Cliffs AP Chemistry, 3rd Edition

Experiment 10 A: Determination of the Equilibrium Constant, Ka, for a Chemical Reaction

Background: Weak acids, upon dissolving in water, dissociate slightly according to the fol-
lowing equation:

HA H O++ 23 EH O+-A

The equilibrium expression for this dissociation is: K
HA

HO A

a

3

=

+-

6

77

@

AA

Most weak acids dissociate 5% or less, that is, 95% or more of the acid remains as HA. The
smaller the value of Ka, the weaker the acid.

If the acid contains more than one ionizable hydrogen, the acid is known as a polyprotic acid.
Specifically, diprotic acids, such as sulfuric acid, H 2 SO 4 , contain two ionizable hydrogens and
triprotics, such as phosphoric acid, H 3 PO 4 , contain three ionizable hydrogens.

In the case of a diprotic acid, the dissociation occurs in two steps and a separate equilibrium
constant exists for each step:

H A H O H O HA K

HA

HO HA

22 3 a (^12)
3
++=E +-
+-
6
77
@
AA
HA H O H O A- K
HA
HO A
a
2
23 2
3 2
-+++=E -
+-
7
77
A
AA
The second dissociation generally occurs to a smaller extent than the first dissociation so that
Ka2is smaller than Kal(H 30 +)^2 ×(A2–) / (H 2 A) = Ka1×Ka2=KΣ. Exceptions are generally bio-
molecules in specific solvent media.
This experiment will determine Kafor a monoprotic weak acid.
A common weak acid, acetic acid, dissociates according to the following equation:
HCHO HO HO CHO 232 ++ 2? 3 +- (^232)
acid base conjugate conjugate
acid base
The equilibrium expression for this dissociation is:
K.
HC H O
HO CHO
a 18 10 M
232
(^32325)
==#
+-

• 6

77

@

AA

If [HC 2 H 3 O 2 ] = [C 2 H 3 O 2 – ], the two terms cancel, resulting in H 3 O+= 1.8 × 10 –5M.

K.

HC H O

HO CHO

a 18 10 M

232

(^32325)
==#

• 
  
  
  
  • (^7) -
    6
    A 7
    @
    A
    Scenario: A student prepared three solutions by adding 12.00, 18.50, and 35.00 mL of a 7.50 ×
    10 –2M NaOH solution to 50.00 mL of a 0.100 M solution of a weak monoprotic acid, HA. The
    Part III: AP Chemistry Laboratory Experiments