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Polyprotic Acids and Bases: Understanding Ka and Kb Values and pH Behavior, Lecture notes of Chemistry

An in-depth look into the behavior of polyprotic acids and bases, focusing on the significance of Ka and Kb values and their impact on pH. It includes relevant reactions, species, and equilibria, as well as problems to test understanding.

Typology: Lecture notes

2021/2022

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21 - POLYPROTIC ACIDS &
BASES
CHEM 251 SDSU
pf3
pf4
pf5

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21 - POLYPROTIC ACIDS &

BASES

CHEM 251 SDSU

POLYPROTIC KA & KB VALUES

H 3 A H 2 A

HA

2-

A

3-

Ka1 Ka2 Ka

Kb3 Kb 2 Kb

The Ka and Kb values are numbered in order of the strength of the

equilibrium to act as an acid or base. The lower the number the

further the reaction shifts to the products.

Ka1 good acid; Kb3 bad base

Ka3 bad acid; Kb1 good base

PH OF A POLYPROTIC

SOLUTION

H 3 A H 2 A

  • HA

2- A

3-

Ka1 Ka2 Ka

The pH of a solution will depend on the primary

form of the acid (base) present

Relevant Reactions:

H 3 A H 2 A

    • H

Species:

H 3 A

Equilibria:

H 3 A + OH

  • H 2 A - HA

2-

  • H

H 2 A

    • OH

      HA 

2- A

3-

  • H

HA

2-

  • OH
  • A

3-

  • H 2 O

NaH 2 A

K 2 HA

Li 3 A

Ka
Ka1 & Ka
Ka2 & Ka
Ka

PROBLEMS

What is the pH of a 36 mM solution of:

- Carbonic acid (H 2 CO 3 )?

  • Sodium carbonate (Na 2 CO 3 )?
  • Sodium bicarbonate (NaHCO 3 )?
H

2

CO

3

⇔ H
+ HCO

3

HCO

3

⇔ H
+ CO

3

2 −

CO

3

2 −

+ H

2

O ⇔ HCO

3

+ OH

36 mM H 2

CO 3

treat like a weak monoprotic acid (HA)

K a

=

H

! "

$

A

− ! "

$

[ HA]

= 4. 46 × 10

− 7

[ HA] =^0.^036 −^ x^ H

! "

$

= A

− ! "

$

= x

  1. 46 × 10

− 7

( x ) ( x )

  1. 036 − x

  2. 6056 × 10

− 8 − 4. 46 × 10

− 7

( x ) =^ x

2

0 = x

2

    1. 46 × 10

− 7

( x ) −^1.^6056 ×^10

− 8

x = H

! "

$

= 1. 26 × 10

− 4 M pH = − log 1. 26 × 10

− 4

( ) =^3.^90

PROBLEMS

What is the pH of a 36 mM solution of:

  • Carbonic acid (H 2 CO 3 )?
  • Sodium carbonate (Na 2 CO 3 )? - Sodium bicarbonate (NaHCO 3 )?

H 2 CO 3 ⇔ H

  • HCO 3

− HCO 3

− ⇔ H

  • CO 3

2 − CO 3

2 −

  • H 2 O ⇔ HCO 3

  • OH