2 1

A H HA H K

− + =

2 2 1 1

] [ ] [ 2

1 1 ] [

+ + +

+ =

H K K H K T

C A H

] [ ] [ ] [

2 2 − − + = HA

A H K

] [ ] [ 1 ] [ ] [ 2 2

2 1 2 2 1 1

1 1 ] [ ] [ 1 1 ] [ ] [ ] [ ] [

+ + + +

+ + =         + + = =

− + − − H K K H T H K K H K T T

C HA H K A H HA C A H C HA 1 1 ] [ ] [ 1 1 ] [ ] [ ] [ ] [

2 2 1 2 2 1

] [ ] [ 2 2 ] [ ] [ 2 2

+ + =         + + = =

+ + + +

− + − − − − K H K K H T H K K H T T

C A H K HA A C HA C A

α1 α2

Diprotic acids: calculations (cont.)

David Reckhow CEE 680 #13 5

2 2 1 1

] [ ] [

1 1

+ + +

+

H K K H K T

C A H ] [

2 0 ≡

α

T

C HA ] [

1 −

≡ α

] [ ] [

2 1

1 1

+ +

+ +

H K K H T

C A ] [

2 2 −

≡ α 1 1

2 2 1 2

] [ ] [

+ +

+ +

K H K K H

 If pH << pK1, or [H+] >> K1  If pK1 << pH << pK2, or K1>> [H+] >> K2  If pK2,<< pH, or K2 >> [H+]

1 [H+]/K1 K2/[H+] K1K2/[H+]2 [H+]/K2 [H+]2/K1K2 K1/[H+] 1 1

Carbonate System (CT=10-3)

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• 12
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• 8
• 6
• 4
• 2

2 4 6 8 10 12 14 pH Log C Log H+ Log H2CO3 Log HCO3- Log CO3-2 Log OH-

David Reckhow CEE 680 #13 6

H+ H2CO3 HCO3

• CO3
• 2

OH-

To next lecture

David Reckhow CEE 680 #13 7

DAR

David Reckhow CEE 680 #13 8

pH

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Log C

• 14
• 13
• 12
• 11
• 10
• 9
• 8
• 7
• 6
• 5
• 4
• 3
• 2
• 1

H+

OH-