 |
 |
4 of 18 |  |
 |
 |
G-4
significant digits in the calculated value of [H3O
+
]). Since [HA]
initial
= 0.50 M, therefore, only 0.60% of HA is
dissociated into hydronium ions and A
-
in the initial solution.
The equivalence point of the titration is the point in a titration where the moles of titrant added are
stoichiometrically equal to the moles of substance being titrated, i.e. the analyte. For a weak acid, the pH of
the equivalence point is greater than pH = 7. The pH of the equivalence point is dependent on the K
a
of the
acid being titrated and on its initial concentration.
The equivalence point must be determined exactly in order to find the K
a
for the weak acid, as well as to select
a suitable indicator for titrations using acid-base indicators. The equivalence point for the titration can be
determined from the pH titration curve. This is done by first drawing an extended straight line through both of
the nearly horizontal regions of the graph before and after the equivalence point. Next, a tangential line (C-D)
is drawn through the actual data points in the vertical portion of the titration curve to intersect the previous two
lines at C and D. Averaging the values of pH at the two intersection points gives the value of the pH at the
equivalence point. The volume of titrant added to reach this point is called the equivalence point volume.
At the equivalence point, the pH of the solution is determined by the reaction of the conjugate base A
-
with
water.
A
-
(aq)
+ H2O
(l)
HA
(aq)
+ OH
-
(aq)
(7)
The concentration of the conjugate base can be determined using the following relationship:
[A
-
]
at equiv. pt.
=
moles A
-
total volume at the equivalence point
(8)
The pH at the equivalence point is then the same as the pH of a solution of A
-
at this concentration. (Note that
in the titration of polyprotic acids, which can donate more than one proton, there will be more than one
equivalence point.)
The value of K
a
for a weak acid can also be determined from the titration curve (see Figure 2). On the
titration curve, the point at which the initial acid has been half neutralized gives the pH of the solution where
the number of moles of A
-
formed through the neutralization reaction is exactly equal to the number of moles of
HA remaining in the solution. In the equilibrium expression
K
a
=
[H3O
+
][A
-
]
[HA]
[HA] will equal [A
-
] at this point, therefore these will cancel out of the relation leaving
K
a
= [H3O
+
]. On the titration curve, the pH of the solution when the acid is half neutralized will therefore
equal the pK
a
value (pK
a
= -log K
a
). This point occurs when one half the volume of NaOH required to reach
