8.2 The Equilibrium of Weak Acids and Bases
Ion product constant for water (Kw), pH & pOH, Acid dissociation constant (Ka),
Equilibrium of weak acids and bases (pp.388-403)
SCH4U Grade 12 University Chemistry
Mr. Dvorsky
We learned in previous lessons that…
 Acids and bases differ in their properties and characteristics.
Arrhenius Theory
Brønsted-Lowry Theory
Contains hydrogen and
ACID
Proton – donor
dissociates in water to form 𝑯+
(𝒂𝒒)
Contains hydroxide and
BASE
Proton - acceptor
dissociates in water to form 𝑶𝑯−
(𝒂𝒒)
 Conjugate acid-base pairs exist at equilibrium where an acid in one reaction differs by one hydrogen
with a base in the reverse reaction [i.e. HCl (acid) & Cl- (conjugate base)]
o If a substance can act as a acid or base, it is termed amphoteric.
 Strong acids: Binary acids [HX(aq) where X = Cl, Br, I], Oxoacids [#O ≥ #H+2]
 Strong bases: Oxides and hydroxides of all Group 1 Metals and Group 2 Metals below beryllium
 Strong acids & bases (& strong electrolytes) completely dissociate in water into their constituent ions.
o Weak acids & bases do NOT completely dissociate in water.
Ion Product Constant for Water (Kw)
Dissociation of pure water (neutral) at 25˚C:
+
−
2𝐻2 𝑂(ℓ) ⇌ 𝐻3 𝑂(𝑎𝑞)
+ 𝑂𝐻(𝑎𝑞)
𝐾𝑐 =
[𝐻3 𝑂+ ][𝑂𝐻 − ]
[𝐻2 𝑂]2
𝐾𝑤 = [𝐻3 𝑂+ ][𝑂𝐻 − ] = 𝐾𝑐 [𝐻2 𝑂]2
𝐾𝑤 = (1.0 × 10−7
ACIDIC Solution
𝑚𝑜𝑙
𝑚𝑜𝑙
) (1.0 × 10−7
) = 1.0 × 10−14
𝐿
𝐿
NEUTRAL Solution
𝑚𝑜𝑙
[𝐻3 𝑂+ ] = 1.0 × 10−7
𝐿
𝑚𝑜𝑙
[𝑂𝐻 − ] = 1.0 × 10−7
𝐿
𝒎𝒐𝒍
[𝑯𝟑 𝑶+ ] > 1.0 × 𝟏𝟎−𝟕
𝑳
𝑚𝑜𝑙
[𝑂𝐻 − ] < 1.0 × 10−7
𝐿
BASIC Solution
𝑚𝑜𝑙
𝐿
𝒎𝒐𝒍
[𝑶𝑯− ] > 1.0 × 𝟏𝟎−𝟕
𝑳
[𝐻3 𝑂+ ] < 1.0 × 10−7
𝑚𝑜𝑙
𝐿
Problem: Determining [𝑯𝟑 𝑶+ ] 𝒂𝒏𝒅 [𝑶𝑯− ] with a strong acid or base & Kw e.g. 0.16


Ba(OH)2
Strong acid/base dissociate completely, so use molar concentration to determine its associated ion
𝑚𝑜𝑙
𝑚𝑜𝑙
[OH-] = 2 ×0.16 𝐿 = 0.32 𝐿
Find the concentration of the other ion using Kw
+
[𝐻3 𝑂 ] =
𝐾𝑤
[𝑂𝐻 − ]
=
1.0×10−14
0.32
𝑚𝑜𝑙
𝐿
= 3.1 × 10−14
Strong base since [OH-] > [𝐻3 𝑂+ ]
𝑚𝑜𝑙
𝐿
pH and pOH scale


developed as a convenient way to represent acidity and basicity
logarithmic scale, based on 10
pH = –log[H3O+]  exponential power of hydrogen ions in moles per liter
pOH = –log[OH-]  exponential power of hydroxide ions in moles per liter
𝐾𝑤 = [𝐻3 𝑂+ ][𝑂𝐻 − ] = 1.0 × 10−14 𝑎𝑡 25°𝐶
∴ pH + pOH = 14
So, [H3O+]=10-pH and [OH-]=10-pOH
Acid Dissociation Constant (Ka)
Weak acids do not completely dissociate when dissolved in water
Depends on:
initial concentration of acid
- amount of acid that dissociates
For weak monoprotic acids:
+
𝐻𝐴(𝑎𝑞) + 𝐻2 𝑂(𝑎𝑞) ⇌ 𝐻3 𝑂(𝑎𝑞)
+ 𝐴−
(𝑎𝑞)
+
−
[𝐻3 𝑂 ][𝐴 ]
𝐾𝑐 =
[𝐻𝐴][𝐻2 𝑂]
But concentration of water is almost constant, so
[𝐻3 𝑂+ ][𝐴− ]
𝐾𝑎 = 𝐾𝑐 [𝐻2 𝑂] =
[𝐻𝐴]
You can determine the value of Ka by measuring the pH of a solution.
Ka > 1 = strong acid
1 < Ka = weak acid
(very weak if Ka < 1×10-16)
Solving Acid-Base Equilibrium Problems
1. Write the balanced chemical equation.
2. Use the chemical equation to set up an ICE table.
3. Let x represent the change in concentration of a substance.
a. If
b. If
[𝐻𝐴]
> 500, x
𝐾𝑎
[𝐻𝐴]
<500, x
𝐾𝑎
is negligible and can be ignored.
may not be negligible and may require quadratic equation to solve.
Percent Dissociation
-
Fraction of acid molecules that dissociate completely with the initial concentration of the acid,
expressed as a percent
Depends on:
a. the value of Ka for the acid
b. Initial concentration of the weak acid
Polyprotic Acids
Polyprotic acids have more than one hydrogen atom that dissociates.
Each dissociation has a corresponding acid dissociation constant.
- Need to divide problems with polyprotic acids
into as many sub-problems as there are hydrogen atoms that dissociate
𝑚𝑜𝑙
𝐿
𝟐−
Problem: Determine pH, [𝑯𝟐 𝑷𝑶−
𝟒 ] 𝒂𝒏𝒅 [𝑯𝑷𝑶𝟒 ] of 3.5



solution of phosphoric acid
From data tables, find that Ka1 = 7.0×10-3 and Ka2 = 6.3×10-8
Write equations of dissociation:
+
−
1st dissociation:
𝐻3 𝑃𝑂4(𝑎𝑞) + 𝐻2 𝑂(𝑎𝑞) ⇌ 𝐻3 𝑂(𝑎𝑞)
+ 𝐻2 𝑃𝑂4(𝑎𝑞)
+
−
2−
2nd dissociation:
𝐻2 𝑃𝑂4(𝑎𝑞)
+ 𝐻2 𝑂(𝑎𝑞) ⇌ 𝐻3 𝑂(𝑎𝑞)
+ 𝐻𝑃𝑂4(𝑎𝑞)
Construct ICE Tables and determine if the dissociation of 𝑯𝟑 𝑷𝑶𝟒(𝒂𝒒) is negligible
1st dissociation:
Concentration
+
3
4(𝑎𝑞)
2 (𝑎𝑞)
3 (𝑎𝑞)
2
(mol/L)
Initial
3.5
~0
Change
-x
+x
Equilibrium
3.5 -x
X
𝐻 𝑃𝑂
𝐾𝑎 =
[𝐻3 𝑃𝑂4(𝑎𝑞) ]
𝐾𝑎
=
3.5
7.0×10−3
+𝐻 𝑂
[𝐻3 𝑂 + ][𝐻2 𝑃𝑂4− ]
7.0 × 10−3 =
[𝐻3 𝑃𝑂4 ]
(𝑥)(𝑥)
3.5−𝑥
(𝑥)(𝑥)
3.5
x=0.16
𝑚𝑜𝑙
𝐿
+
−
𝐻3 𝑃𝑂4(𝑎𝑞) + 𝐻2 𝑂(𝑎𝑞) ⇌ 𝐻3 𝑂(𝑎𝑞)
+ 𝐻2 𝑃𝑂4(𝑎𝑞)
So…
Equilibrium
3.5
2nd dissociation:
Concentration
(mol/L)
Initial
Change
Equilibrium
=
0.16
6.3×10−8
0.16
-y
0.16 -y
0.16
+y
0.16+y
[𝐻3 𝑂 + ][𝐻𝑃𝑂42− ]
0.16−𝑦
(𝑦)(0.16)
0.16
y=6.3 × 10−8
+ 𝐻2 𝑂(𝑎𝑞) ⇌
𝐿
−8
, [𝑯𝑷𝑶𝟐−
𝟒 ]=6.3 × 10
𝑚𝑜𝑙
𝐿
+
𝐻3 𝑂(𝑎𝑞)
0.16
𝑚𝑜𝑙
0
+y
y
(𝑦)(0.16+𝑦)
0.16
+
𝑝𝐻 = − log[𝐻3 𝑂(𝑎𝑞)
] = − log[0.16] = 0.80
∴ [𝑯𝟐 𝑷𝑶−
𝟒 ] = 𝟎. 𝟏𝟔
0.16
= 2.5 × 106 > 500, so y is probably negligible compared to 0.16𝑚𝑜𝑙
𝐿
−
𝐻2 𝑃𝑂4(𝑎𝑞)
Equilibrium
6.3 × 10−8 =
[𝐻2 𝑃𝑂4− ]
so.. 6.3 × 10−8 =
So…
0.16
−
+
2−
𝐻2 𝑃𝑂4(𝑎𝑞)
+ 𝐻2 𝑂(𝑎𝑞) ⇌ 𝐻3 𝑂(𝑎𝑞)
+ 𝐻𝑃𝑂4(𝑎𝑞)
𝐾𝑎 =
𝐾𝑎
0
+x
x
= 5000> 500, so x is probably negligible compared to 3.5𝑚𝑜𝑙
𝐿
so.. 7.0 × 10−3 =
[𝐻2 𝑃𝑂−
4]
−
+ 𝐻 𝑃𝑂4(𝑎𝑞)
⇌𝐻 𝑂
𝑚𝑜𝑙
𝐿
, pH=0.80
2−
+ 𝐻𝑃𝑂4(𝑎𝑞)
6.3 × 10−8