Figure 3
Ka= 10–pKa= 10–5= 1 × 10 –5
K
A
HA OH
H
H
K
K
A
OH
A
W
2
===- :
+
_
^ _
_
_
_
_
i
h i
i
i
i
i
K
OH
KA
A
W
= - 2
_
_
i
i
- What is the pH at the equivalence point?
The pH at the equivalence point appears to be around 9. Therefore, (OH–) ≈ 10 –5M
..
.. ..
A mL mL ..
mL M mL M
43 2 5 2 M at equiv point
43 2 0 103 5 2 0 100
0 0812
##
= +
^^
i
hh
.
Ka Kw.
OH
A
110
1 10 0 0812
5 2 812 10
14
6
: 2
#
#
==- =#
_
_
_
_ ^
i
i
i
i h
- Why is the equivalence point not a pH of 7?
The neutralization reaction is HA OH++--&A H O 2 , where HA is the weak acid and A–is
its conjugate base. This assumes that HA is a monoprotic acid. Because the principal
product of this reaction is the weak base, A–, the resulting solution will be basic with a pH
greater than 7.
- Determine the EMaof the solid acid.
43 2. mL#0 103. 1000 mol OHmL =0 00445. mol OH dispensed from buret
52. mL#0 100. 1000 mol HmL =0 00052. mol H
+ +
used in back-titration
(0.00445 – 0.00052) mol = 0.00393 mol OH–actually used to neutralize
pH
Volume of NaOH added (mL)
0 19.1 38.2
5
9
equivalence pt.
at half-equivalence pt., pH = pKa
Laboratory Experiments
