Review for Solving pH Problems:
-
Acid Ionization:
HA + H2O ↔ A + H3O
+
-
CH3COOH + H2O ↔ CH3COO + H3O
+
Base Ionization:
+
-
B + H2O ↔ BH + OH
+
CH3NH2 + H2O ↔ CH3NH3 + OH
1) Strong Acid – complete dissociation
+
[H ] is equal to original [HA]
2) Strong Base – complete dissociation
-
[OH ] is equal to original [MOH]
Where M represents a cation.
Bases derived from group II metals will have the formula M(OH)2 and will yield two hydroxides.
3) Weak acid – when only weak acid is initially present.
Solved by examining the equilibrium of acid ionization.
-8
What is the pH of a solution of 0.100 M HOCl? The Ka of HOCl is 3.5 x 10 .
+
Set up a reaction table for the following ionization: HOCl ↔ H + OCl
initial
change
final
HOCl
0.100
-x
0.100 –x
→
Solve by using the equilibrium expression: 𝐾𝑎 =
+
-
H
0
+x
x
OCl
0
+x
x
𝐻 + [𝑂𝐶𝑙 − ]
[𝐻𝑂𝐶𝑙 ]
= 3.5 𝑥 10−8
𝑥 [𝑥]
[0.100 ]
The assumption is made that x << than 0.1 M due to the small Ka.
-5
Solving for x: x = 5.92 x 10 M and the pH is 4.23.
4) Weak Base – when only weak base is initially present
Solved by examining the equilibrium of base ionization.
-4
What is the pH of a 1.0 M methylamine solution. The Kb of CH3NH2 is 4.38 x 10 .
+
Set up a reaction table for the following ionization: CH3NH2 + H2O ↔ CH3NH3 + OH
initial
change
final
CH3NH2
1.0
-x
1.0 –x
→
Solve by using the equilibrium expression: 𝐾𝑏 =
+
CH3NH3
0
+x
+x
𝐶𝐻3 𝑁𝐻3+ 𝑂𝐻 −
𝐶𝐻3 𝑁𝐻2
OH
0
+x
+x
=
𝑥 𝑥
1.0
The assumption is made that x << than 1.0 M due to the small Kb.
Solving for x: x = 0.0209 M = [OH ] and the pOH is 1.68. The pH is 12.3
-
= 4.38 𝑥 10−4
5) Salts – A salt is added. A salt is the conjugate base of a weak acid with a cation or a conjugate acid of a weak base
with an anion.
a) conjugate base and cation
-4
-
-11
What is the pH of a 0.30 M NaF solution? Ka (HF) = 7.2 x 10 Kb (F ) = 1.4 x 10
The first thing is to determine what is in solution. In this case, it is the base, F . Set up a reaction table for base
ionization, which is the reaction of base with water: F + H2O ↔ HF + OH
-
initial
change
final
F
0.30
-x
0.30 –x
→
Solve by using the equilibrium expression: 𝐾𝑏 =
HF
0
+x
x
𝐻𝐹 𝑂𝐻 −
𝐹−
OH
0
+x
x
=
𝑥 𝑥
0.30
-
= 1.40 𝑥 10−11
The assumption is made that x << than 0.30 M due to the small Kb.
-6
Solving for x: x = 2.05 x 10 M = [OH ] and the pOH is 5.69. The pH is 8.31
b) conjugate acid and anion
+
-4
-11
What is the pH of a 0.5 M CH3NH3Cl solution. Kb (CH3NH3 ) = 4.38 x 10 Ka (CH3NH2) = 2.23 x 10
+
The first thing is to determine what is in solution. In this case, it is the acid, CH 3NH3 . Set up a reaction table for
+
+
acid ionization, which is the reaction of acid with water: CH 3NH3 + H2O ↔ CH3NH2 + H3O
+
initial
change
final
CH3NH3
0.5
-x
0.5 –x
→
Solve by using the equilibrium expression: 𝐾𝑎 =
CH3NH2
0
+x
x
H3O
0
+x
x
𝐻3 𝑂 + [𝐶𝐻3 𝑁𝐻2 ]
[𝐶𝐻3 𝑁𝐻3+ ]
+
= 2.23 𝑥 10−11 =
𝑥 [𝑥]
[0.500 ]
The assumption is made that x << than 0.5 M due to the small Ka.
-6
Solving for x: x = 3.379 x 10 M and the pH is 5.47.
6) Common ion solution, commonly known as a buffer solution
Remember that if more than one acid or an acid and a base are present in the same solution. The
component with the largest equilibrium constant for the ionization with water (Ka or Kb) will be most
important in setting the pH and should be used to determine the pH.
a) a weak acid in the presence of a salt that contains the conjugate base
-5
What is the pH of a solution 0.50 M CH3COOH and 0.30 M NaCH3COOH? Ka(CH3COOH) = 1.8 x 10
+
Set up a reaction table for the acid ionization of CH3COOH: CH3COOH + H2O ↔ CH3COO + H3O . This time
however, the concentration of conjugate base is not zero.
initial
change
final
CH3COOH
0.50
-x
0. 50 –x
→
Solve by using the equilibrium expression: 𝐾𝑎 =
CH3COO
0.30
+x
0.30 +x
-
𝐻3 𝑂 + [𝐶𝐻3 𝐶𝑂𝑂 − ]
[𝐶𝐻3 𝐶𝑂𝑂𝐻 ]
The assumption is made that x << than 0.3 M due to the small Ka.
-5
Solving for x: x = 3.0 x 10 and the pH is 4.52
H3O
0
+x
x
+
= 1.8 𝑥 10−5 =
𝑥 [0.3]
[0.500 ]
NOTE: A common ion solution is a buffer solution and this can be easily solved by using the Henderson-Hasselbach
equation.
-
pH = pKa + log([A ]/[HA]
Substituting values in: pH = 4.74 + log(0.30/0.50) = 4.52
b) a weak base in the presence of a salt that contain the conjugate acid
-5
What is the pH of a solution that contains 0.25 M NH3 (Kb = 1.8 x 10 ) and 0.40 M NH4Cl?
+
Set up a reaction table for the base ionization of NH3 with water : NH3 + H2O ↔ OH + NH4 . This time
however, the concentration of conjugate acid is not zero.
+
NH3
→
NH4
OH
initial
0.25
0.40
change
-x
+x
+x
final
0.25
0.40
x
Solve by using the equilibrium expression: 𝐾𝑏 =
𝑁𝐻4+ 𝑂𝐻 −
𝑁𝐻3
=
𝑥 0.4
0.25
= 1.80 𝑥 10−5
The assumption is made that x << than 0.25 M due to the small K b.
-5
Solving for x: x = 1.1 x 10 M = [OH ] and the pOH is 4.95. The pH is 9.05
This problem could also be solved using the Henderson-Hasselbach equation. Note that the pKa of the conjugate acid (5.5
-10
x 10 in this case) must be used and that it should still be base over acid in the log function.
+
-10
Ka for NH4 = Kw/Kb = 5.56 x 10
-
pH = pKa + log([A ]/[HA]
Substituting values in: pH = 9.25 + log(0.25/0.0.40) = 9.05
7) Additions to solution – in each of these cases, there are two steps to the solution.
First one considers the stoichiometric reaction. Second, one considers an equilibrium
Because volumes change, amounts must be converted from concentration to moles. Values in reaction table are
+
thus in moles. When finished, to find [OH ] or [H ], moles must be divided by new volume.
a) strong base to a weak acid
What is the pH of 50.0 mL 0.1 M CH3COOH after 10.0 mL of 0.1 M NaOH has been added?
-5
Ka = 1.8 x 10
CH3COOH + OH → CH3COO + H2O
Reaction: The strong acid will cause the reaction to go to completion. The limiting reagent here is the hydroxide ion.
initial
change
final
CH3COOH
0.005
- 0.001
0.004
-
OH
0.001
-0.001
0
→
CH3COO
0
+0.001
0.001
-5
At this point, there is an acid and a base in solution. The Ka for the acid is 1.8 x 10 while the Kb for the conjugate base is
-10
5.56 x 10 . Since the equilibrium constant for the acid is largest, the acid ionization with water, as opposed to the base
+
ionization with water is used to determine the pH. CH 3COOH + H2O ↔ CH3COO + H3O
The solution for this will be first shown using a reaction table. Values are entered as concentrations: 0.004 mol/0.06 L
and 0.001 mol/0.06L.
Equilibrium with a reaction table:
-
+
CH3COOH
↔ CH3COO
H3O
initial
0.0667
0.01667
0
change
-x
+x
+x
final
0.0667 –x
0.01667 +x
x
x is small compared to acid concentration and can be dropped.
-5
-5
+
Ka = 1.8 x 10 = (x)(0.01667)/(0.0667)
x = 7.20 x 10 = [H ] and pH is 4.14
Equilibrium using the Henderson-Hasselback equation:
Since this is a mixture of an acid and the conjugate base, which is a buffer, this could be solved using the Henderson
Hasselbach equation.
pH = pKa + log([A ]/[HA] = 4.74 + log(0.001/0.060)/(0.004/0.06) = 4.74 – 0.602 =4.14
b) strong acid to a weak base
What is the pH of 100 mL 0.050 M NH3 after 20 mL of 0.10 M HCl has been added.
-5
Kb = 1.8 x 10
+
NH3 + HCl → NH4 + Cl
Once again, because volumes change, amounts must be converted from concentration to moles. Values in reaction table
+
are thus in moles. When finished, to find [OH ] or [H ], moles must be divided by new volume. Cl need not be included as
it is a spectator ion and is the anion of a strong acid.
Reaction: The strong base will cause the reaction to go to completion. In this case, the limiting reagent is the acid.
NH3
0.005
-0.002
0.003
initial
change
final
+
H
0.002
-0.002
0
→
+
-
NH4
0
+ 0.002
0.002
Cl
+
-10
At this point, there is an acid and a base in solution. The Ka for the conjugate acid (NH4 ) is 5.56 x while the Kb for the
-5
base is 1.80 x 10 . Since the equilibrium constant for the base is largest, the base ionization with water, as opposed to
+
the acid ionization with water is used to determine the pH. NH3 + H2O ↔ NH4 + OH
The solution for this will be first shown using a reaction table. Values are entered as concentrations: 0.003 mol/0.120 L
and 0.002 mol/0.120 L.
NH
0.025
-x
0.025 –x
initial
change
final
-5
→
+
NH4
0.01667
+x
0.01667 + x
-
OH
0
+x
x
-
-5
Kb = 1.80 x 10 = (x)(0.01667)/(0.025) and x = [OH ] = 2.7 x 10 pOH = 4.57 and pH = 9.42
Equilibrium using the Henderson-Hasselback equation:
Since this is a mixture of an acid and the conjugate base, which is a buffer, this could be solved using the Henderson
Hasselbach equation. Remember to use acid terms.
pH = pKa + log([A ]/[HA] = 9.25 + log(0.025)/(0.01667) = 9.25 +0.176 = 9.43
c) strong base to a buffer solution
A problem of this type would have a reaction followed by an equilibrium.
The reaction would be solved as described in 6 above.
The equilibrium would then be solved as in 7, part a above.
d) strong acid to a buffer solution
A problem of this type would have a reaction followed by an equilibrium.
The reaction would be solved as described in 6 above.
The equilibrium would then be solved as in 7, part b above.