Chapter 15 – Titration Problems Example and Assignment
59. The titration of 100.0 mL of 0.200 M HC2H3O2 (Ka = 1.8 x 10-5) by 0.100 M KOH. Calculate the pH after the following volumes
of KOH have been added.
1st you must notice that KOH is a strong base ( KOH  K1+ + OH1-), so we have 0.100
mol
L
OH1-, multiplying by the volume in
mL will give us mmol of OH1-.
2nd you must notice that HC2H3O2 is a weak acid ( HC2H3O2
⇌ H1+ + C2H3O21-), so we have 0.200
mol
L
HC2H3O2, and we would
have to do an ICE table solving for “x” to determine the concentration of H 1+ or C2H3O21-. Multiplying by the volume in mL will
give us mmol of the species we want.
a.
0.0 mL of KOH added
Only a weak acid is present so.
Ka 
b.
HC2H3O2 + H2O ⇌ H3O1+ + C2H3O21initial
0.200
0
0
change
-x
+x
+x
equilibrium 0.200-x
0+x
0+x
[ H  ][C2 H 3O2  ]
[ x][ x]
[ x2 ]


[ HC2 H 3O2 ]
[0.200  x] [0.200]
 1.9 103  x  [ H  ],
 pH  2.72
50.0 mL of KOH added …. the added OH1- will react with the best acid present, HC2H3O2.
To do the acid/base rxn need to know how much acid and base is present
100.0 mL x 0.200
50.0 mL x 0.100
mol
L
mol
L
HC2H3O2 = 20.0 mmol HC2H3O2
OH1- = 5.00 mmol OH1-
Before
Change
After
Only a weak acid is present so.
Ka 
HC2H3O2 +
20.0 mmol
-5.00 mmol
15.0 mmol
OH1
5.00 mmol
-5.00 mmol
-0 mmol
HC2H3O2 + H2O
initial
15.0
0
change
-x
+x
equilibrium 15.0-x
x
[ H  ][C2 H 3O2  ]
[ x][ x]
[ x][5.00  x]


[ HC2 H 3O2 ]
[0.200  x]
[15.0  x]
C2H3O21- + H2O
0
+5.00 mmol
+5.00 mmol
⇌ H3O1+ + C2H3O215.00
+x
5.00 + x
 5.4 105  x  [ H  ],
 pH  4.26
or since this is a buffer use
pH  pK a  log
c.
[C2 H 3O2  ]
[5.00mmol /150mL]
  log(1.8 105 )  log
 4.26
[ HC2 H 3O2 ]
[15.0mmol /150mL]
pH = pKa since [HC2H3O2] = [C2H3O21-]
d. Buffer again …
Before
Change
After
HC2H3O2 +
20.0 mmol
-15.00 mmol
5.0 mmol
OH1
15.00 mmol
-15.00 mmol
-0 mmol
C2H3O21- + H2O
0
+15.00 mmol
+15.00 mmol
Chapter 15 – Titration Problems Example and Assignment
a buffer again so …..
pH  pK a  log
e. Equivalence pt…
Before
Change
After
[C2 H 3O2  ]
[15.00mmol ]
  log(1.8 105 )  log
 5.22
[ HC2 H 3O2 ]
[5.0mmol ]
HC2H3O2 +
20.0 mmol
-20.00 mmol
0
OH1
20.0 mmol
-20.0 mmol
0
C2H3O21- + H2O
0
+20.0 mmol
+20.0 mmol
only the weak base is present so this is the pH of a weak base problem. use Kb = Kw/Ka
pOH = 5.21
pH = 8.79
f. after the before/change/after only hydroxide and conjugate base is present. at this point only the strong base is used to determine
the pH. 5.00mmol OH- in 350.0 mL = 0.014 M… leads to pOH = 1.85, pH = 12.15
ASSIGNMENT
titrate 100.0 mL 0.150 M HNO2 (Ka is in chapter 14) is titrated with 0.200 M KOH
Determine the pH after the following additions of KOH
0 mL, 20.0 mL, 40.0 mL, 60.0 mL, 100.0 mL
Chapter 15 – Titration Problems Example and Assignment