Republic of the Philippines
ISABELA STATE UNIVERSITY
Echague, Isabela
COLLEGE OF ARTS AND SCIENCES
SENIOR HIGH SCHOOL-GENERAL CHEMISTRY 2
Chemical Equilibrium
Learning Outcomes
At the end of this module, the students will be able to:
1. Explain several chemical equilibria and perform related calculations
2. Relate the equilibrium with thermodynamics
3. Understand the Le chatellier principle and its application in equilibrium
4. Perform calculations associated with solubility product constant
5. Understand the principle of Acid-Base equilibra
6. Properly select appropriate buffer systems for a given pH requirement and describe
step by step procedure for buffer preparation
Learning Content
Chemical equilibrium provides a foundation not only for chemical analysis, but also for other
subjects such as biochemistry, geology, and oceanography. This chapter introduces equilibria
for different types of reaction.
In chemical equilibrium, the rate of forward reaction is equal to the rate of the reverse reaction.
The concentration of the reactants and products no longer change with time in a state of
equilibrium. Note that a chemical equilibrium is a dynamic process, it means that the forward
and reverse processes are proceeding continuously.
Equilibrium constant
For the hypothetical chemical reaction:
aA+bB⇌cC+dD
the equilibrium constant is defined as:
[𝐶]𝑐 [𝐷]𝑑
𝐾𝑐 =
[𝐴]𝑎 [𝐵]𝑏
The notation [A] signifies the molar concentration of species A. An alternative expression for
the equilibrium constant involves partial pressures:
𝐾𝑝 =
𝑃𝐶𝑐 𝑃𝐷𝑑
𝑃𝐴𝑎 𝑃𝐵𝑏
Kp is used when temperature s constant and where partial pressure (P) of gas is directly related
to concentration. Pressure can be derived from molar concentration using the ideal gas law,
PV=nRT :
Chemical Equilibrium
40 | J B A J
where
𝑛
𝑉
𝑚𝑜𝑙
𝑛
𝑃
=
𝑉
𝑅𝑇
= molar concentration ( )
𝐿
P= partial pressure of gas , atm
R = universal gas constant, 0.08206 L*atm/ (mol*K)
T = temperature, K
Rules in Writing Equilibrium Constant
1. The K expression for the opposite direction of a reversible reaction is the reciprocal of
the original equilibrium constant.
2. The value of K depends on how the equilibrium equation is balanced.
3. Concentrations of reacting species in aqueous solutions are expressed in moles/liter
while in gaseous phase in moles/literor in atmospheres.
4. Concentrations of pure solids, pure liquids and solvents are not included in equilibrium
constant expressions.
5. K constants are dimensionless and the balanced equation and T are specified when
determining the value of K.
Let’s try!
Write the equilibrium constant expression for the reaction
CaH2(s)+2H2O(g)⇌Ca(OH)2(s)+2H2(g)
The following are the equilibrium constant expression for the above reaction:
𝐾𝑝 =
2
𝑃𝐻
2
2
𝑃𝐻
2𝑂
[𝐻 ]2
𝐾𝑐 = [𝐻 2
2
2 𝑂]
Observe that the gas-phase species H2O and H2 appear in the expression but the solids CaH 2
and Ca(OH)2 do not appear as the rule suggested.
Determination of Kp from Kc and vice versa
Consider the reaction:
The equilibrium expressions are written as follows:
Using the ideal gas law, the partial pressures of A and B can be calculated as follows:
Chemical Equilibrium
41 | J B A J
Then,
or simply,
where Δn= no. of moles gas in product side – no. of moles of gas in reactant side
Let’s try!
At equilibrium, the pressure of the reaction mixture:
CaCO3(s) ↔ CaO(s) + CO2(g)
is 0.105 atm at 350oC. Calculate Kp and Kc for this reaction.
Solution: Based from the problem, the given value is the pressure of the reaction which is the
partial pressure of CO2, so we can calculate the Kp of the reaction.
𝐾𝑝 = 𝑃𝐶𝑂2
𝐾𝑝 = 0.105
Using the equation Kp = Kc (RT)Δn
Kp = Kc (RT)Δn
0.105 = Kc (0.08206)(350 + 273.15)1
Kc =
0.105
(0.08206)(350 + 273.15)1
𝐾𝑐 = 2.05 𝑥10−3
Comprehension Check 1.
Ammonium Carbamate (NH4CO2NH2) decomposes as follows:
NH4CO2NH2(s) ↔ 2NH3(g) + CO2(g)
starting with only the solid, it is found that when the system reaches equilibrium at 40oC, the
total gas pressure (NH3 and CO2) is 0.363 atm. Calculate the Kp. Show your complete
solution.
Chemical Equilibrium
42 | J B A J
Factors Affecting Equilibrium
Le Chatelier’s principle states that if an external stress is applied to a system at equilibrium, the
system adjusts in such a way that the stress is partially offset as it tries to reestablish
equilibrium. If an equilibrium system is subjected to a change in temperature, pressure, or
concentration of a reacting species, the system will respond with a new equilibrium state. Le
Chatelier’s principle can be used to predict the behavior of a system due to changes in pressure,
temperature, or concentration.
Change in Concentration
•
•
•
If concentration of either the reactant or product is increased, the system will react to
consume the added reactant or product.
If reactant or product decreased, system will produce it.
Increasing the concentration of reactants will drive the reaction to the right, while
increasing the concentration of products will drive the reaction to the left.
For example:
1. Consider the equilibrium reaction:
𝑁2 (𝑔) + 3 𝐻2 (𝑔) ↔ 2 𝑁𝐻3 (𝑔)
Changes
in
equilibrium
Addition of N2
Removal of H2
Addition of NH3
the Analysis
Increased
reactant
Equilibrium shift
in
the
amount Shift to the right (favors
formation of more product of
favors forward reaction)
Decreased in the amount of Shift to the left (favors
reactant
formation of more reactant or
favors reverse reaction)
Increased in the amount of Shift to the left (favors
product
formation of more reactant or
favors reverse reaction)
2. Consider the equilibrium reaction:
𝐶𝑎𝐶𝑂3 (𝑠) ↔ 𝐶𝑎𝑂 (𝑠) + 𝐶𝑂2 (𝑔)
Changes
in
the Analysis
Equilibrium shift
equilibrium
Addition of CaCO3
Increased in the amount No change
reactant,
however,
solid
reactant is not considered in
the equilibrium constant
Removal of CO2
Decreased in the amount of Shift to the right (favors
product
formation of more product of
favors forward reaction)
Addition of CO2
Increased in the amount of Shift to the left (favors
product
formation of more reactant or
favors reverse reaction)
Chemical Equilibrium
43 | J B A J
Change in volume and pressure
•
•
•
•
•
If V is decreased, P will increase
To reduce P, system will shift in the direction that reduces the number of moles of gas
When the volume of an equilibrium mix of gases is REDUCED, a net change will occur
in the direction that produces FEWER MOLES OF GAS.
When the volume of an equilibrium mix of gases is INCREASED, a net change will
occur in the direction that produces MORE MOLES OF GAS.
for a gas-phase reaction in which the number of moles of gas on both sides of the
equation are equal, the system will be unaffected by changes in pressure, since Δn=0.
For example:
1. Consider the equilibrium reaction:
2.
𝑁2 (𝑔) + 3 𝐻2 (𝑔) ↔ 2 𝑁𝐻3 (𝑔)
Changes
in
the Analysis
Equilibrium shift
equilibrium
The volume of the reaction volume of an equilibrium mix Shift to the right (favors
vessel is doubled
of gases is INCREASED
formation of more product of
favors forward reaction)
(no of moles of gas reactant >
No. of moles of gas product)
Increase the total pressure Since
P
is
inversely Shift to the left (favors
of the system by a factor 2 proportional to V, there formation of more reactant or
volume is decreased by a favors reverse reaction)
factor of 2.
(no of moles of gas reactant >
No. of moles of gas product)
Change in Temperature
The system will respond to the change in temperature depending of the heat of reaction:
• Endothermic: Reactants + heat↔ products
• Exothermic: Reactants ↔ products + heat
For an exothermic reaction, heat is a product. Therefore, increasing the temperature will shift
the equilibrium to the left, while decreasing the temperature will shift the equilibrium to the
right.
Example:
1. Consider the equilibrium reaction:
𝑁2 (𝑔) + 3 𝐻2 (𝑔) ↔ 2 𝑁𝐻3 (𝑔) ∆𝐻 = −92
𝑘𝐽
𝑚𝑜𝑙
Changes
in
the Analysis
Equilibrium shift
equilibrium
Temperature is increased Since ΔH is negative, the Shift to the left (favors
by 10oC.
reaction is said to be formation of more reactant or
favors reverse reaction)
Chemical Equilibrium
44 | J B A J
Temperature is decreased to
0oC.
exothermic and heat is in
product side
Since ΔH is negative,
reaction is said to
exothermic and heat is in
product side
the
the Shift to the right (favors
be formation of more product of
the favors forward reaction)
Presence of Catalyst
Catalysts increase the rates of both forward and reverse reactions. Catalyst increases the rate at
which equilibrium is achieved, but does not change the composition of the equilibrium mixture.
Figure 1. Visual comparison of catalyzed and uncatalyzed reactions
Comprehension Check 2.
Consider the reaction below and determine the equilibrium shift upon imposing the
following changes:
𝑘𝐽
𝐻2 (𝑔) + 𝐼2 (𝑔) ↔ 2 𝐻𝐼 (𝑔) ∆𝐻 = +51
𝑚𝑜𝑙
Changes in the equilibrium
Analysis
Addition of I2
Removal of HI
Doubling the volume of the
reaction vessel
Decreasing the total pressure of
the system by a factor of 4
Increasing the temperature by
25oC.
Presence of catalyst
Chemical Equilibrium
Equilibrium shift
45 | J B A J
Manipulating the Equilibrium Constant
We will use two useful relationships when working with equilibrium constants. First, if we
reverse a reaction’s direction, the equilibrium constant for the new reaction is simply the
inverse of that for the original reaction. For example, the equilibrium constant for the reaction
[𝐴𝐵2 ]
𝐴 + 2𝐵 ↔ 𝐴𝐵2
𝐾1 =
[𝐴][𝐵]2
is the inverse of that for the reaction
𝐴𝐵2 ↔ 𝐴 + 2𝐵
𝐾2 =
[𝐴][𝐵]2
1
=
𝐾1
[𝐴𝐵2 ]
Second, if we add together two reactions to obtain a new reaction, the equilibrium constant
for the new reaction is the product of the equilibrium constants for the original reactions.
[𝐴𝐶]
𝐴 + 𝐶 ↔ 𝐴𝐶
𝐾1 =
[𝐴][𝐶]
[𝐴𝐶2 ]
𝐴𝐶 + 𝐶 ↔ 𝐴𝐶2
𝐾2 =
[𝐴𝐶][𝐶]
[𝐴𝐶]
[𝐴𝐶2 ]
[𝐴𝐶2 ]
𝐴 + 2𝐶 ↔ 𝐴𝐶2
𝐾3 = 𝐾1 𝐾2 =
∗
=
[𝐴][𝐶] [𝐴𝐶][𝐶] [𝐴][𝐶]2
Let’s try!
Calculate the equilibrium constant for the reaction
2A + B ↔ C + 3D
given the following information
Rxn 1: A + B ↔ D
Rxn 2: A + E ↔ C + D + F
Rxn 3: C + E ↔ B
Rxn 4: F + C ↔ D + B
The overall reaction is given as
K1 = 0.40
K2 = 0.10
K3 = 2.0
K4 = 5.0
Rxn 1 + Rxn 2 – Rxn 3 + Rxn 4
If Rxn 3 is reversed, giving
𝑅𝑥𝑛 5: 𝐵 ↔ 𝐶 + 𝐸
𝐾5 =
1
1
=
= 0.50
𝐾3 2.0
Then the overall reaction is
Rxn 1 + Rxn 2 + Rxn 5 + Rxn 4
and the overall equilibrium constant is
𝐾𝑜𝑣𝑒𝑟𝑎𝑙𝑙 = 𝐾1 ∗ 𝐾2 ∗ 𝐾5 ∗ 𝐾4 = 0.40 ∗ 0.10 ∗ 0.50 ∗ 5.0 = 0.10
Chemical Equilibrium
46 | J B A J
Comprehension Check 3.
Consider the reactions below:
(1) A(g) + B(g) ↔ 2C(g)
(2) 2 C(g) ↔ D(g) + E(g)
Kc1 = 4.00 x 10-3
Kc2 = 1.00 x 102
Evaluate the following operations:
Predicting the Direction of a Reaction
RELATIVE VALUES
IMPLICATIONS
Qc = Kc
The initial concentrations are equilibrium concentrations. The system
is already at equilibrium; there will be no net reaction in either
direction.
Qc < Kc
The ratio of initial concentration of products to reactants is too small.
To reach equilibrium, reactants must be converted to products. The
system proceeds in the forward direction.
Qc > Kc
The ratio of initial concentration of products to reactants is too large.
To reach equilibrium, products must be converted to reactants. The
system proceeds in the reverse direction.
Calculating Equilibrium Concentrations
For ease of calculation, setting-up of Initial-Change-Equilibrium (I-C-E) concentrations table
is recommended.
Example 1: Kc for the reaction of hydrogen and iodine to produce hydrogen iodide
H2(g)+ I2(g)↔ 2HI(g)
is 53.4 at 430oC. What will the concentration be at equilibrium if we start with 0.240M
concentration of both H2 and I2.
Chemical Equilibrium
47 | J B A J
H2(g)
Initial
Concentration (M)
+
I2(g)
↔
2HI(g)
H2(g)
I2(g)
2HI(g)
0.240
0.240
0
Change in
Concentration (M)
Equilibrium
Concentration (M)
Note that at start of the reaction there is no HI formed yet which is why a zero (0) concentration is written.
For the change in concentration, we will assign x as the consumed and produced concentrations
for reactants and product respectively. And using the balance equation, the following are
written in the table:
H2(g)
I2(g)
2HI(g)
Initial
Concentration (M)
0.240
0.240
0
Change in
Concentration (M)
-x
-x
+2x
0.240 - x
0.240-x
2x
Equilibrium
Concentration (M)
Using the equilibrium concentration and the equilibrium constant expression and value, we can
derive the value of x:
[𝐻𝐼]2
𝐾𝑐 =
[𝐻2 ][𝐼2 ]
[2𝑥]2
53.4 =
[0.240 − 𝑥][0.240 − 𝑥]
Evaluating the expression, we will arrive with a quadratic equation:
49.4𝑥 2 − 25.632𝑥 + 3.06584 = 0
Using quadratic equation, the value of x is 0.187 M. Therefore, the equilibrium concentrations
are:
Equilibrium
Concentration (M)
Chemical Equilibrium
H2(g)
I2(g)
2HI(g)
0.240 – 0.187=
0.053M
0.240 – 0.187=
0.053M
2 (0.187) =
0.374M
48 | J B A J
Example 2:
Kc for the reaction of hydrogen and iodine to produce hydrogen iodide
H2(g)+ I2(g)↔ 2HI(g)
Is 53.4 at 430oC. What will the concentration be at equilibrium if the starting concentrations
are as follows: [H2]= 0.00623M, [I2]=0.00414M, and [HI]= 0.0424M
Since all compounds involved in the equilibrium have their initial concentration we need to
calculate the value of Qc to determine the direction of the reaction:
[𝐻𝐼]2
[0.0424]2
𝑄𝑐 =
=
= 69.70
[𝐻2 ][𝐼2 ] [0.00623][0.00414]
Since Qc>Kc, the direction of the reaction is towards the reactant or the reverse reaction,
meaning we need to reduce the concentration of the product to achieve equilibrium.
Setting-up of ICE table:
H2(g)
Initial Concentration
(M)
2HI(g)
0.00623
0.00414
0.0424
+x
+x
-2x
(0.00623+x)M
=0.00676 M
(0.00414+x)M =
=0.00467 M
(0.0424-2x)M
=0.0414 M
Change
in
Concentration (M)
Equilibrium
Concentration (M)
I2(g)
Calculation:
[𝐻𝐼]2
[𝐻2 ][𝐼2 ]
[0.0424 − 2𝑥]2
53.4 =
[0.00623 + x][0.00414 + x]
𝐾𝑐 =
53.4 =
[0.0424 − 2𝑥]2
[0.00623 + x][0.00414 + x]
𝒙 = 5.3 𝑥10−3
Comprehension Check 4.
The Kc for the reaction of hydrogen and iodine to produce hydrogen iodide,
H2(g) + I2(g) ↔ 2HI(g)
is 54.3 at 430oC. Calculate the equilibrium concentrations of H 2, I2, and HI if the initial
concentrations are [H2] = [I2] = 0M, and HI = 0.525 M. Show your complete solution.
Chemical Equilibrium
49 | J B A J
Equilibrium Constants for Chemical Reactions
Several types of reactions are commonly used in analytical procedures, either in preparing
samples for analysis or during the analysis itself. The most important of these are precipitation
reactions, acid–base reactions, complexation reactions, and oxidation–reduction reactions.
Precipitation Reactions (Solubility Product Constant, Ksp)
A precipitation reaction occurs when two or more soluble species combine to form an insoluble
product that we call a precipitate. The most common precipitation reaction is a metathesis
reaction, in which two soluble ionic compounds exchange parts. When a solution of lead nitrate
is added to a solution of potassium chloride, for example, a precipitate of lead chloride forms.
We usually write the balanced reaction as a net ionic equation, in which only the precipitate
and those ions involved in the reaction are included. Thus, the precipitation of PbCl2 is written
as:
2+
−
𝑃𝑏(𝑎𝑞)
+ 2𝐶𝑙(𝑎𝑞)
↔ 𝑃𝑏𝐶𝑙2 (𝑠)
In the equilibrium treatment of precipitation, however, the reverse reaction describing the
dissolution of the precipitate is more frequently encountered.
2+
−
𝑃𝑏𝐶𝑙2 (𝑠) ↔ 𝑃𝑏(𝑎𝑞)
+ 2𝐶𝑙(𝑎𝑞)
The equilibrium constant for this reaction is called the solubility product, Ksp, and is given as
Ksp = [Pb2+][Cl–]2 = 1.7 * 10–5
Note that the precipitate, which is a solid, does not appear in the Ksp expression. We most
commonly use the solubility product to find the concentration of one ion when the
concentration of the other is known or fixed by some means. For example, what is the
Pb2+concentration of in equilibrium with 0.10 M Cl- in a solution of KCl containing excess,
undissolved PbCl2(s)? To answer this question, we rearrange equation, Ksp = [Pb2+][Cl–]2
[𝑃𝑏 2+ ] =
𝐾𝑠𝑝
1.7 𝑥 10−5
=
= 1.7 𝑥10−3 𝑀
[𝐶𝑙− ]2
(0.10)2
Because PbCl2 is slightly soluble, additional Cl- obtained from PbCl2 is almost negligible
compared with 0.10 M Cl-
Common Ion Effect
For the ionic solubility reaction
+
−
𝐴𝑔𝐶𝑙(𝑠) ↔ 𝐴𝑔(𝑎𝑞)
+ 𝐶𝑙(𝑎𝑞)
the product [𝐴𝑔+ ][𝐶𝑙 − ] is constant at equilibrium in the presence of excess solid AgCl. If the
concentration of were increased by adding another source of Ag+, such as AgNO3, or Cl-, as
such NaCl, then the concentration of must decrease so that the product remains constant. In
other words, less AgCl(s) will dissolve if or is already present from some other source. This
application of Le Chatelier’s principle is called the common ion effect. A salt will be less
soluble if one of its constituent ions is already present in the solution.
Chemical Equilibrium
50 | J B A J
Comprehension Check 5.
1. Calculate the equilibrium concentration of Ag+ in equilibrium with 0.10 M SCN- in a
solution of NaSCN containing excess, undissolved AgSCN(s). Ksp AgSCN = 1.1 x 10-12
2. Calculate the solubility of Pb(IO3)2 in 1.0 x 10–4 M Pb(NO3)2. Ksp of Pb(IO3)2 =2.5 x 1013
Acid-base reactions
There are three working definitions for acids and bases: Arrhenius, Bronsted-Lowry and Lewis
definition. For equilibrium reaction, we will be using the definition of Bronsted and Lowry
which defines acid as proton donors and base as proton acceptor. Note that these definitions
are interrelated. Defining a base as a proton acceptor means an acid must be available to provide
the proton. For example, in reaction below acetic acid, CH3COOH, donates a proton to
ammonia, NH3, which serves as the base.
−
𝐶𝐻3 𝐶𝑂𝑂𝐻(𝑎𝑞) + 𝑁𝐻3(𝑎𝑞) ↔ 𝐶𝐻3 𝐶𝑂𝑂(𝑎𝑞)
+ 𝑁𝐻4+(𝑎𝑞)
Acetic acid
Ammonia
Acetate ion Ammonium ion
When an acid and a base react, the products are a new acid and base. These new products are
called conjugate acid and conjugate base. For example, acetate ion is the conjugate base of
acetic acid while ammonium ion is the conjugate acid of ammonia.
Strong and Weak Acids
The reaction of an acid with its solvent (typically water) is called an acid dissociation reaction.
Acids are divided into two categories based on the ease with which they can donate protons to
the solvent. Strong acids, such as HNO3, almost completely transfer their protons to the solvent
molecules.
+
𝐻𝑁𝑂3(𝑎𝑞) + 𝐻2 𝑂(𝑙) → 𝐻3 𝑂(𝑎𝑞)
+ 𝑁𝑂3−(𝑎𝑞)
In this reaction H2O serves as the base. The hydronium ion, H3O+, is the conjugate acid of H2O,
and the nitrate ion is the conjugate base of HNO3. It is the hydronium ion that is the acidic
species in solution, and its concentration determines the acidity of the resulting solution. We
have chosen to use a single arrow (→) in place of the double arrows (↔) to indicate that we
treat HNO3 as if it were completely dissociated in aqueous solutions. A solution of 0.10 M
HNO3 is effectively 0.10 M in H3O+ and 0.10 M in NO3–. In aqueous solutions, the common
strong acids are hydrochloric acid (HCl), hydroiodic acid (HI), hydrobromic acid (HBr), nitric
acid (HNO3), perchloric acid (HClO4), and the first proton of sulfuric acid (H2SO4).
Weak acids, of which aqueous acetic acid is one example, cannot completely donate their acidic
protons to the solvent. Instead, most of the acid remains undissociated, with only a small
fraction present as the conjugate base.
−
𝐶𝐻3 𝐶𝑂𝑂𝐻(𝑎𝑞) + 𝐻2 𝑂(𝑙) ↔ 𝐻3 𝑂 +(𝑎𝑞) + 𝐶𝐻3 𝐶𝑂𝑂(𝑎𝑞)
Chemical Equilibrium
51 | J B A J
The equilibrium constant for this reaction is called an acid dissociation constant, Ka, and is
written as
[𝐻3 𝑂 + ][𝐶𝐻3 𝐶𝑂𝑂− ]
𝐾𝑎 =
= 1.75 𝑥 10−5
[𝐶𝐻3 𝐶𝑂𝑂𝐻]
Note that the concentration of H2O is omitted from the Ka expression because its value is so
large that it is unaffected by the dissociation reaction (see rules in writing equilibrium
constant expression).
Strong and Weak Bases
Just as the acidity of an aqueous solution is a measure of the concentration of the hydronium
ion, H3O+, the basicity of an aqueous solution is a measure of the concentration of the hydroxide
ion, OH–. The most common example of a strong base is an alkali metal hydroxide, such as
sodium hydroxide, which completely dissociates to produce the hydroxide ion.
+
−
𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝑁𝑎(𝑎𝑞)
+ 𝑂𝐻(𝑎𝑞)
Weak bases only partially accept protons from the solvent and are characterized by a base
dissociation constant, Kb. For example, the base dissociation reaction and base dissociation
constant for the ammonia are
−
𝑁𝐻3(𝑎𝑞) + 𝐻2 𝑂(𝑙) ↔ 𝑁𝐻4+(𝑎𝑞) + 𝑂𝐻(𝑎𝑞)
𝐾𝑏 =
[𝑂𝐻− ][𝑁𝐻4+ ]
= 1.75 𝑥 10−5
[𝑁𝐻3 ]
Autoprotolysis
Water undergoes self-ionization, called autoprotolysis, in which it acts as both an acid and a
base:
+
−
𝐻2 𝑂(𝑙) + 𝐻2 𝑂(𝑙) ↔ 𝐻3 𝑂(𝑎𝑞)
+ 𝑂𝐻(𝑎𝑞)
with equilibrium constant,
𝐾𝑤 = [𝐻3 𝑂 + ][𝑂𝐻 − ] = 1.0 𝑥 10−14
Table 1 shows how Kw varies with temperature. Its value at 25.0oC is 1.01 x 10-14.
Table 1. Temperature dependence of Kw.
Chemical Equilibrium
52 | J B A J
Let’s try!
Calculate the pH of water at 35oC.
Solution The stoichiometry of autoprotolysis of water tells us that H3O+ and OH- are produced
in a 1:1 molar ratio. Their concentrations must be equal. Assigning each concentration x, we
can write
𝐾𝑤 = [𝐻3 𝑂 + ][𝑂𝐻 − ] = 2.07 𝑥 10−14
[𝑥][𝑥] = 2.07 𝑥 10−14
√𝑥 2 = √2.07 𝑥 10−14
𝑥 = 1.44 𝑥10−7
+
The concentrations of H3O and OH are both 1.44 x 10-7 M in pure water at 35oC. Using the
formula,
pH = -log [H3O+]
pH= -log [1.44 x 10-7]
pH= 6.84
The pH of water at 35oC is 6.84 which is slightly acidic than the expected neutral pH pf 7.0
Comprehension Check 6.
1. Write the dissociation constant expression for the following reaction:
+
a. 𝑁𝐻4+(𝑎𝑞) + 𝐻2 𝑂(𝑙) ↔ 𝐻3 𝑂(𝑎𝑞)
+ 𝑁𝐻3(𝑎𝑞)
+
−
b. 𝐻3 𝑃𝑂4(𝑎𝑞) + 𝐻2 𝑂(𝑙) ↔ 𝐻3 𝑂 (𝑎𝑞) + 𝐻2 𝑃𝑂4(𝑎𝑞)
2. Calculate the pH of water at 100oC.
Relation between Ka and Kb
A most important relation exists between Ka and Kb of a conjugate acid-base pair in aqueous
solution. We can derive this result with the acid HA and its conjugate base A -.
When the reactions are added, their equilibrium constants are multiplied to give
Relation between Ka and Kb for a conjugate pair:
Kw = Ka * Kb
Chemical Equilibrium
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Let’s try!
1. The Ka for acetic acid is 1.75 x 10 -5. Find the Kb for acetate ion.
Solution: Using the equation Kw = Ka * Kb
𝐾𝑤 1.00 𝑥 10−14
𝐾𝑏 =
=
= 5.71 𝑥 10−10
𝐾𝑎
1.75 𝑥 10−5
Comprehension Check 7.
1. Calculate the Kb for chloroacetate ion if the Ka for chloroacetic acid is 1.36 x 10 -3.
2. The Kb for for methylamine is 4.47 x 10-4. Find the Ka for methylammonium ion.
Buffers
Buffers are defined as solutions that tend to resist changes in pH as the result of dilution or
the addition of small amounts of acids or bases.
Consider the dissociation of a weak acid, HA in water:
According to Le Chatelier principle, addition of acid to the system leads the reaction of H3O+
and A- to form HA; therefore, the net formation of H 3O+ is less than it might have been if there
were no A-. In contrast, addition of base like NaOH causes the dissociation of the acid and
minimizing the decline in H3O+ concentration. The behavior of a buffer system can be
quantified using the dissociation of weak acid and taking negative logarithms of both side of
the equation:
−𝑙𝑜𝑔 𝐾𝑎 = −log[𝐻3 𝑂+ ] − log
[𝐴− ]
[𝐻𝐴]
Rearranging the equation gives
−𝑙𝑜𝑔 [𝐻3 𝑂+ ] = −log 𝐾𝑎 + log
[𝐴− ]
[𝐻𝐴]
From the definition pH, pH=-log [H3O+] and pKa= -log Ka, therefore
𝑝𝐻 = pKa + log
[𝐴− ]
[𝐻𝐴]
This equation is known as the Henderson-Hasselbalch equation (HHE) that relates the pH of
weak acid solution to the concentration and pKa of the acid. We can also rewrite the equation
as,
𝑝𝐻 = pKa + log
Chemical Equilibrium
[𝑠𝑎𝑙𝑡]
[𝑎𝑐𝑖𝑑]
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In the event that [HA] = [A-], then pKa would be equal to the pH of the solution. This
relationship enables the calculation of the composition of buffers that have a specified pH and
the pH range over which buffering occurs. The useful buffering range is ~pKa ± 1 pH units.
Examples:
1. Calculate the pH of a 0.20 M acetate buffer containing 0.1 M acetic acid and 0.1 M sodium
acetate given that the pKa of acetic acid is 4.76.
Solution: Given: [CH3COOH] = 0.1 M; [CH3COO-] = 0.1 M; pKa= 4.76
Substituting all the given to the HHE:
𝑝𝐻 = 4.76 + 𝑙𝑜𝑔
0.1
= 4.76 + 0.00 = 4.76
0.1
2. What weight of sodium acetate (FW =82.03 g/mol) must be added to 500.0 mL of 2.00 M
acetic acid to produce a buffer solution that has a pH of 4.00? Ka=1.75 x 10-5.
Given:
[CH3COOH] = 2.00 M ; VCH3COOH = 500 mL = 0.500 L
molCH3COOH= [CH3COOH] VCH3COOH = (2.00 M) (0.500L)= 1 mol
pH = 4.00
Ka =1.75 x 10-5 ; pKa= -log Ka = -log (1.75 x 10-5) = 4.76
Using the HHE, we can calculate for the mol of CH3COO𝑝𝐻 = pKa + log
[𝐶𝐻3 𝐶𝑂𝑂− ]
[𝐶𝐻3 𝐶𝑂𝑂𝐻]
4.00 = 4.76 + log
𝑚𝑜𝑙𝐶𝐻3 𝐶𝑂𝑂 −
1
4.00 − 4.76 = log
−0.76 = log
𝑚𝑜𝑙𝐶𝐻3 𝐶𝑂𝑂−
1
𝑚𝑜𝑙𝐶𝐻3 𝐶𝑂𝑂 −
1
; get the antilog of both side
10−0.76 =
𝑚𝑜𝑙𝐶𝐻3 𝐶𝑂𝑂 −
1
𝑚𝑜𝑙𝐶𝐻3 𝐶𝑂𝑂 − = 0.174 𝑚𝑜𝑙
From the mol of 𝐶𝐻3 𝐶𝑂𝑂 − , we can now calculate the mass of sodium acetate (NaCH 3COO):
Mass 𝐶𝐻3 𝐶𝑂𝑂𝑁𝑎= 0.174 mol (82.03g/mol) = 14.27 g
Comprehension Check 8.
Calculate for the mass of NaH2PO4 and Na2HPO4 needed for the preparation of 100 mL of
0.10 M phosphate buffer with pH of 7.0. Ka2 for phosphoric acid is 6.32 x 10-8.
Chemical Equilibrium
55 | J B A J
Teaching and Learning Activities
The following activities will be implemented:
a. Lecture Discussion
b. Exercises/ Comprehension Check
c. Module
d. Video Presentation
Recommended learning materials and resources for supplementary reading.
The following materials and resources are recommended as supplementary reading:
1. Chapter 6: Chemical Equilibrium of Harris’s Quantitative Chemical Analysis (8 th ed.)
2. Chapter 6: Equilibrium Chemistry of Harvey’s Modern Analytical Chemistry.
Flexible Teaching Learning Modality (FTLM) adopted
The following modality will be used interchangeably for the implementation of the course:
1. Edmodo
2. Google Meet
3. Module
Assessment Task
Please see embedded comprehension check in the learning content.
References
Harris, D. C. (2010). Qunatitative Chemical Analysis (8th ed.). New York: W. H. Freeman and
Company.
Harvey, D. (2000). Modern Analytical Chemistry. United States of America: The McGraw-Hill
Companies, Inc.
Skoog, D. A., West, D. M., Holler, F. J., & Crouch, S. R. (1996). Fundamentals of analytical
chemistry. 8th. International Student, Ed.
Submission Policy:
All activities must be written in a clean paper (you can use back page of a scrap paper), scan,
save with a file name format Name.Chapter#.Exercise#(e.g. J.Abucay.Chapter 2.CC1) and
submit to your output in our on or before the set deadline.
Chemical Equilibrium
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