(e) Given: 0.500 liter = volume
035. M=+6@NaC H O 232 "Na+ C H O 232 -
[HA] = 0.50 M
1.5 g LiOH added to solution
.
..
g LiOH
g LiOH
mole LiOH mole LiOH
1
15
23 95
#^1 =0 063
Restatement: pH =?
Step 1: Write a balanced equation expressing the reaction betweeen acetic acid and lithium
hydroxide.
HA + OH–→A–+ H 2 O
Step 2: Create a chart that expresses initial and final concentrations (at equilibrium) of the
species.
Species Initial Concentration Final Concentration
HA 0.50 M..
050. .M
0 500
-=0 063 0 375
A– 0.35 M 035. +=0 5000 063.. 0 475.M
Step 3:Write an equilibrium expression for the ionization of acetic acid.
.
.
.
.
HA
HA H
HM
174 10
0 374
048
137 10
a^5
5
#
#
== =
=
++
+
K
6
77
6
7 6
7
@
AA
@
A @
A
Step 4:Solve for the pH.
pH = −log[H+] = −log (1.37 × 10 –5) = 4.86
Part II: Specific Topics