Worksheet 23 – Strong Acid/Strong Base Titrations A. Initial pH This ...

Worksheet 23 – Strong Acid/Strong Base Titrations
A. Initial pH
This is always determined based solely on the initial concentration of the acid or
base being titrated. Every mole of acid or base will produce one mole of H 3 O + or
OH - . Exception: Bases formed with Group II cations will form two moles of OH - for
every mole of base, e.g. Ca(OH) 2 .
pH = -log[H 3 O + ] (starting with strong acid)
or pOH = -log[OH - ] and pH = 14 – pOH (starting with strong base)
B. pH during the titration
Neutralization of 1 mole of a strong acid (H + ) will produce one mole of water.
H + + OH - H 2 O
Neutralization of 1 mole of a strong base (OH - ) will produce 1 mole of water.
OH - + H + H 2 O
The pH is determined by calculating the number of moles of unreacted acid or base
remaining in the solution.
This is then divided by the total volume of the solution to get the concentration of
the remaining acid/base.
C. Equivalence point
If there are equal moles of H + and OH - , the only remaining species (other than the
spectator ions) is water. The pH of a water solution is 7.00. At the equivalence
point in titrations of strong acids and bases, the solution will be neutral.
D. Above the equivalence point
Above the equivalence point, any added strong acid or base will determine the pH.
The excess H + or OH - divided by the volume can be used to determine the pH of the
solution.

1. Consider the titration of 50.0 mL of 2.0 M HNO 3 with 1.0 M KOH.
At each step of the titration...
a) write a reaction to show the initial reaction upon mixing
b) construct an ICE table to represent the reaction
(Should you use concentration or moles?)
c) determine the major species present after the reaction is complete
d) calculate the pH of the solution
Complete these steps for the addition of 0.0 mL, 25.0 mL, 50.0 mL, 90.0 mL,
100.0 mL, 110.0 mL and 150.0 mL of 1.0 M KOH.
0.0 mL KOH
a) HNO 3 + H 2 O NO 3 - + H 3 O +
b) [HNO 3 ] [NO - 3 ] [H 3 O + ]
Initial 2.0 0 0
Change -2.0 +2.0 +2.0
Equil. 0 2.0 2.0
This ICE table uses units of concentration because the volume is fixed
c) The major species present at equilibrium are H 2 O, NO 3 - and H 3 O +
d) pH is calculated using the concentration of [H 3 O + ]
+
H O = 2.0 M
[ ]
3
pH = −log
( 2.0) = −0.
30

25.0 mL KOH
a) The major reaction (ignoring spectator ions) is a neutralization:
H + + OH - H 2 O
b) This ICE table uses units of moles because the volume changes when the
KOH is added
+
+
2.0 mol HNO3
1mol H
+
mol H = 0.050 L ×
×
= 0.100 mol H
1L 1mol HNO
3
-
-
1.0 mol NaOH 1.0 mol OH
mol OH = 0.025 L ×
×
=
1L
1.0 mol NaOH
0.025 mol OH
[H + ] [OH - ] [H 2 O]
Initial 0.100 0.025
Change -0.025 -0.025
Equil. 0.075 0
c) The major species present at equilibrium are H 2 O, K + , NO 3 - and H +
d) pH is calculated using the concentration of [H + ]
+ 0.075 moles
[ H ] =
= 1.0
0.050 + 0.025 mL
pH = − log 1.0 =
( ) 0
-
50.0 mL KOH
a) H + + OH - H 2 O
b) This ICE table uses units of moles because the volume changes when the
KOH is added
-
-
1.0 mol NaOH 1.0 mol OH
-
mol OH = 0.050 L ×
×
= 0.050 mol OH
1L 1.0 mol NaOH
[H + ] [OH - ] [H 2 O]
Initial 0.100 0.050
Change -0.050 -0.050
Equil. 0.050 0
c) The major species present at equilibrium are H 2 O, K + , NO 3 - and H +
d) pH is calculated using the concentration of [H + ]
+ 0.050 moles
[ H ] =
= 0.50
0.050 + 0.050 mL
pH = −log
0.50 = 0.
( ) 30

90.0 mL KOH
a) H + + OH - H 2 O
b) This ICE table uses units of moles because the volume changes when the
KOH is added
-
-
1.0 mol NaOH 1.0 mol OH
-
mol OH = 0.090 L ×
×
= 0.090 mol OH
1L 1.0 mol NaOH
[H + ] [OH - ] [H 2 O]
Initial 0.100 0.090
Change -0.090 -0.090
Equil. 0.090 0
c) The major species present at equilibrium are H 2 O, K + , NO 3 - and H +
d) pH is calculated using the concentration of [H + ]
+ 0.090 moles
[ H ] =
= 0.0714
0.050 + 0.090 mL
pH = −log
0.0714 = 1.
( ) 15
100.0 mL KOH
a) H + + OH - H 2 O
b) This ICE table uses units of moles because the volume changes when the
KOH is added
-
-
1.0 mol NaOH 1.0 mol OH
-
mol OH = 0.100 L ×
×
= 0.100 mol OH
1L 1.0 mol NaOH
[H + ] [OH - ] [H 2 O]
Initial 0.100 0.100
Change -0.100 -0.100
Equil. 0 0
c) The major species present at equilibrium are H 2 O, K + , NO 3
-
d) The solution has been completely neutralized, and the pH is 7

110.0 mL KOH
a) H + + OH - H 2 O
b) This ICE table uses units of moles because the volume changes when the
KOH is added
-
-
1.0 mol NaOH 1.0 mol OH
-
mol OH = 0.110 L ×
×
= 0.110 mol OH
1L 1.0 mol NaOH
[H + ] [OH - ] [H 2 O]
Initial 0.100 0.110
Change -0.100 -0.100
Equil. 0 0.010
c) The major species present at equilibrium are H 2 O, K + , NO 3 - and OH -
d) pH is calculated using the concentration of [OH - ]
− 0.010 moles
[ OH ] =
= 0.0625
0.050 + 0.110 mL
pOH = − log 0.0625 = 1.20
pH = 14 − pOH
( )
= 12.8
150.0 mL KOH
a) H + + OH - H 2 O
b) This ICE table uses units of moles because the volume changes when the
KOH is added
-
-
1.0 mol NaOH 1.0 mol OH
-
mol OH = 0.150 L ×
×
= 0.150 mol OH
1L 1.0 mol NaOH
[H + ] [OH - ] [H 2 O]
Initial 0.100 0.150
Change -0.100 -0.100
Equil. 0 0.050
c) The major species present at equilibrium are H 2 O, K + , NO 3 - and OH -
d) pH is calculated using the concentration of [OH - ]
− 0.050 moles
[ OH ] =
= 0.25
0.050 + 0.150 mL
pOH = − log 0.25 = 0.602
( )
pH = 14 − pOH
= 13.4

2. Create a titration curve by plotting the pH of the solution (y-axis) vs the volume of KOH
added (x-axis) for the titration described in question #1.
pH of a strong acid/strong base titration
pH
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
-1
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150
Volume KOH added (mL)
3. Create a titration curve by plotting the pH of the solution (y-axis) vs the volume of KOH
added (x-axis) for the titration of HNO 2 from the previous worksheet. How does the
curve for a strong acid titration compare to that of a weak acid titration?
pH of a weak acid/strong base titration
pH
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150
Volume KOH added (mL)
The strong acid/strong base curve starts at a lower pH because it begins as a strong acid solution.
Beyond the equivalence point, both curves have the same pH values because the pH depends on the
concentration of the excess KOH. Since the same base was used for both titrations, they have the
same behavior after 100 mL has been added to either acid solution. The equivalence point of the
strong acid/strong base is at pH = 7, but the equivalence point of the weak acid/strong base is NOT
at pH = 7. The weak acid/strong base titration curve has a buffer zone, while the strong
acid/strong base titration curve does not. (There’s no buffer region in the strong acid/strong base
titration because there is no conjugate weak acid/weak base pair present)