This problem involves the concept of multiple equilibria. The dissociation constants given in
the example are related to the following reactions:

H 3 PO 4 (aq) ?H+(aq) + H 2 PO 4 – (aq) K 1 = 7.0 × 10 –3

H 2 PO 4 – (aq) ?H+(aq) + HPO 4 2–(aq) K 2 = 6.0 × 10 –8

HPO 4 2–(aq) ?H+(aq) + PO 4 3–(aq) K 3 = 5.0 × 10 –13

For multiple equilibria dissociation constants (such as polyprotic acids), Kfor the overall reac-
tion is the product of the equilibrium constants for the individual reactions. Therefore,

.

KK K K

HPO

HHPO

HPO

H HPO

HPO

HPO

HPO

HPO

210 10 2 10 10

4

4

4

4

4

3

(^42422)
(^12334)
2
2
34

==

• --
  ++


• 
  

  2
  2

  
  

• 
  
  
  
  • 
    
    
    
    • 
      
    
    
    
    
  
  
  
  


• 
  

33 --

^

_ _

_

_ _

_

_ _

^

_ _

h

i i

i

i i

i

i i

h

i i

which is the equilibrium constant for the sum of three individual reactions:

H 3 PO 4 (aq) ?3 H+(aq) + PO 4 3–(aq)

7. A buffer is found to contain 0.35 M NH 3 (Kb= 1.8 ⋅l0–5) and 0.20 M NH 4 Cl. What

would be the mathematical expression for Kbin terms of 7ANH 4 +, [OH–], and [NH 3 ]?

A...

. xx
18 10 020
5 035

=

• ^^+ hh

B...

..xx

18 10 020

5 035 020

# =

• ^^- hh

C..

.

.

x

xx

18 10

035

5 020

# =

• 
  


• 
  
  
  
  • ^
  
  
  
  

^^

h

hh

D...

. x
18 10 035
5 020

- =

^^hh

E...

..xx

18 10 035

5 020 035

# =

• ^^+-hh

Answer: C

Step 1:Get a picture of the solution in equilibrium. Ammonia (NH 3 ) is a weak base. NH 3 reacts
with water in accordance with the following equilibrium equation:

NH 32 ()aq++H O(),?NH+- 4 ()aq OH()aq

Ammonium chloride is soluble in water (see the solubility rules on page 116). Therefore, the
concentration NH 4 +and that of Cl–are both 0.20 M.

Part II: Specific Topics