Chemistry-Chem-01_150916_01

© Art Traynor 2011
Chemistry
Chemical Kinetics
Definitions
Mechanics
– n. Physics. 1. The branch of physics addressed to describing the
action of forces on bodies and with motion comprised of kinetics,
statics, and kinematics. 2. The theoretical and practical application
of this science to machinery, mechanical appliances, etc.
3. The technical aspect or working part; mechanism; structure.
4. Routine or basic methods, procedures, techniques, or details.
Webster’s Encyclopedic
Unabridged Dictionary
of the English Language
( Pg. 1193 )
Dynamics
– n. Physics. 1. The branch of mechanics addressed to describing the motion
and equilibrium of phenomenological systems under the action of some force
or plurality of forces ( principally external to the system ).
2. The motivating or driving forces, physical or moral, in any field.
3. The pattern or history of growth, change, and development in any field.
4. Variation and gradation in the volume of musical sound.
5. Psychodynamics.
( Pg. 610 )

© Art Traynor 2011
Chemistry
Chemical Kinetics
Definitions
Kinetics
– n. Physics. The branch of Mechanics addressed to
describing the actions of forces in producing or
changing the motion of masses.
( Pg. 1057 )
Kinematics
– n. Physics. 1. The branch of Mechanics addressed to
describing pure motion , without references to the
masses upon which it acts. 2. The theory of
mechanical contrivance describing the conversion of
one form of motion into another.
( Pg. 1057 )
Statics
– n. Physics. The branch of Mechanics addressed to
describing bodies at rest or forces in equilibrium.
( Pg. 1861 )

© Art Traynor 2011
Chemistry
Chemical Kinetics
Definitions
Kinetic
– adj. 1. Pertaining to motion. 2. Caused by motion.
3. Characterized by movement.
From the Greek kīnētikós , kīnē-
kīneîn ( to move ) + -tikos
( Pg. 1057 )

© Art Traynor 2011
Chemistry
Definitions
Mechanics
The study of the relationships between
Force
Matter
Motion
DynamicsKinematics
Mechanics
Describes Motion
Relating Motion
to Causes
Velocity
Acceleration
Vector Quantities
* Magnitude
* Direction
Displacement
Time
Average Velocity
Instantaneous Velocity
Force
Mass
Newton’s Laws
Gravity / Elastic
Normal / Friction (Drag)
Tension
Inertia
Weight
Chemical Kinetics
Statics
Kinetics
Bodies at Rest
Forces in Equilibrium

© Art Traynor 2011
Chemistry
Chemical Kinetics
Reaction Metrics
Reaction Metrics
Chemical reactions
can be principally characterized / distinguished by their :
Section 12.3, (Pg. 543)
Zumdahl
Stoichiometry
Energetics ( Thermodynamics )
Rate
Chemical Spontaneity
Refers to the inherent tendency
for a reaction to occur
Implying nothing about
the Speed or pace of a
reaction
Reaction Rates invoke a
notion of ( economic )
time-yield optimization
Reaction Mechanism
Refers to the series of steps
( e.g. Transition States )
by which a reaction progresses

© Art Traynor 2011
Chemistry
Chemical Kinetics
Definition
Chemical Kinetics
The study of rates of chemical reactions
including an account of the dynamic precipitants
influencing:
the speed of a chemical reaction
the reaction's mechanism
its transition states,
reduced to mathematical form ( models )
Reaction Dynamic Precipitants
Concentrations
Temperature
Catalysts
Wiki: “Chemical Kinetics”

© Art Traynor 2011
Chemistry
Chemical Kinetics
Reaction Rate
Reaction Rates
The change in Concentration
of a Reactant or Product
of a Chemical Reaction
per unit Time
Section 12.1, (Pg. 545)
Zumdahl
Concentration of Species A at Time ti + 1 – Concentration of Species A at Time ti
ti + 1 – ti
Rate =
∆ [ A ]
∆ t
Rate =
Notation indicating the
Concentration of Species A
canonically stated in Moles
per liter or solution Molarity
Excluding reverse reactions
a simple model of reaction rates ( excluding other dynamic precipitants )
restricts reaction rate dependence to the concentration of reactant species

© Art Traynor 2011
Chemistry
Chemical Kinetics
Reaction Rate
Reaction Law
A mathematical formula
modeling the dependence of a chemical reaction rate
upon the product of the concentration of a reactant species,
indexed by a discrete order,
and in proportion to a reaction constant
Section 12.1, (Pg. 549)
Zumdahl
Rate = – k [ A ]n
Where :
k = reaction rate proportionality “ constant ”
A = reactant species , e.g. A = NO2
n = order of the reactant
As the reaction proceeds the
reactant species are consumed
and thus a negative sign is
associated with a reactant’s
consumption rate
A rate law expressed as a
function of reactant species
concentration, not time, is
known as a Derivative Rate Law
(DRL)
A rate law expressed as a
function of time, not reactant
species concentration, is known
as an Integrated Rate Law (IRL)
Multi-species
Form Rate = – k [ A ]
n
[ B ]
m
[ C ]
p Section 12.1, (Pg. 567)
Zumdahl
Solutions must be found for k, and
the indices to all reactant species
to fully specify the applicable rate
law for the reaction

© Art Traynor 2011
Chemistry
Chemical Kinetics
Rates Laws
Method of Initial Rates (MOIR)
“ Initial ” Rates for a given reaction are determined experimentally by
measuring the progress of the reaction from as close as t = 0 as possible
Section 12.3, (Pg. 553)
Zumdahl
For n-reactants, n+1 observations will be necessary wherein the
concentration of only one reactant species can be altered (with the other
reactant species concentrations held constant) so as to determine each
species effects upon the pace of the reaction and a species Order assigned
to that reactant.

 Determination of a reactant
species order can only be
determined by observation and
cannot be analytically derived
n = 2 Reactant
Species
N+1 = 3
Observations
Changes in
Reactant Species
Concentrations

© Art Traynor 2011
Chemistry
Chemical Kinetics
Rate Law Determination
Reaction Law Problem Solving Technique ( PST )
Where: Rate = – k [ A ]
n
[ B ]
m
Inspection of the observation data between successive experiments will
reveal one species exhibiting a change in solution concentration while its
co-reactant concentration remains constant.
①
Example: Find the rate law for the reaction
→(NH4 ) (aq) +
+
(NO2 ) (aq)
–
N2 (g) + 2H2O(l )
Section 12.3, (Pg. 553)
Zumdahl
Species
Concentration
Increases
Species
Concentration
Constant
The species with changed concentration ( i.e. the ∆-species ) relative to its constant
concentration co-reactant will enable the Order of the ∆-Species for the reaction to be
determined by an evaluation of the ratio of observed reaction rates
Using problem set one
numbers 7 & 8
Applying MOIR

© Art Traynor 2011
Chemistry
Chemical Kinetics
n
[ B ]
m
Having identified NO2 as the ∆-species for our first calculation we designate
rate equation conventional variable names to the respective reactant
species to prepare for the first ∆-Species Order ( DSO ) calculation.
Example: Find the rate law for the reaction Section 12.3, (Pg. 553)
Zumdahl
②
A = NH4
+
NO2
–
B =
Using problem set one
numbers 7 & 8
→(NH4 ) (aq) +
+
(NO2 ) (aq)
–
N2 (g) + 2H2O(l )
Applying MOIR

© Art Traynor 2011
Chemistry
Chemical Kinetics
n
[ B ]
m
Zumdahl
③ We now set up for our first reactant ∆-Species Order ( DSO ) calculation.
DSOA = =
∆-R(i+1)
∆-R(i )
m
Ar(i+1)
Ar(i )
m
( i = 1 )
→(NH4 ) (aq) +
+
(NO2 ) (aq)
–
N2 (g) + 2H2O(l )
Applying MOIR

© Art Traynor 2011
Chemistry
Chemical Kinetics
n
[ B ]
m
Zumdahl
We now set up for our first reactant ∆-Species Exponential Product ( DSEP )
calculation.
DSEPA = =
[ ∆-R ](i+1)
[ ∆-R ](i )
[ A ](i+1)
[ A ](i ) ( i = 1 )
④
→(NH4 ) (aq) +
+
(NO2 ) (aq)
–
N2 (g) + 2H2O(l )
Applying MOIR

© Art Traynor 2011
Chemistry
Chemical Kinetics
n
[ B ]
m
Zumdahl
We now equate our expressions for DSO and DSEP and solve for the
unknown exponent which will give us the Order for the Reactant under
consideration
( bA )m
→(NH4 ) (aq) +
+
(NO2 ) (aq)
–
N2 (g) + 2H2O(l )
⑤
DSEPADSOA =
= pA
We repeat this process of finding DSO, DSEP, equating the expressions, and the
solving for the unknown exponent for each Reactant which will allow us to then
state our rate law for the reaction
Applying MOIR

© Art Traynor 2011
Chemistry
Chemical Kinetics
n
[ B ]
m
Zumdahl
Having solved for the Orders of the participating Reactants, we can now make
a preliminary statement of the Derivative Rate Law ( DRL ) for the reaction
→(NH4 ) (aq) +
+
(NO2 ) (aq)
–
N2 (g) + 2H2O(l )
⑥
DRL = – k [ A ]
n
[ B ]
m
At this point we have all the information we need to solve for the remaining
unknown, the Reaction Rate Constant ( RRC ) denoted k in the DRL equation
Applying MOIR

© Art Traynor 2011
Chemistry
Chemical Kinetics
n
[ B ]
m
Zumdahl
We now calculate the RRC for the reaction
→(NH4 ) (aq) +
+
(NO2 ) (aq)
–
N2 (g) + 2H2O(l )
RRC = k =
⑦
IR (i ) M · s
– 1
[ A ](i ) [ B ](i )
Variables
Reaction Rate
Constant
Initial Rate
( ith observation )
IR (i )
k
Units
Molarity M =
Moles of Solute
Liters of Solution
Section 11.1, (Pg. 500)
Zumdahl
Applying MOIR
Note that there are no
indices in the divisor
terms of the RRC
calculation

© Art Traynor 2011
Chemistry
Chemical Kinetics
Integrated Rate Law ( IRL )
Section 12.3, (Pg. 553)
ZumdahlAn equation describing the progress of a chemical reaction is
a differential equation of the form
Rate = – k [ A ]
n
Which is simply to otherwise say:
= – k [ A ]
nd [ A ]
dt
Given a reaction rate,
dependent only upon
concentration we can
discover an expression
dependent upon time by
integration
Supplied in the “ Rate = “ form, the Leibniz notation
representing the derivative can simply be
substituted into the expression
= – k [ A ]
nd [ A ]
dt
In the Leibniz derivative form it’s easy to see the
expression as one for which the Method of
Separation of Variables (MOSOV) can be applied
and an Antiderivative found
dt
1
= – k [ A ]
n
· dt
d [ A ]
dt
dt
1
·
Following the “ Ratio of Infinitesimals ”
interpretation, the LHS differential ratio can
be algebraically manipulated thus
“ separating ” (i.e. isolating) like variables
to either side of the ( differential ) equation
d [ A ] = – k [ A ]
n
· dt With variables separated and accompanied by their
corresponding differentials, the integral operator is
next applied to find the Antiderivative expression
∫ ∫Differential
Terms
The Derivative
Term

© Art Traynor 2011
Chemistry
Chemical Kinetics
Integrated Rate Law ( IRL )
Section 12.3, (Pg. 553)
ZumdahlAn equation describing the progress of a chemical reaction is
a differential equation of the form
Rate = – k [ A ]
n
We solve the equation by integration, supplying an expression in the
process for the time-dependency of the reaction:
Given a reaction rate,
dependent only upon
concentration we can
discover an expression
dependent upon time by
integration
d [ A ] = – k [ A ]
n
∫ ∫
d [ A ] = – k [ A ]
n
∫ ∫

© Art Traynor 2011
Chemistry
Reaction Equilibria
Definitions
Reaction Equilibria
Chemical reactions may be broadly characterized by their
equilibrium state
UC Davis ChemWiki:
“ Reversible v. Irreversible Reactions”
Irreversible Reaction
Reversible Reaction
→C6H12O6 (s) + O2 (g ) CO2 (g ) +2H2O(g )
⇌+ CO(aq) H2 (g) +2CO2 (g )
Bidirectional Harpoon
Indicates Reaction May
Progress in Either Direction
( Left/Right )
Unidirectional Arrow
Indicates Reaction
Progresses only in
One Direction (Left)
Example: Combustion of Glucose
2H2O(g )
Example: Steam & Carbon Monoxide
at high temperature

© Art Traynor 2011
Chemistry
Reaction Equilibria
Definitions
Irreversible Reaction
 Reactants convert to products according to the
Stoichiometric Reaction Ratios ( SRR ) of their
balanced reaction equation
One of the reaction species may function as
a Limiting Reagent should the reaction
abundance of that species be less than that
required by the SRR

Section 3.9, (Pg. 110)
Zumdahl
 At “ completion ” only products will remain – no reactants

© Art Traynor 2011
Chemistry
Reaction Equilibria
Definitions
Reversible Reaction
 Reactants convert to Products according to the
Stoichiometric Reaction Ratios ( SRR ) of their
balanced reaction equation
One of the reaction species may function as
a Limiting Reagent should the reaction
abundance of that species be less than that
required by the SRR

Section 3.9, (Pg. 110)
Zumdahl
 The reaction system “ completes ” at an equilibrium
composed of both Products and Reactants
Reactant concentrations are never zero
Species Forward Rate = Reverse Rate
Reaction Constituent concentrations assume a constant value
( no change without exogenous perturbation )


© Art Traynor 2011
Chemistry
Chemical Equilibrium
Definitions
Chemical Equilibrium
A reaction state in which:
 Both Reactants and Products are present
 Concentrations of the Reaction constituents have no tendency
to change with Time
Wiki: “ Chemical Equilibrium ”
Forward
Reaction Rate
Reverse
Reaction Rate
=
→ ←
Dynamic Equilibrium
A reaction state, characteristic of a Reversible System, in which: Wiki: “ Dynamic Equilibrium ”
 The ratios of reaction constituents cease to change
 Reaction constituents nevertheless may continue to combine but at
equivalent rates so the system composition assumes a Steady State
Reaction Constituents = Reactants + Products
 This equivalency is expressed as an Equilibrium Constant ( K ),
equating the relative concentrations of the reaction constituents

© Art Traynor 2011
Chemistry
Chemical Kinetics
Definition
Chemical or Reaction Kinetics
The branch of Chemistry addressed to describing
rates of chemical processes
Wiki: “ Chemical Kinetics”
 Concentrations of Reaction Constituents
Law of Mass Action ( LOMA )
An account of the influence of several key factors on
the rate or speed at which a reaction proceeds is
provided by consideration of the following :
 Reaction Mechanism
 Transition States

© Art Traynor 2011
Chemistry
1865
Empire of France
9 December 1748 – 6 November 1822
Claude Louis Berthollet
Region of Rhône-Alpes
Department of Haute-Savoie
1803
Reversible Reactions
The observation that some chemical
reactions are reversible set the stage for
the description of Chemical Equilibrium
Chemical Kinetics
History
Law of Mass Action
Described Chemical Equilibrium as an
equivalency between forward & reverse
reaction rates deriving expressions for
rate constants
Kingdom of Norway
District of Østlandet
Amt & City of Oslo (Christiania)

© Art Traynor 2011
Chemistry
Chemical Kinetics
Law of Mass Action
Law of Mass Action ( LOMA )
A mathematical model accounting for the behaviors of Solutions in
Dynamic Equilibrium positing that the rate of a chemical reaction is
proportional to the product of the masses of the reactants
Section 13.2, (Pg. 598)
Zumdahl
⇌+jA kB +lC mD
For a reaction of the form :
Kx =
An Equilibrium Expression with Equilibrium Constant ( K ) is given by :
[ C ]
l
[ D ]
m
[ A ]
j
[ B ]
k
Products
Reactants
→ Note well that the RC concentration product terms in the
Equilibrium Expression rational are INDEXED BY
THEIR SPECIES STOICHOIMETRIC COEFFICIENTS
Reaction Constituents = Reactants + Products

© Art Traynor 2011
Chemistry
Chemical Kinetics
Law of Mass Action
The following corollaries extend LOMA
Section 13.2, (Pg. 600)
Zumdahl
 Reverse Reaction Equilibrium Expression is the Reciprocal of the Forward Reaction
 Scalar Product of Stoichiometric Coefficients is also Scaled to the Concentration Indices
 Units of “ K ” are Determined by the Concentration Indices ( varying by Reaction )

© Art Traynor 2011
Chemistry
Chemical Kinetics
Law of Mass Action
Section 13.2, (Pg. 598)
Zumdahl
 Reverse Reaction Equilibrium Expression is the Reciprocal of the Forward Reaction
K =
[ C ]l
[ D ]m
[ A ]
j
[ B ]
k
⇌
Forward Reaction Reverse Reaction
→ ←
=
[ A ] j
[ B ]k
[ C ]
l
[ D ]
m
1
K

© Art Traynor 2011
Chemistry
Chemical Kinetics
Law of Mass Action
Section 13.2, (Pg. 598)
Zumdahl
 Any scalar applied to the reaction species Stoichiometric
coefficients is similarly scaled to the concentration indices
Kn =
[ C ]nl
[ D ]nm
[ A ]
nj
[ B ]
nk
Forward Reaction
→
⇌+jA kB +lC mD
For a reaction of the form :
n
⇌+njA nkB +nlC nmD

© Art Traynor 2011
Chemistry
Chemical Kinetics
Law of Mass Action
Section 13.2, (Pg. 601)
Zumdahl
 Each set of observationally derived equilibrium concentrations
① = { [ N2 ]
n
, [ H2 ]
m
, [ NH3 ]
p
} , … , = = { [ N2 ]
n
, [ H2 ]
m
, [ NH3 ]
p
}
represents an Equilibrium Position, the cardinality of which is unbounded
①
②
③
n
 Notwithstanding the infinitude of possible
Equilibrium Positions, there exits a unique
Equilibrium Constant solution “ K ” for a
reaction system where all key factors ( such as
temperature, etc. ) are held constant

© Art Traynor 2011
Chemistry
Chemical Kinetics
Equilibrium Expressions
Equilibrium Constant for Pressure ( Kp )
The Ideal Gas Law ( IGL ) supplies an expression relating the
abundance of a gaseous solution to its characteristic pressure
( at a constant Temperature ):
Section 13.3, (Pg. 602)
Zumdahl
VP = nRT
= RT
n
V
1
V
VP
1
Variables
Volume V
P
Molar Cardinality
R
n
Pressure
Universal Gas
Constant
TTemperature
P = RT
n
V
Solving the IGL for Pressure, it becomes
readily apparent that a rational representing
substance Concentration is implicated in the
expression
[ G ] =
n
V
P = RT [ G ]
1
RT
[ G ] =
P
RT
This gaseous concentration term in turn supplies
a substitution variable we may apply to the
LOMA equation (thus relating K to Kp )

© Art Traynor 2011
Chemistry
Chemical Kinetics
Section 13.3, (Pg. 602)
Zumdahl
Variables
Volume V
P
Molar Cardinality
R
n
Pressure
Universal Gas
Constant
TTemperature
RT [ G ] =
P
RT
Formulae
Ideal Gas Law
( IGL ) PV = nRT
[ G ]Gaseous
Concentration
RT
1
RT [ G ] = P
Before introducing the gaseous substance
Concentration term into LOMA, a final
manipulation is necessary to explicitly
state a solution for Pressure ( as a
product of gaseous concentration, UGC,
and Temperature ).
P = [ G ] RT

© Art Traynor 2011
Chemistry
Chemical Kinetics
Section 13.3, (Pg. 602)
Zumdahl
Variables
Volume V
P
Molar Cardinality
R
n
Pressure
Universal Gas
Constant
TTemperature
[ G ]Gaseous
Concentration
⇌+jA kB +lC mD
K =
[ C ]
l
[ D ]
m
[ A ]
j
[ B ]
k
Products
Reactants
→
P = [ G ] RT
Note well that the RC concentration
product terms in the Equilibrium
Expression rational are
INDEXED BY THEIR SPECIES
STOICHOIMETRIC COEFFICIENTS
General Reversible Reaction EquationPressure Solution for Gaseous Concentration
Kp =
( PC
l
) ( PD
m
)
→
( PA
j
) ( PB
k
)
( [ G ]C ·RT )l
( [ G ]D ·RT )m
( [ G ]A ·RT )
j
( [ G ]B ·RT )
k
( [ G ]C
l
)( RT )
l
( [ G ]D
m
)( RT )
m
Pressure is first substituted into LOMA
The Pressure Solution for Gaseous
Concentration is then substituted into
the transformed LOMA expression
UGC and Temperature are then factored
( [ G ]A
j
)( RT )
j
( [ G ]D
k
)( RT )
k

© Art Traynor 2011
Chemistry
Chemical Kinetics
Section 13.3, (Pg. 602)
Zumdahl
Variables
Volume V
P
Molar Cardinality
R
n
Pressure
Universal Gas
Constant
TTemperature
[ G ]Gaseous
Concentration
⇌+jA kB +lC mD
K =
[ C ]
l
[ D ]
m
[ A ]
j
[ B ]
k
Products
Reactants
→
P = [ G ] RT
Kp =
( PC
l
) ( PD
m
)
→
( PA
j
) ( PB
k
)
( [ G ]C
l
)( RT )l
( [ G ]D
m
)( RT )m
UGC and Temperature are then factored
( [ G ]A
j
)( RT )
j
( [ G ]D
k
)( RT )
k
( [ G ]C
l
)( [ G ]D
m
)( RT )
l
( RT )
m
The Associative property is deployed
to collect like terms
( [ G ]A
j
)( [ G ]D
k
)( RT )
j
( RT )
k

© Art Traynor 2011
Chemistry
Chemical Kinetics
Section 13.3, (Pg. 602)
Zumdahl
Variables
Volume V
P
Molar Cardinality
R
n
Pressure
Universal Gas
Constant
TTemperature
[ G ]Gaseous
Concentration
⇌+jA kB +lC mD
K =
[ C ]
l
[ D ]
m
[ A ]
j
[ B ]
k
Products
Reactants
→
P = [ G ] RT
Kp =
( PC
l
) ( PD
m
)
→
( PA
j
) ( PB
k
)
( [ G ]C
l
)( [ G ]D
m
)( RT )l
( RT )m
The Associative property is deployed
to collect like terms
( [ G ]A
j
)( [ G ]D
k
)( RT )
j
( RT )
k
( [ G ]C
l
)( [ G ]D
m
)( RT )
l + m
( [ G ]A
j
)( [ G ]D
k
)( RT )
j +
k
The expression is further simplified
by the application of the Property
of Logarithmic Multiplication

© Art Traynor 2011
Chemistry
Chemical Kinetics
Section 13.3, (Pg. 602)
Zumdahl
Variables
Volume V
P
Molar Cardinality
R
n
Pressure
Universal Gas
Constant
TTemperature
[ G ]Gaseous
Concentration
⇌+jA kB +lC mD
K =
[ C ]
l
[ D ]
m
[ A ]
j
[ B ]
k
Products
Reactants
→
P = [ G ] RT
Kp =
( PC
l
) ( PD
m
)
→
( PA
j
) ( PB
k
)
( [ G ]C
l
)( [ G ]D
m
)( RT )l + m
( [ G ]A
j
)( [ G ]D
k
)( RT )
j + k
( [ G ]C
l
)( [ G ]D
m
)
( [ G ]A
j
)( [ G ]D
k
)
( RT )
l + m
( RT )
j + k
·
The Distributive Property is
applied to isolate the RT terms

© Art Traynor 2011
Chemistry
Chemical Kinetics
Section 13.3, (Pg. 602)
Zumdahl
Variables
Volume V
P
Molar Cardinality
R
n
Pressure
Universal Gas
Constant
TTemperature
[ G ]Gaseous
Concentration
⇌+jA kB +lC mD
K =
[ C ]
l
[ D ]
m
[ A ]
j
[ B ]
k
Products
Reactants
→
P = [ G ] RT
Kp =
( PC
l
) ( PD
m
)
→
( PA
j
) ( PB
k
)
( [ G ]C
l
)( [ G ]D
m
)
( [ G ]A
j
)( [ G ]D
k
)
( RT )l + m
( RT )
j + k
·
The rational collecting the
concentration terms is, on
inspection, revealed to b
mathematically identical to “ K ”
permitting substitution
( RT )
l + m
( RT )
j + k
K ·
The rational collecting RT terms
may also be further simplified by
applying the index difference
property for logarithmic division

© Art Traynor 2011
Chemistry
Chemical Kinetics
Section 13.3, (Pg. 602)
Zumdahl
Variables
Volume V
P
Molar Cardinality
R
n
Pressure
Universal Gas
Constant
TTemperature
[ G ]Gaseous
Concentration
⇌+jA kB +lC mD
K =
[ C ]
l
[ D ]
m
[ A ]
j
[ B ]
k
Products
Reactants
→
P = [ G ] RT
Kp =
( PC
l
) ( PD
m
)
→
( PA
j
) ( PB
k
)
( RT )l + m
( RT )
j + k
K ·
The rational collecting RT terms may also be
further simplified by applying the index
difference property for logarithmic division
K · ( RT )
( l + m ) – ( j + k )
K · ( RT )
∆ n
Let ∆ n = ( l + m ) – ( j + k )
and elegance ensues
Kp =

© Art Traynor 2011
Chemistry
Le Châtelier’s Principle
A utility virtue of LCP is its value in predicting the effect of a
change in any of the independent variables on the overall
equilibrium response of the system
Le Châtelier’s Principle ( LCP )

Also know simply as “ The Equilibrium Law”, LCP asserts that every
Equilibrium Position in a reaction system is defined by the solution set of
unique values of Concentration, Temperature, Volume, and Pressure for
the participating species of the system
Whenever a reaction system in equilibrium is perturbed by a
change in these determining factors the system will respond to
vitiate the input disturbance

Any change in the status quo of a reaction system precipitates
an opposing reaction in the products of that system

Wiki: “ Le Châtelier’s Principle ”
Chemical Kinetics
LCP is akin to an equation in four
independent variables
Sometimes referred to as
Homeostasis in the Life Sciences

© Art Traynor 2011
Chemistry
A utility virtue of LCP is its value in predicting the effect of a
change in any of the independent variables on the overall
equilibrium response of the system

Also know simply as “ The Equilibrium Law”, LCP asserts that every
Equilibrium Position in a reaction system is defined by the solution set of
unique values of Concentration, Temperature, Volume, and Pressure for
the participating species of the system
Used to manipulate the outcomes of reversible reactions
Suggestive of manipulations of inputs and conditions to
optimize the desired Yield of a reaction system
 Wiki: “ Le Châtelier’s Principle ”
Chemical Kinetics
Has a direct analog in Economics
to Price Equilibrium Systems
Has Engineering analogs in
system design ( e.g. shear pins,
zinc-ing a propeller, etc. )

© Art Traynor 2011
Chemistry
Concentration

LCP reckons reaction system responses in any of the following reaction
system variables :
Chemical Kinetics
LCP is akin to an equation in four
independent variables
Homeostasis in the Life Sciences
Temperature
Volume
Pressure

© Art Traynor 2011
Chemistry
Concentration

LCP reckons reaction system responses in any of the following reaction
system variables :
Chemical Kinetics
System shifts toward the
direction that Consumes Energy
( i.e. BRE LHS )
Temperature
Unlike some other reaction system inputs , changes in
Temperature will affect the Equilibrium Constant KT

Temperature changes may thus be gauged by treating
energy as a reaction system participant

n Reaction is Exothermic ( evolving Heat ) – Heat as Product
Exogenous Heat increase shifts equilibrium to the BRE LHS
( increasing Reactant species proliferation )
o
Reactant Species Increase solution proliferation
Products Species Decrease solution proliferation
Value of Equilibrium Constant KT Decreases
Section 13.3, (Pg. 627)
Zumdahl

© Art Traynor 2011
Chemistry
B (aq )
Reactants
A (aq )
+
C (aq)
Products
D(aq)
+→
Chemical Kinetics
LCP reckons reaction system responses in reaction system variables :
Temperature
 Reaction is Exothermic ( evolving Heat ) – Heat as Product
ProductsReactants
①
Section 13.3, (Pg. 627)
Zumdahl
② Exogenous increase in heat
Reaction is exothermic①
The system will always shift to consume
exogenous energy input

© Art Traynor 2011
Chemistry
B (aq )
Reactants
A (aq )
+
C (aq)
Products
D(aq)
+
Chemical Kinetics
Temperature
ProductsReactants KT
②
①
Exogenous increase in heat
②
Section 13.3, (Pg. 627)
Zumdahl
The exothermic character of the forward
reaction only permits additional energy
dissipation by the addition of Reactant
species.
Thus the reverse reaction is the
only avenue available for added
energy dissipation

© Art Traynor 2011
Chemistry
B (aq )
Reactants
A (aq )
+
C (aq)
Products
D(aq)
+
Chemical Kinetics
Temperature
②
①
②
Section 13.3, (Pg. 627)
Zumdahl
Thus an exothermic reaction when perturbed
by additional energy will cause the reaction
to favor the reverse reaction ( absorbing the
added energy in the less favorable reverse
direction)
→
→

Products are thus consumed, and
reactants are then proliferated
into solution

© Art Traynor 2011
Chemistry
B (aq )
Reactants
A (aq )
+
C (aq)
Products
D(aq)
+
→
→
Chemical Kinetics
Temperature
→
②
①
②
 
Section 13.3, (Pg. 627)
Zumdahl
The equilibrium shift “ to the left ” ( i.e. in the
direction of the reverse reaction ) with an
added energy perturbation in an exothermic
system implies that the value of K T must
diminish

© Art Traynor 2011
Chemistry
B (aq )
Reactants
A (aq )
+
C (aq)
Products
D(aq)
+
→
→
Chemical Kinetics
Temperature
→
②
①
②
 
Section 13.3, (Pg. 627)
Zumdahl
Another perspective considers the nature of
the reaction ( i.e. exothermic ) as an
unknown, providing a value for K T and
asking how an exogenous energy
perturbation will affect the equilibrium
position of the system

© Art Traynor 2011
Chemistry
B (aq )
Reactants
A (aq )
+
C (aq)
Products
D(aq)
+
→
→
Chemical Kinetics
Temperature
→
②
①
②
 
Section 13.3, (Pg. 627)
Zumdahl
The diminished K T indicates a decreased
capacity for the system to absorb energy
which will be dissipated via the less
favorable reverse reaction

© Art Traynor 2011
Chemistry
B (aq )
Reactants
A (aq )
+
C (aq)
Products
D(aq)
+
→
→
Chemical Kinetics
Temperature
→ Reactant Species Increase
Products Species Decrease
Value of Constant K T Decreases
Given that a reaction is exothermic①
②
①
An exogenous increase in heat
②
Precipitates the following three
responses in the reaction system

 
Section 13.3, (Pg. 627)
Zumdahl

© Art Traynor 2011
Chemistry
Chemical Kinetics
Problem Solving Technique ( PST )
Problem Determination Precis (PDP)
Problem Set 2, Number 6
 A 0.682 g sample of ICl ( g )
 is placed in a 625 mL reaction vessel
 at 682 K.
A 0.682 g sample of ICl ( g ) is placed in a 625 mL reaction vessel at
682 K. When equilibrium is reached for the reaction below 0.0383 g
of I2 is found in the mixture. Find Kc for this reaction.
 When equilibrium is reached for the reaction below
 0.0383 g of I2 is found in the mixture.
 Find Kc for this reaction.
Indicative of a possible time
dependent relationship
⇌ I2 (g ) +2Cl2 (g )2ICl (g )
Indicates formula necessary
for solution
Indicates units must be
reckoned in terms of a
Concentration metric

© Art Traynor 2011
Chemistry
Chemical Kinetics
Balanced Reaction Equation (BRE)
⇌ I2 (g ) +2Cl2 (g )2ICl (g )
⋀ Problem Data Set – Explicit (PDS-E)
Time Dependent Data – Explicit (TDD-E)
ICl (g )
I2 (g )
Cl2 (g )
Species Abundance (Units = g)
0.6820 g ( t = i )
0.0383 g ( t = e )
0.6820 g ( t = i )
ti (initial)
time ( t )
0 t
t = i
te (equilibrium)
∆ t
Boundary
t = e

© Art Traynor 2011
Chemistry
Chemical Kinetics
⇌ I2 (g ) +2Cl2 (g )2ICl (g )
Formulae
mL → L X mL
1
·
1 L
1000 m L
Data Set Statics - Explicit (DSS-E)
625.0 mL
0.625 mL
Vessel Volume
( V mL )
Vessel Volume
( V L )
①
682.0 KTemperature
( K )
①
There’s a strong possibility that we
will need to express volume in Liters
as the concentration formulae will
otherwise reflect a scaling error

© Art Traynor 2011
Chemistry
Chemical Kinetics
Problem Determination Objective (PDO)
⋀
Problem Data Set –
Explicit (PDS-E)
Find Kc for this reaction.
K =
[ C ]
l
[ D ]
m
[ A ] j
[ B ]k
Formulae
LOMA
②
Variable Assignment Table – (VAT)
ICl (g )
I2 (g )
Cl2 (g )
A
C
D
Reactant(s)
Reaction Species
Product(s)
[ R ] [ P ]
Boundary⇌
=
=
=
j
l
m
2
1
1
=
=
=

© Art Traynor 2011
Chemistry
Chemical Kinetics
Problem Determination Precis (PDP) Formulae
g → mol UCF
X g
1
·
1 mol
X amu (g)
mol → M UCF
X mol
X L
[ X ] =
③
④
⇌ I2 (g ) +2Cl2 (g )2ICl (g )
Problem Data Set – Implicit (PDS-E)
Molar Conversion Table (MCT)
I
I2
Cl2
A
C
D
Species
Cl
Sub
script
1
1
2
2
EAMU SAMU
126.90x
= 162.35
35.45x
162.35
Species
Abundance
0.6820 g ( t = i )
0.0383 g ( t = e )
g → mol UCF [X] = mol → M UCF③ ④
g ( t = e )
LOCOM
③
Σ MP = Σ MR

© Art Traynor 2011
Chemistry
Chemical Kinetics
As noted by inspection in the PDP,
the LOMA calculation requires a
Concentration metric and one
moreover by which volumetric
proportions can be evaluated ( i.e.
Molarity considers substance
abundance partitioned by volume )
g → mol UCF
X g
1
·
1 mol
X amu (g)
mol → M UCF
X mol
X L
[ X ] =
③
④
⇌ I2 (g ) +2Cl2 (g )2ICl (g )
I
I2
Cl2
A
C
D
Species
Cl
Sub
script
1
1
2
2
EAMU SAMU
126.90x
= 162.35
35.45x
162.35
Species
Abundance
0.6820 g ( t = i )
0.0383 g ( t = e )
g → mol UCF [X] = mol → M UCF
0.00420080074 mol ( t = i ) 0.00672128118 M ( t = i )
0.00015090621 mol ( t = i ) 0.00024144996 M ( t = i )
③ ④
Thus we restate our TDD-E in terms
of Molarity ( requiring the calculation
of Moles from each species AMU –
found in the Periodic Table ) using
the MCT and the DSS-E Statics
provided

© Art Traynor 2011
Chemistry
Chemical Kinetics
The remaining unknown
g → mol UCF
X g
1
·
1 mol
X amu (g)
mol → M UCF
X mol
X L
[ X ] =
③
④
⇌ I2 (g ) +2Cl2 (g )2ICl (g )
I
I2
Cl2
A
C
D
Species
Cl
Sub
script
1
1
2
2
EAMU SAMU
126.90x
= 162.35
35.45x
162.35
Species
Abundance
0.6820 g ( t = i )
0.0383 g ( t = e )
g → mol UCF [X] = mol → M UCF
0.00420080074 mol ( t = i ) 0.00672128118 M ( t = i )
0.00015090621 mol ( t = e ) 0.00024144996 M ( t = e )
③ ④
g ( t = e ) mol ( t = e ) M ( t = e )

© Art Traynor 2011
Chemistry
Chemical Kinetics
⇌ I2 (g ) +2Cl2 (g )2ICl (g )
K =
[ C ]l
[ D ]m
[ A ]
j
[ B ]
k
Formulae
Grams to Moles
UCF
X g
1
·
1 mol
X amu (g)
Moles to Molarity
X mol
X L
[ X ] =
0 t
[ A ]
j
[ B ]
k
t = i t = e
[ A ]
j
[ B ]
k
[ C ]
l
[ D ]
m
∆ t

© Art Traynor 2011
Chemistry
Acid
– adj. 1. Of, relating to, or being an acid; having the reactions characteristic
of an acid. 2. (of salts and esters) Derived by partial
exchange of replaceable hydrogen ( i.e. Protonation ).
From the Latin acidus or acēre ( sour )
Merriam-Webster dot com
Acids & Bases
Definitions
Akin to the disassociation
phenomena of ionic solutions
Base
– n. 1. Any of various typically water-soluble and bitter tasting compounds
that in solution have a pH greater than 7, are capable of reacting
with an acid to form a salt, and are molecules or ions able to take up a
proton from an acid or able to give up an unshared pair of electrons to
an acid.
Merriam-Webster dot com
Alkalis – slippery to the feel
( Alkalinity )

© Art Traynor 2011
Chemistry
Definitions
Acid
Wiki: “ Acid”
 Sour Taste ( among comestible species )
A chemical substance characterized by the following properties:
 Affinity to react with Bases and Group I – II metals to form salts
 Exhibits a pH of less than seven
Section 14.1, (Pg. 638)
Zumdahl
Acids & Bases
Base
Wiki: “ Base”
 Bitter Taste ( among comestible species )
A chemical substance characterized by the following properties:
 Affinity to react with Acids to form salts
 Exhibits a pH of greater than seven
Section 14.1, (Pg. 638)
Zumdahl
 Catalyze certain reactions ( i.e. Base Catalysis )
 Accept protons from any proton donor
 Include partially/completely displaceable OH – ions
( e.g. hydroxides of the Group I & II metals )
 Functions in solution to mobilize (donate) protons

© Art Traynor 2011
Chemistry
Acids & Bases
Definitions
Deprotonation
The liberation of a proton ( i.e. mobilizing an H+ cation ) from a molecule,
typically in solution, by the attractive force of a solvent species’ unbound/lone
pair ( which can also be regarded equivalently as the donation of electrons from
the solvent-base ).
–HA (aq)
Base
( Solvent )
Acid
( Solute )
Wiki: “ Deprotonation”
H (aq)
+
H2O(l )+
Deprotonation

© Art Traynor 2011
Chemistry
Acids & Bases
Definitions
–HA (aq)
→
→
Base
( Solvent )
Acid
( Solute )
Conjugate
Acid
H (aq)
+
H2O(l )+ (H3 O) (aq)
+
Deprotonation
Protonation
The accretion of a proton ( i.e. a mobilized H+ cation ) to an atom, molecule, or
ion, thus forming a Conjugate Acid.
Wiki: “ Protonation”
Protonation
Base
( Solvent )
H (aq)
+
H2O(l )+

© Art Traynor 2011
Chemistry
Dissociation
Dissociation
A process by which molecules ( or ionic compounds such as salts, or other
complexes ) separate ( i.e. disassociate ) reversibly into lesser constituent
components such as atoms, ions, or radicals.
Acids & Bases
Wiki: “ Dissociaton
(chemistry)”
Acid Dissociation
In aqueous solution, a covalent bond between an acid’s electronegative
atom binding a characteristic hydrogen atom is broken by Heterolytic
Fission, yielding a proton H
+
( i. e. Deprotonation ) and an anion
( i.e. the Conjugate Base ).
Wiki: “ Disssociaton
(chemistry)”
The analytic solution expressing this relationship is given by
⇌HA (aq ) +H (aq)
+
A (aq)
–
Section 13.2, (Pg. 639)
Zumdahl
Note that the solvent ( typically
water ) is usually excluded
from this expression

© Art Traynor 2011
Chemistry
Dissociation
Acids & Bases
Acid Dissociation
In aqueous solution, a covalent bond between an acid’s electronegative
atom binding a characteristic hydrogen atom is broken by Heterolytic
Fission, yielding a proton H
+
( i. e. Deprotonation ) and an anion ( i.e.
the Conjugate Base ).
Wiki: “ Disassociaton
(chemistry)”
⇌HA (aq ) +H (aq)
+
A (aq)
–
Section 13.2, (Pg. 639)
Zumdahl
BaseDeprotonationAcid
Conjugate
Acid
HA (aq ) H2O(l )+ ⇌ (H3 O) (aq)
+
A (aq)
–
+
Conjugate
Base
+ H2O(l )
+
–
Note that the solvent ( typically
water ) is usually excluded
from this expression

© Art Traynor 2011
Chemistry
Acid
Wiki: “ Acid”
 Arrhenius
There are three principal models accounting for
the characteristic properties of an Acid:
Acids & Bases
Models
Section 14.1, (Pg. 638)
Zumdahl
 Brønsted – Lowry
 Lewis

© Art Traynor 2011
Chemistry
Acid
Wiki: “ Acid”
 Arrhenius
Acids & Bases
Models
Section 14.1, (Pg. 638)
Zumdahl
 Lewis
Acids are regarded as substances which increase the
concentration of hydrogen cations ( more precisely,
Hydronium cations ) in solution
Judge an Acid by its solution
behavior, not the content of its
character
+

© Art Traynor 2011
Chemistry
Acid
Wiki: “ Acid”
 Arrhenius
Acids & Bases
Models
Section 14.1, (Pg. 638)
Zumdahl
 Lewis
Acids are regarded as substances which act in
solution as proton ( i.e. Hydrogen Cation ) donors
or those which tend to Deprotonate
The BL expansion of the Arrhenius Definition
subsumes species such as alcohols and amines
( containing OH or NH fragments ) into the
chemical category of acids


© Art Traynor 2011
Chemistry
Acid
Wiki: “ Acid”
 Arrhenius
Acids & Bases
Models
Section 14.1, (Pg. 638)
Zumdahl
 Lewis
Acids are regarded as substances which act in
solution as electron pair receptors in formation of
a covalent bond
The Lewis expansion of the Arrhenius Definition
subsumes species such as metallic cations, and other
electron-deficient exotica ( i.e. BF3 and AlCl3 ),
alcohols and amines into the chemical category of acids


© Art Traynor 2011
Chemistry
Acid
Wiki: “Conjugate Acid”
Acids & Bases
Models
Section 14.1, (Pg. 639)
Zumdahl
Judge an Acid by its solution
character
The resultant species formed by the deprotonation of a proton donor
( i.e. BL acid ) in solution.
HA (aq )
BaseDisassociation
( Deprotonation )
H2O(l )+
Acid
⇌ (H3 O) (aq)
+
Conjugate
Acid
A (aq)
–
+
Conjugate
Base

© Art Traynor 2011
Chemistry
Acid
Acids & Bases
Models
Section 14.1, (Pg. 639)
Zumdahl
Species Conjugate
HA (aq ) (H3 O) (aq)
+
Acid
Conjugate
Acid
H2O(l ) A (aq)
–
Acids
Bases
Base Conjugate
Base

© Art Traynor 2011
Chemistry
Acid
Acids & Bases
Models
Section 14.1, (Pg. 639)
Zumdahl
Species Conjugate
HA (aq )
Acid
A (aq)
–
Acids
Bases
Conjugate
Base
Conjugate Acid–Base Pair
The conjugate base is what
remains of the Forward Reaction
Acid ( FRA ) after deprotonation

© Art Traynor 2011
Chemistry
Acid
Acids & Bases
Models
Section 14.1, (Pg. 639)
Zumdahl
Species Conjugate
HA (aq ) (H3 O) (aq)
+
Acid
Conjugate
Acid
H2O(l ) A (aq)
–
Acids
Bases
Base Conjugate
Base

© Art Traynor 2011
Chemistry
Acid
Acids & Bases
Models
Section 14.1, (Pg. 639)
Zumdahl
In the BL acid scheme, the forward reaction base and its reverse reaction
conjugate are, at equilibrium, in a balanced tension in attraction for the
free proton.
Base
H2O(l )
A (aq)
–
Conjugate
Base
H (aq)
+
⇌

© Art Traynor 2011
Chemistry
Acid
Acids & Bases
Models
Section 14.1, (Pg. 639)
Zumdahl
If an H20 solution species represents a much stronger base than the A–
species, the equilibrium contest for the proton will resolve more in favor of
forward reactant species proliferation and most of the dissolved acid will
be represented by the anion species
Base
H2O(l )
A (aq)
–
Conjugate
Base
H (aq)
+
⇌
A (aq)
–
→

© Art Traynor 2011
Chemistry
Acid
Acids & Bases
Models
Section 14.1, (Pg. 639)
Zumdahl
Base
H2O(l )
A (aq)
–
Conjugate
Base
H (aq)
+
⇌
Here the “ free ” proton favors
the Hydronium form
If an A– solution species represents a much stronger base than the H20
species, the equilibrium contest for the proton will resolve more in favor of
reverse reactant species proliferation and most of the acid dissolved at
equilibrium will be present in solution as the HA species.
HA (aq )
Acid
→

© Art Traynor 2011
Chemistry
Acid
Acids & Bases
Models
Section 14.1, (Pg. 639)
Zumdahl
HA (aq )
Base
Disassociation
( Deprotonation )
H2O(l )
+
Acid
⇌
(H3 O) (aq)
+
Conjugate
Acid
A (aq)
–
+
Conjugate
Base
Disassociation
( Deprotonation )
+
–

© Art Traynor 2011
Chemistry
Acid
Acids & Bases
Models
Section 14.1, (Pg. 639)
Zumdahl
HA (aq )
Base
Disassociation
( Deprotonation )
H2O(l )
+
Acid
(H3 O) (aq)
+
Conjugate
Acid
A (aq)
–
+
Conjugate
Base
Disassociation
( Deprotonation )
+
–
→
→
→

© Art Traynor 2011
Chemistry
Coincidence with Oxygen
Acids & Bases
Acid Species Characteristics
Section 14.1, (Pg. 642)
ZumdahlMost Acids include Oxygen in their molecular composition and are
referred to as Oxyacids
 Most Oxyacids feature the ( characteristic ) acidic Hydrogen cation
( i.e. proton ) bonded directly to an Oxygen atom
Organic Acids are typical of the Oxyacids – characteristically weak
– and featuring the Carboxyl Group as a key structural constituent

F9
Uus117
I53
Br35
At85
Cl17
⑦
–C
O
O – H
One exception is the family of Hydrohalic acids of the form HX
where X represents an element drawn from the Group 7 Halogens

 The strong affinity of Oxygen for electrons is posited to explain the
“ Hydrogen Power ” reaction characteristics of acids ( which the
Hydrohalic acids emulate with their equally strong electron affinity )
Nauli Lecture Notes 10/xx/15

© Art Traynor 2011
Chemistry
Forward Dissociation Acid Species ( FDAS )
Acids & Bases
Acid Species Identification
Section 14.1, (Pg. 639)
Zumdahl
Acidic Species Indicia Table ( ASIT )
HNO3 (aq) ⇌ NO3
–
( aq ) + H3 O+
(aq)H2 O(aq)+
2.4 x 10
1
1.0 x 10
– 14
Constants
Kw = 1.0 x 10– 14
mol
2
· L
– 2
AIPC
≫
Kw ≪ Ka ( by thirteen-plus orders of magnitude! ) ∴ the Nitric Acid is the
FDAS and the disassociation will proceed to yield nearly all Product
BRE
Acid Dissociation
Constant
( ADC ) K a, w
The FDAS for a given reaction BDE can be identified by evaluation of
the species congruence with the following criteria
Example:
FDAS
The Forward Dissociation Acidic Species ( FDAS ) will include
Hydrogen which will be “ donated ” to its conjugate complement
The FDAS will exhibit a greater Acid Dissociation Constant ( ADC ) Ka
value than the other species in the Forward Dissociation expression
Acids are regarded as substances which act in solution as electron
pair receptors in formation of a covalent bond
①
②
③
④ The strength of the FDAS will exhibit an inverse proportional
magnitude to its CBS

© Art Traynor 2011
Chemistry
Forward Dissociation Acid Species ( FDAS )
Acids & Bases
Acid Species Identification
Section 14.1, (Pg. 639)
Zumdahl
The Forward Dissociation Acidic Species ( FDAS ) will include
The FDAS will exhibit a greater Acid Dissociation Constant ( ADC ) Ka
value than the other species in the Forward Dissociation expression
①
②
③
④
HNO3 (aq) ⇌ NO3
–
( aq ) + H3 O+
(aq)H2 O(aq)+
2.4 x 10
1
1.0 x 10
– 14
Constants
Kw = 1.0 x 10– 14
mol
2
· L
– 2
AIPC
≫
Kw ≪ Ka ( by thirteen-plus orders of magnitude! ) ∴ the Nitric Acid is the
FDAS and the dissociation will proceed to yield nearly all Product
BRE
Acid
Disassociation
Constant
( ADC ) K a, w
The FDAS for a given reaction BDE can be identified by evaluation of
the species congruence with the following criteria
Example:
FDAS
5.0 x 10
– 16
a Kb value
1.0 x 10
0
≪
CBS
The strength of the FDAS will exhibit an inverse proportional

© Art Traynor 2011
Chemistry
Law of Mass Action
Law of Mass Action - Acids ( LOMA-A )
Section 14.1, (Pg. 639)
Zumdahl
Ka =
[ H3O
+
] [ A
–
]
[ HA ]
Acids & Bases
A mathematical model accounting for the behaviors of Solutions in
Dynamic Equilibrium positing that the rate of a chemical reaction is
proportional to the product of the masses of the reactants
Variables
Hydrogen Ion H+
HA
Conjugate Base
Acid Species
(with Hydrogen)⇌HA (aq ) +H (aq)
+
A (aq)
–
A–The analytic equilibrium Acid Dissociation Constant ( ADC ) is given by:
Products
Reactants
→
[ H
+
] [ A
–
]
[ HA ]
=

© Art Traynor 2011
Chemistry
Water – Acid & Base
Amphoteric Substance
Section 14.2, (Pg. 643)
Zumdahl
Acids & Bases
A solution substance capable of assuming the properties of either an
Acid or Base depending on the oxidation state of the oxide species solute
Wiki: “Amphoterism”
Autoionization of Water
The spontaneous reaction between water molecules whereby one H2O
donates a proton ( Deprotonates – acting as an Acid ) and another acts
as a receptor ( Protonates – acting as a Base ) yielding a Hydronium
Cation and a Hydroxide Anion in aqueous solution
⇌HA (aq ) +H (aq)
+
A (aq)
–
H2O(l )+ ⇌ (H3 O) (aq)
+
(OH) (aq)
–
+H2O(l )
+ ⇌ +
Section 14.2, (Pg. 643)
Zumdahl
Wiki: “Self-Ionization
of Water”

© Art Traynor 2011
Chemistry
Acids & Bases
H2O(l )+ ⇌ (H3 O) (aq)
+
(OH) (aq)
–
+H2O(l )
+ ⇌ +
Section 14.2, (Pg. 644)
Zumdahl
of Water”
Ka =
[ H3O
+
] [ A
–
]
[ HA ]
Aqueous Ion Product Constant ( AIPC )
Kx =
[ C ]
l
[ D ]
m
[ A ]
j
[ B ]
k
→
Analytic Equilibrium Constant
or
Products
Reactants
→
[ H
+
] [ A
–
]
[ HA ]
Acid Dissociation Constant
Kw =
[ H3O+
] [ OH –
]
[ H2O ]2
The activity of water as solvent in
a very dilute solution is assumed
to be unity however, thus vitiating
the divisor term
Kw = [ H3O
+
] [ OH
–
]

© Art Traynor 2011
Chemistry
Acids & Bases
H2O(l )+ ⇌ (H3 O) (aq)
+
(OH) (aq)
–
+H2O(l )
+ ⇌ +
Section 14.2, (Pg. 644)
Zumdahl
of Water”
Aqueous Ion Product Constant ( AIPC )
Kw =
[ H3O
+
] [ OH
–
]
[ H2O ]2
It can be shown by use of an ICE
table substitution that water auto-
ionization solution species are
equivalent in concentration at 25° C
Kw = [ H3O
+
] [ OH
–
]
Acid Dissociation Constant
( x )· ( x )
( x )2
→
= ( x )· ( x ) = 1.0 x 10– 14
mol
2
· L
– 2
The activity of water as solvent in
a very dilute solution is assumed
to be unity however, thus vitiating
the divisor term
( x )· ( x )
1
→
[ H3O
+
] = x = [ OH
–
] = 1.0 x 10– 7
Kw = [ H + ] [ OH –
]

© Art Traynor 2011
Chemistry
Acidic Strength
Acid Strength
HA (aq ) H2O(l )+ ⇌ (H) (aq)
+
A (aq)
–
+
Acids & Bases
“ Left ” Hand Side “ Right ” Hand Side
The Strength of an acid refers to its ability or tendency to relinquish a
Hydrogen Cation ( or Proton H+ ) in solution.
 Strong Acid
An Acid that completely ionizes (disassociates) in solution
100 %
50 %
0 %
100 %
50 %
0 %
One Mole of a Strong Acid HA dissolves

© Art Traynor 2011
Chemistry
Acidic Strength
Acid Strength
HA (aq ) H2O(l )+ ⇌ (H) (aq)
+
A (aq)
–
+
Acids & Bases
 Strong Acid
100 %
50 %
0 %
100 %
50 %
0 %
+
–
–
–
–+
+
–
+
+
+
+
–
+
–
+
+
+
+ +
–
+
+
+
+
+
+
+
+
One Mole of H
+
( as Hydronium Cation H30
+
)
One Mole of the Conjugate Base A
–

© Art Traynor 2011
Chemistry
Acidic Strength
Acid Strength
HA (aq ) H2O(l )+ ⇌ (H) (aq)
+
A (aq)
–
+
Acids & Bases
 Strong Acid
100 %
50 %
0 %
100 %
50 %
0 %
+
–
–
–
–+
+
–
+
+
+
+
–
+
–
+
+
+
+ +
–
+
+
+
+
+
+
+
+ –
–
+
+ –
–
+
+ – – –
– –
–
–
+ + +
+ +
–+
+
–
+
+
–
+
+
– –
–
–
+
+
+
+ +
–
+
+ +
–
+
+
+
+ –
+
–
One Mole of H
+
( as Hydronium Cation H30
+
)
One Mole of the Conjugate Base A
–
Essentially none of the un-ionized acid HA remains

© Art Traynor 2011
Chemistry
Acidic Strength
Acid Strength ( Molecular )
Acids & Bases
 Strong Acid ( Qualitative Characteristics )
An Acid that completely ionizes ( Disassociates ) in solution
One Mole of a Strong Acid HA dissolves One Mole of H
+
( as
Hydronium Cation H3 0
+
) and One Mole of the Conjugate Base A
–
leaving essentially none of the un-ionized acid HA as an intact species

The stronger the acid the more readily it will Deprotonate ( forms H
+
)
n Disassociation Equilibrium lies far to the “ right ” of the forward reaction
An Acid that yields an inversely Weak Conjugate Base
n Where a Weak CB refers to the affinity of the CB for a Proton ( H
+
cation )
n A much weaker CB than water

© Art Traynor 2011
Chemistry
Acidic Strength
Acids & Bases
 Strong Acid ( Quantitative Characteristics )
Ka measures the strength of an acidic molecule whereas pH
measures the strength of an aqueous acidic solution, which for
a strong acid must be less than that of the Hydronium cation:

Acid Dissociation Constant (ADC ) denoted Ka is greater
than the ADC of the Hydronium Cation:

Ka > 1.0 for Ka ( H
3
O
+
) = 1.0
pKa < – 1.74 for pKa ( H
3
O
+
) = – 1.74
UCSB PDF: “Table of Acids”
Wiki: “Acid Strength”

© Art Traynor 2011
Chemistry
Acidic Strength
Acids & Bases
 Weak Acid ( Qualitative Characteristics )
An Acid that incompletely ionizes ( Disassociates ) in solution
One Mole of a Strong Acid HA dissolves One Mole of H
+
( as
Hydronium Cation H3 0
+
) and One Mole of the Conjugate Base A
–
leaving essentially none of the un-ionized acid HA as an intact species

The stronger the acid the more readily it will Deprotonate ( forms H+
)
n Disassociation Equilibrium lies slightly to the “ right ” of the forward reaction
An Acid that yields an inversely Strong Conjugate Base
n Where a Strong CB refers to the affinity of the CB for a Proton ( H
+
cation )
n A much stronger CB than water

© Art Traynor 2011
Chemistry
Weak Acid Dissociation
Acids & Bases
Weak Acid Dissociation
Section 14.1, (Pg. 649)
ZumdahlA weak acid dissociation is characterized by the following:
 Incomplete Ionization
A significant amount of acid [HA] will remain in solution once the
dissociation has achieved equilibrium

A Weak Acid will not “ run to completion ” or dissociate nearly one-hundred
percent as is the case with a Strong Acid solution
n Disassociation Equilibrium lies to the “ left ” of the forward reaction

© Art Traynor 2011
Chemistry
B (aq )
Reactants
A (aq )
+
C (aq)
Products
D(aq)
+
ProductsReactants
Acids & Bases
Example:
Sample Exercise 14.4 – assume reaction temperature is increased
from 25 ° C to 60° C , ( Kw rises to 1 x 10 – 13 ) , using
Le Châtelier’s principle , predict whether the reaction
2 H2O(l ) ⇌ (H3 O) (aq)
+
(OH) (aq)
–
+
is exothermic or endothermic
Section 14.2, (Pg. 645)
Zumdahl
?
Given that an equilibrium reaction
system experiences an exogenous
increase in heat
①
①
② Which causes Kw to rise
Kw
→
The system will always shift to
consume exogenous energy input
②

© Art Traynor 2011
Chemistry
B (aq )
Reactants
A (aq )
+
C (aq)
Products
D(aq)
+
ProductsReactants
Acids & Bases
Example:
2 H2O(l ) ⇌ (H3 O) (aq)
+
(OH) (aq)
–
+
Section 14.2, (Pg. 645)
Zumdahl
?
Given that an equilibrium reaction
system experiences an exogenous
increase in heat
①
①
② Which causes Kw to elevate ( increase )
Kw
→
exogenous energy input causing :
Product Species Increases
Reactant Species Decrease
②
→
→



© Art Traynor 2011
Chemistry
B (aq )
Reactants
A (aq )
+
C (aq)
Products
D(aq)
+
ProductsReactants
Acids & Bases
Example:
2 H2O(l ) ⇌ (H3 O) (aq)
+
(OH) (aq)
–
+
Section 14.2, (Pg. 645)
Zumdahl
Exogenous increase in heat①
①
② Causes Kw to elevate ( increase )
Kw
→
exogenous energy input :
②
→
→


The elevated K w indicates an increased
capacity for the system to absorb energy
Coupled with the increase in Products ( and
further depletion of Reactants ) suggest the
system is Endothermic

© Art Traynor 2011
Chemistry
pH
Acids & Bases
Scaled absolutely with a minimal value of zero ( most Acidic )
and a maximum value of 14 ( most Basic ).
Acid Strength ( Solution ) pH
 A measure of the acid/base character of an aqueous solution

A Cologrithm ( base 10, or the negative of the Log ) of the solution
Concentration ( more precisely Activity ) of the solvated Hydronium Cation.

“ p ” can be considered mathematically as representing a Cologrithm
Operator as in p ( [ H+ ] ) where the concentration of Hydrogen
Cation assumes the argument of the operator

Wiki: “ PH ”
Wiki: “ Cologrithm ”

© Art Traynor 2011
Chemistry
pH
Acids & Bases
While the “ H ” refers most unambiguously to Hydrogen, there
are many interpretations of the use of “ p ” rather than a more
conventional mathematical Cologrithm operator :
 The simple answer is that in the 1909 article introducing the scale
author Søren Peder Lauritz Sørensen referred to it as a capital “ H ”
subscript to a lower case “ p ”

“ power ” of Hydrogen
So why “ p” ?
“ potenz” German for “ power ”
“ puissance” French for “ power ”
“ pondus hydrogenii ” Latin for “ quantity of hydrogen”
②
Ka =
[ H
+
] [ A
–
]
[ HA ]
Wiki: “ PH ”

© Art Traynor 2011
Chemistry
pH
Acids & Bases
pX = colog 10 ([ X ])
log 10
[ X ]
1
pX =
log 10 (1 ) – log 10 ([ X ])pX =
0 – log 10 ([ X ])pX =
– log 10 ([ X ])pX =
logb = logb p – logb q
p
q( )
Property of Logarithms:
Quotient / Difference


Wiki: “ PH ”

© Art Traynor 2011
Chemistry
pH
Acids & Bases
– log 10 ([ H+ ])pH =


Wiki: “ PH ”
– log 10 ([ X ])pX =
– log 10 ([ OH
–
])pOH =
– log 10 ([ X ])pX =

© Art Traynor 2011
Chemistry
pH
Acids & Bases
Example: Exercise 27 (a), Pg. 689,
Zumdahl
27 (a) : Calculate [ H+ ] for human blood at pH = 7.41
pH = – log 10 ([ H+ ]) Analytic Solution
7.41 = – log 10 ([ H+ ]) Substitution
10
– 7.41
= 10
log10
([ H+ ])
Evaluate the Equivalent Antilog Expression
3.8 x 10– 8
= [ H+ ]
7.41 = – log 10 ([ H+ ]) ( –1 ) Migrate the Negative from RHS to LHS
Formulae
pH = – log 10 ([ H+ ])pH
①

© Art Traynor 2011
Chemistry
Solution Strength pH
Acids & Bases
Hydrogen Cation / Hydronium Concentration
The pX4[ X ] formula can be restated to provide an explicit solution for
[ X ] ( where a value for pX has been given/determined ).

Wiki: “ PH ”
pX = – log 10 ([ X ])
pX = – log 10 ([ X ]) ( –1 )
– pX = log 10 ([ X ])
10
– pX
= 10
log 10 ([ X ])
10
– pX
= [ X ]
[ X ] = 10 – pX
Formulae
pX = – log 10 ([ X ])pX
①
[ X ] for pX Formula ( [X ]4pX )

© Art Traynor 2011
Chemistry
Solution Strength pH
Acids & Bases
Hydrogen Cation / Hydronium Concentration
Using the analytic [X ]4pX formula, simple substitution will yield formulae
for H
+
and OH
–
concentrations given by their pH and pOH values.

Wiki: “ PH ”
Formulae
[ H+ ] for pH Formula ( [H+]4pH )
①
[ X ] = 10
– pX
[ H+ ] = 10
– pH
[ X ] = 10 – pX
[ X ]4pX
[ OH – ] for pOH Formula ( [OH – ]4pOH )
[ X ] = 10
– pX
[ OH –
] = 10 – pOH
Hydroxide Anion Concentration

© Art Traynor 2011
Chemistry
pH
Acids & Bases


Wiki: “ PH ”
Formulae
pH = – log 10 ([ H+ ])pH
①
Σp(H+OH)
②
14.0 = pH + pOH
14.0 = pH + pOH
pH = – log 10 ([ H+ ])①
②
14.0 = – log 10 ([ H+ ]) + pOH
pOH = 14.0 + log 10 ([ H+ ])
pOH = 14.0 – log10([ H+
])pOH(Σp–pH)
③

© Art Traynor 2011
Chemistry
Acids & Bases
Kw for [ H+ ][ OH – ] Product Formula ( Kw4 [H+][OH –
] )
A useful identity, that of the product yielding Kw

can be employed to determine the concentration of a solution knowing only the pH value
Kw = [ H+ ] [ OH
–
] = 1.0 x 10– 14
If pH is given, [ H+ ] can be determined from the [H+]4pH formula
With [ H+ ] thus determined, [ OH
–
] can be found by solving
the Kw equation explicitly for the remaining, single unknown

Ion Product Constant Kw

© Art Traynor 2011
Chemistry
Acids & Bases
[ OH – ] for Kw [ H+ ]Quotient Formula ( [OH –
]4Kw[H+]Q )
Kw = [ H+ ] · [ OH
–
]
Formulae
Kw = [ H+
] · [ OH–
]Kw4[H
+
][OH
–
]
①
Constants
Kw = 1.0 x 10 – 14
mol
2
· L
– 2IOPC or Kw
①
①
[H+ ]
1
Kw = [ H+ ] · [ OH
–
]
[ H+ ]
Kw
= [ OH
–
]
[ OH
–
] =
[ H+ ]
Kw

© Art Traynor 2011
Chemistry
Formulae
Acids & Bases
Significant Formulae
 pH = – log 10 ([ H+ ])
[ H+ ] = 10
– pH

Kw = [ H3O
+
] [ OH
–
]
Kw = [ H + ] [ OH –
]


14.0 = pH + pOH
pOH = 14.0 + log 10 ([ H+ ])


[ OH
–
] =
[ H+ ]
Kw

pOH = – log 10 ([ OH
–
])

© Art Traynor 2011
Chemistry
Formulae
Acids & Bases
Significant Formulae ( Cont’d )
 Ka =
[ H3O
+
] [ A
–
]
[ HA ]
[ H
+
]e [ A
–
]e
[ HA ]e
[ HA ]e · Ka = [ H
+
]e [ A
–
]e
 Ka, b =
Kw
Kb, a
Kw = Ka · Kb
It’s not clear how it got established, but in Common
Ion problems it seems, it’s perfectly OK to use the
Ka for one species to reckon the Kb for another
so long as the dissociations share that same ion!
Kb =
[ BH
+
] [ OH
–
]
[ B ]

↔
↔
[ BH
+
] [ A
–
]
[ B ]

© Art Traynor 2011
Chemistry
Formulae
Acids & Bases
Significant Reactions

 ⇌ +( BH) (aq)
+
B (aq) + H2O(l ) (OH) (aq)
–
HA (aq ) H2O(l )+ ⇌ (H) (aq)
+
A (aq)
–
+
2 H2O(l ) ⇌ (H3 O) (aq)
+
(OH) (aq)
–
+
Strong Acid
Ka > 1.0 , pKa < – 1.74
Strong Base
Ionization of Water

© Art Traynor 2011
Chemistry
Constants
Acids & Bases
Significant Constants
 [ H3O
+
] = [ OH
–
] = 1.0 x 10– 7
 [ H+ ] = [ OH
–
] = 1.0 x 10– 7
 Kw = [ H3O+
] [ OH –
] = 1.0 x 10– 14
mol
2
· L
– 2
 Kw = [ H+ ] [ OH
–
] = 1.0 x 10– 14
mol
2
· L
– 2

© Art Traynor 2011
Chemistry
Acidic Solution Metrics
Percent Dissociation
Section 14.5, (Pg. 655)
Zumdahl
Weak Acid Equilibrium
Also known as Percent Ionization, for any unique equilibrium position in
a weak acid solution, the relative dissociation of the forward-reaction
acid species can be expressed as a ratio of the equilibrium Cation
concentration and the initial Weak Acid concentration.
PDHA =
[ H+ ]e
[ HA ]i
x 100
As a Weak Acid becomes increasingly dilute, its PD will increase
accordingly.
This is superficially confusing
because “ dissociation ” has a
connotation of a “ decrease ” in
concentration which inhibits a
proper understanding of an
“ increasing decrease ”

© Art Traynor 2011
Chemistry
Base
 Arrhenius
the characteristic properties of an Base:
Acids & Bases
Models
Section 14.6, (Pg. 659)
Zumdahl
 Lewis

© Art Traynor 2011
Chemistry
Base
 Arrhenius
Acids & Bases
Models
 Lewis
Bases are regarded as those substances which increase
the concentration of hydroxide anions in solution
Judge an Base by its solution
character
the characteristic properties of an Base: Section 14.6, (Pg. 659)
Zumdahl
–

© Art Traynor 2011
Chemistry
Wiki: “ Acid”
 Arrhenius
Acids & Bases
Models
Section 14.1, (Pg. 638)
Zumdahl
 Lewis
Bases are regarded as substances which act in
solution as proton ( i.e. Hydrogen Cation )
acceptors or those which tend to Protonate
The BL expansion of the Arrhenius Definition
subsumes species such as ammonia and amines
( which though not containing OH nevertheless
cause OH to proliferate a solution ) into the
chemical category of Bases

Base
the characteristic properties of an Base:

© Art Traynor 2011
Chemistry
Strong Base
Acids & Bases
Analogous to a Strong Acid, a Strong Base is a substance which tends to dissociate
completely in solution into a characteristic cation and a Hydroxide Anion
Section 14.6, (Pg. 659)
Zumdahl
Basic Strength
 A substance which proliferates OH
–
anions in solution
 A proton acceptor ( by the Brønsted – Lowry definition )
 Cation Affinities
Hydroxides of Group I elements form Strong Bases
Hydroxides of Group II elements also form Strong Bases
n Thus the legacy designation “ Alkaline ” Earth Elements
n GII Hydroxides mobilize two moles of OH
–
anions into
solution for every singular mole of Alkaline Earth Metal
①
H1
Na11
Li3
K19
Rb37
Cs55
Fr87 Mg12
Be4
Ca20
Sr38
Ba56
Ra88
②
n Thus the legacy designation “ Alkali ”

© Art Traynor 2011
Chemistry
Strong Base
Acids & Bases
Analogous to a Strong Acid, a Strong Base is a substance which tends to dissociate
completely in solution into a characteristic cation and a Hydroxide Anion
Section 14.6, (Pg. 659)
Zumdahl
Basic Strength
Law of Mass Action – Bases ( LOMA-B )
Kb =
[ BH
+
] [ OH
–
]
[ B ]
Variables
Hydrogen Ion OH –
B
Conjugate Base
( with Hydrogen )
Base Species
HB+The analytic equilibrium Basic Dissociation Constant ( BDC ) is given by:
Products
Reactants
→
Section 14.6, (Pg. 661)
Zumdahl
⇌ +( BH) (aq)
+
B (aq) + H2O(l ) (OH) (aq)
–

© Art Traynor 2011
Chemistry
Acids & Bases
Kw for Ka · Kb Product Formula ( Kw4Ka·Kb )
A useful identity, that of the product yielding Kw

can be employed to determine either Ion Product Constant ( Acid or Base ) when the other
is a known quantity ( via the constant value of Kw = 1.0 x 10– 14
mol
2
· L
– 2
)
Kw = Ka · Kb
Kb
1
Solving the Kw4Ka·Kb formula explicitly for Ka ↔ Kb
Kw = Ka · Kb
Kb
Ka =
Kw
Ka
1
Kw = Ka · Kb
Ka
Kb =
Kw
↔
↔
Section 14.6, (Pg. 671)
Zumdahl

© Art Traynor 2011
Chemistry
Problems
Calculate percent dissociation of 1.0 M benzoic acid, which has a Ka = 6.3 x 10– 5
.
What about % dissociation of a 10.0 M benzoic acid? What is the relationship of
% dissociation to initial concentration of a weak acid?
Acids & Bases
a. Calculate percent dissociation of 1.0 M benzoic acid, which has a Ka = 6.3 x 10– 5
.
b. What about % dissociation of a 10.0 M benzoic acid?
c. What is the relationship of % dissociation to initial concentration of a weak acid?
Restate the PDP into component subparts, so as to address each in turn①

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
a. Calculate percent dissociation of 1.0 M benzoic acid, which has a Ka = 6.3 x 10– 5
.
 Calculate percent dissociation
Problem Determination Precis (PDP) – Subpart ( a )
The PDP will always include some indication of the PDO which can be teased
out of the PDP by parsing its contents ( sentence fragments, clauses ).
②
PD Percent Disassociation ( Canonical Form )
 of 1.0 M benzoic acid PD(HA) Species Identification
 which has a Ka = 6.3 x 10– 5 PD(HA)4K “for” (i.e. ‘given’ ) a K-value
Note that the BRE is not
given (explicit) and must
be reckoned implicitly

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
Problem Solution Path (PSP) – PD(HA)4K
1. Write the BRE’s – identify all participating species in the solution.
2. Identify the species that will yield the greatest profusion of H+.
Section 14.5, (Pg. 651)
Zumdahl
Step Three
3. Write the Equilibrium Expression for the dominant species
4. Tabulate the known species concentrations noting carefully any time dependence
Step Two
Step One
Step Four
i.e. : TDD-E
i.e. : BRE
i.e. : Formulae
The PSP for PD(HA)4K is outlined explicitly by Zumdahl at pg 651.③

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
5. Construct an ICE table to assign variables to unknown concentrations.
Section 14.5, (Pg. 651)
Zumdahl
Step Five
i.e. : ICET
The PSP for PD(HA)4K is outlined explicitly by Zumdahl at pg 651.③
6. Substitute the values obtained from the ICE table into the ADC
LOMA-A formula and solve for any remaining unknowns
( recalling the Weak Acid assumption [HA]i – x ≈ [HA]i )
which will supply all terms necessary to compute:
Step Six
Formulae
Ka =
ADC
LOMA-A
[H+
] [ A –
]
[ HA ]
 [H+ ]x expressing Concentration of Solution Cation
PD HA =
[H+
] e
[ HA ] i
x 100PDHA
 pH expressing Solution Acidity / Alkalinity
 PDHA expressing Percentage Dissociation of Weak Acid PD(HA)4K

© Art Traynor 2011
Chemistry
PSP Step 1 – Balanced Reaction Equation (BRE)
⇌C6 H5 COOH(aq ) H (aq)
+
C6 H5 CO2 (aq )+
2 H2O(l ) ⇌ (H3 O) (aq)
+
(OH) (aq)
–
+
Problems
Acids & Bases
Ka = 6.3 x 10 – 5
( t = e )
Kw = 1.0 x 10 – 14
( t = e )
Kw ≪ Ka ( by nine orders of magnitude! )
∴ the Benzoic Acid reaction is the only
dissociation that need be considered
To further proceed along the PSP we will need to identify our BRE’s – in this
case we were only provided with the acidic species name (no chemical
formula provided)…we will need to look up (via external reference) or derive it
④
Although not provided
explicitly in the PDP, the
Aqueous Ion Product
Constant ( AIPC ) is easy
enough to recall by memory

© Art Traynor 2011
Chemistry
PSP Step 1 – Balanced Reaction Equation (BRE) ⋀ Problem Data Set – Explicit (PDS-E)
+
C6 H5 CO2 (aq )+
Ka = 6.3 x 10 – 5
( t = e )
Problems
Acids & Bases
time ( t )
0 t
t = i t = e
1.0 M ( t = i )
0.6820 g ( t = i )
ti (initial)
time ( t )
te (equilibrium)
∆ t
Boundary
C6 H5 COOH(aq )
H (aq)
+
C6 H5 CO2 (aq )
0.6820 g ( t = i )
We now populate our TDD-E table noting unknown quantities ( most likely to
be supplied by an ICE table in subsequent evolution of the PSP ).
⑤
It is perhaps characteristic of the
canonical form of PD(HA)4K
PDP’s that we will find ourselves
completely bereft of reaction
product concentration data?

© Art Traynor 2011
Chemistry
Initial → Change → Equilibrium Table ( ICET )
Problems
Acids & Bases
DH+SBRE C6 H5COOH(aq)
1.0 M
⇌ H+
(aq) + C6 H5CO2
–
(aq)
I ( Initial )
– 1 xC ( Change )
E = I + C
E ( Equilibrium )
[ HA ] [ H+
] [ A–
]
+ 1 x + 1 x
0.0 0.0
1.0 – 1 x x x
Formulae
Ka =
ADC
LOMA-
A1
[ H +
] [ A –
]
[ HA ]
PD HA =
[H+
] e
[ HA ] i
x 100PDHA
①
②
Noting that a PD(HA)4K PDP obviously implicates resort to the
ADC/LOMA-A formula, we further note by inspection that we
will need values ( ti ≠ te ) for [HA]i and [H+]e to calculate PD Ha
which an ICE table will supply
⑥
I Can’t remember what
DH+SBRE stands for?
D dissociation (of)
H+
S species
B balanced
R reaction
E equation

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
Problem Determination Objective (PDO) – PD(HA)4K
Inspection and the results of the ICE table reveal that we will
have to manipulate the ADC LOMA-A1 formula to solve for the
unknown quantity
⑦
[ HA ]· Ka =
②
[ H
+
] [ A
–
]
[ HA ] 1
[ HA ]
Ka =
[ H+
] [ A–
]
[ HA ]
[ HA ]· Ka = [ H
+
] [ A
–
] The product of the Weak
Acid concentration and Ka
is equal to the product of
the Ion concentrations
[ HA ]·Ka = [ H+
]·[A–
]
ADC
LOMA-A2
Formulae
③
③

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
We now populate our ADC / LOMA-A formula with the value(s)
from our ICE table and PDS-E to determine our values for
[ HA ] so as to calculate PD Ha
[ HA ]· Ka = [ H
+
] [ A
–
[ HA ]·Ka = [ H+
]·[A–
]
ADC
LOMA-A2
Formulae
③
③
ICE Table Values
[ HA ] = 1.0 M
[ H
+
] = x
[ A
–
] = x
PDS-E Table Value(s)
Ka = 6.3 x 10 – 5
( t = e )
⑧
1.0 M · 6.3 x 10 – 5
= ( x )· ( x )
6.3 x 10 – 5
= x2

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
We now solve for the unknown
x2 = 6.3 x 10 – 5
⑨
x2 = 6.3 x 10 – 5
x = 7.9 x 10 – 3
[ H
+
] = x = 7.9 x 10 – 3
[ A
–
] = x = 7.9 x 10 – 3
[ H
+
] = x = [ A
–
]

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
6.3 x 10 – 5
( t = e ) =
[ H
+
] [ C6H5CO2
–
]
( x )· ( x )
( 1.0 – x )
→
x2
1.0 M
[ C6H5COOH ]
x2 = 6.3 x 10 – 5
( t = e )
Note that the change in
the divisor subtrahend
quantity is considered to
be de minimus and thus
the difference in the
divisor can be considered
to be equivalent solely to
the minuend[ H+
] [ A–
] = 6.3 x 10 – 5
( t = e ) = x2
[ H
+
] = [ A
–
] = x = 7.9 x 10 – 3
( t = e )
②
[ H
+
] [ A
–
]
[ HA ]
→
We could alternatively have solved for the unknown using the
ADC / LOMA-A1 formula which more intuitively reveals the
characteristic Weak Acid vanishing difference in the divisor
[ HA ] i – x ≈ [ HA ] i
⑨
Formulae
Ka =
ADC
LOMA-
A1
[ H +
] [ A –
]
[ HA ]
②
→

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
PDHA =
[H+ ]e
[ HA ]i
x 100
Section 14.5, (Pg. 655)
Zumdahl
Finally we solve the PDHA equation to satisfy our PDO
①
PDHA =
7.9 x 10 – 3
( t = e )
( 1.0 M )
100
1
· = 0.7937%
10

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
b. What about % dissociation of a 10.0 M benzoic acid ?
 What about % dissociation
Problem Determination Precis (PDP) – Subpart ( b )
②
 of a 10.0 M benzoic acid ? PD(HA) Species Identification
PD(HA)4K “for” (i.e. ‘given’ ) a K-value

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
All else is the same as for subpart (a) with the exception of the changed
( increased ) concentration of weak acid at t = i allowing us to jump to
step 5 of the subpart (a) PSP solution series .
③
PSP Step 1 – Balanced Reaction Equation (BRE) ⋀ Problem Data Set – Explicit (PDS-E)
+
C6 H5 CO2 (aq )+
Ka = 6.3 x 10 – 5
( t = e )
time ( t )
0 t
t = i t = e
10.0 M ( t = i )
0.6820 g ( t = i )
ti (initial)
time ( t )
te (equilibrium)
∆ t
Boundary
C6 H5 COOH(aq )
H (aq)
+
C6 H5 CO2 (aq )
0.6820 g ( t = i )It is perhaps characteristic of the
canonical form of PD(HA)4K
PDP’s that we will find ourselves
completely bereft of reaction
product concentration data?

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
Problem Solution Path (PSP) – PD(HA)4K Formulae
Ka =
ADC
LOMA-
A1
[ H +
] [ A –
]
[ HA ]
PD HA =
[ HA ] e
[ HA ] i
x 100PDHA
①
②
We simply substitute the increased Weak Acid concentration
into the ICE table and progress to the next step in the PSP
④
Initial → Change → Equilibrium Table ( ICET )
DH+SBRE C6 H5COOH(aq)
10.0 M
⇌ H+
(aq) + C6 H5CO2
–
(aq)
I ( Initial )
– 1 xC ( Change )
E = I + C
E ( Equilibrium )
[ HA ] [ H+
] [ A–
]
+ 1 x + 1 x
0.0 0.0
10.0 – 1 x x x

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
We now populate our ADC / LOMA-A formula with the value(s)
from our ICE table and PDS-E to determine our values for
[ HA ] so as to calculate PD Ha
[ HA ]· Ka = [ H
+
] [ A
–
[ HA ]·Ka = [ H+
]·[A–
]
ADC
LOMA-A2
Formulae
③
③
ICE Table Values
[ HA ] = 10.0 M
[ H
+
] = x
[ A
–
] = x
PDS-E Table Value(s)
Ka = 6.3 x 10 – 5
( t = e )
10.0 M 6.3 x 10 – 5
= ( x )· ( x )
63.0 x 10 – 5
= x2
⑤

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
We now solve for the unknown
x2 = 63.0 x 10 – 5
x2 = 63.0 x 10 – 5
x = 25.1 x 10 – 3
[ H
+
] = x = 25.1 x 10 – 3
[ A
–
] = x = 25.1 x 10 – 3
[ H
+
] = x = [ A
–
]
⑥

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
PDHA =
[ HA ]e
[ HA ]i
x 100
Section 14.5, (Pg. 655)
Zumdahl
Finally we solve the PDHA equation to satisfy our PDO
①
PDHA =
25.1 x 10 – 3
( t = e )
( 10.0 M )
100
1
· = 2.51%
⑦

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
c. What is the relationship of % dissociation to initial concentration of a weak acid?
Problem Determination Precis (PDP) – Subpart ( c )
This subpart calls for a comparative characterization of the results of the first
two subparts best exhibited by a tabulation of the results obtained
②
PDO Comparative Results Table ( CRT )
C6 H5COOH(aq)
10.0 MI ( Initial )
E ( Equilibrium )
[ HA ]
10.0 – 1x
Subpart (a) Subpart (b)
1.0 M
1.0 – 1x
PD HA 2.51%0.79 %
∴ the results demonstrate that Percent
Dissociation for a Weak Acid increases
with increased initial Weak Acid
solution concentration
PDO Comparative Results Table ( CRT )
Section 14.5, (Pg. 655)
Zumdahl

© Art Traynor 2011
Chemistry
Problems
Problem Determination Precis ( PDP ) & Problem Determination Objective ( PDO )
Problem Set 3, Number 4, subpart ( a )
Which is the stronger Base species of the following pairs :
Acids & Bases
Cl –
(aq) ( Chloride )
[ A –
]
NO 2
–
( aq) ( Nitrite )
Conjugate Base Species
( CBS )
Un-Ionized Acid
[ HA ]
HCl (aq)
( Hydrochloric Acid )
HNO 2
–
( aq)
( Nitrous Acid )
Relative Strength
of CBS Water
H2O
H2O
Acid
Disassociation
Constant
( ADC ) Ka
1.3 x 10
6
7.2 x 10
–4
Acid
Disassociation
Constant
( ADC ) Ka
Un-Ionized Acid
[ HA ]
H3O (aq)
( Hydronium )
H3O (aq)
( Hydronium )
1.0 x 10
– 14
mol2
· L– 2
>
>
Constants
Kw = 1.0 x 10– 14
mol
2
· L
– 2
AIPC
1.0 x 10
– 14
mol
2
· L
– 2

© Art Traynor 2011
Chemistry
Problems
Which is the stronger Base species of the following pairs :
Acids & Bases
HNO 2 (aq) ( Nitrous Acid )
[ H +
]
Conjugate Acid
( CA )
NO 2
–
( aq)
( Nitrite )
Relative
Strength of CA Water
H2O
Acid
Disassociation
Constant
( ADC ) Ka
7.2 x 10
–4
Conjugate Base
[ A –
]
OH–
(aq)
( Hydroxide )
<
Problem Set 3, Number 4, subpart ( b )
Problem Determination Precis ( PDP ) & Problem Determination Objective ( PDO )
1.0 x 10
– 14
mol2
· L– 2
Constants
Kw = 1.0 x 10– 14
mol
2
· L
– 2
AIPC

© Art Traynor 2011
Chemistry
Problems
Predict the net direction of each reaction at equilibrium and whether K > 1 or K < 1
Acids & Bases
a. Predict the net direction of each reaction at equilibrium
b. and whether K > 1 or K < 1
Restate the PDP into component subparts, so as to address each in turn①
HNO3 (aq) ⇌ NO3
–
( aq ) + H3 O+
(aq)H2 O(aq)+
HF (aq) ⇌ F–
( aq ) + H3 O+
(aq)H2 O(aq)+
H2 O(aq) ⇌ H3 O+
(aq) + HCO3
–
(aq)H2 CO3 (aq)+
HCOOH(aq) ⇌ HCOO–
( aq ) + HCN(aq)CN
–
( aq )+

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
 Calculate percent dissociation
②
 of 1.0 M benzoic acid PD(HA) Species Identification
 which has a Ka = 6.3 x 10– 5 PD(HA)4K “for” (i.e. ‘given’ ) a K-value

© Art Traynor 2011
Chemistry
The Forward Disassociation Acidic Species ( FDAS ) will include
The FDAS will exhibit a greater Acid Disassociation Constant ( ADC ) Ka
value than the other species in the Forward Disassociation expression
①
②
③
④
HNO3 (aq) ⇌ NO3
–
( aq ) + H3 O+
(aq)H2 O(aq)+
2.4 x 10
1
1.0 x 10
– 14
Constants
Kw = 1.0 x 10– 14
mol
2
· L
– 2
AIPC
≫
Kw ≪ Ka ( by thirteen-plus orders
of magnitude! ) ∴ the Nitric Acid
is the FDAS and the disassociation
will proceed to yield nearly all
Product ( HAk > 1 )
BDE
Acid
Disassociation
Constant
( ADC ) K a, b, w
FDAS = HA
5.0 x 10
– 16
a Kb value
1.0 x 10
0
≪
CBS = A
–
Problems
Acids & Bases
Consult the ASI table to identify the FDAS and
compare K a, b, w values
③
HA (aq ) ⇌ (H) (aq)
+
A (aq)
–
+
“ Left ” Hand
Side ( LHS )
“ Right ” Hand Side ( RHS )
100 %
50 %
0 %
100 %
50 %
0 %

© Art Traynor 2011
Chemistry
①
②
③
④
7.2 x 10– 4
1.0 x 10
– 14
Constants
Kw = 1.0 x 10– 14
mol
2
· L
– 2
AIPC
≫
Kw ≪ Ka ( by ten orders of
magnitude! ) ∴ the Hydrofluoric
Acid is the FDAS and the
disassociation will remain mostly
in Reactant species
( HAk < 1 )
BDE
Acid
Disassociation
Constant
( ADC ) K a, b, w
1.4 x 10
– 11
a Kb value
1.0 x 10
0
≪
Problems
Acids & Bases
③
HA (aq ) ⇌ (H) (aq)
+
A (aq)
–
+
“ Left ” Hand
Side ( LHS )
100 %
50 %
0 %
100 %
50 %
0 %
HF (aq) ⇌ F–
( aq ) + H3 O+
(aq)H2 O(aq)+
FDAS = HA CBS = A
–
But the formation of Hydronium is Not favored so does the forward reaction proceed
at all? We’d need to know concentrations to say (wouldn’t we)? The disparity
between the FDAS and the CAS is four orders of magnitude…

© Art Traynor 2011
Chemistry
①
②
③
④
4.2 x 10
– 7
Constants
Kw = 1.0 x 10– 14
mol
2
· L
– 2
AIPC
Kw ≪ Ka ( by seven orders of
magnitude! ) ∴ the Carbonic Acid is
the FDAS and the disassociation
will remain mostly in Reactant
species ( HAk < 1 )
BDE
Acid
Disassociation
Constant
( ADC ) K a, b, w
2.4 x 10
– 8
a Kb value
1.0 x 10
0
Problems
Acids & Bases
③
HA (aq ) ⇌ (H) (aq)
+
A (aq)
–
+
“ Left ” Hand
Side ( LHS )
100 %
50 %
0 %
100 %
50 %
0 %
But the formation of Hydronium is Not favored so does the forward reaction proceed
at all? We’d need to know concentrations to say (wouldn’t we)? The disparity
between the FDAS and the CAS is only one order of magnitude…
H2 O(aq) ⇌ H3 O+
(aq) + HCO3
–
(aq)H2 CO3 (aq)+
1.0 x 10
– 14
≪ ≫
FDAS = HA CBS = A
–

© Art Traynor 2011
Chemistry
①
②
③
④
2.5 x 10
– 5
Constants
Kw = 1.0 x 10– 14
mol
2
· L
– 2
AIPC
KHA ≪ KA ( by one order of
magnitude! ) ∴ the Formic Acid is
the FDAS and the disassociation
will proceed to yield nearly all
Product ( HAk > 1 )
BDE
Acid
Disassociation
Constant
( ADC ) K a, b, w
4.0 x 10
– 10
a Ka value
5.6 x 10– 11
Problems
Acids & Bases
③
HA (aq ) ⇌ (H) (aq)
+
A (aq)
–
+
“ Left ” Hand
Side ( LHS )
100 %
50 %
0 %
100 %
50 %
0 %
It seems the formation of Hydrocyanic Acid is favored so the forward reaction will
proceed? We’d need to know concentrations to know how much (wouldn’t we)?
1.8 x 10
– 4
≪≫
HCOOH(aq) ⇌ HCOO–
( aq ) + HCN(aq)CN
–
( aq )+
FDAS = HA CBS = A
–

© Art Traynor 2011
Chemistry
Problems
Determine whether a solution containing the species
below is neutral, acidic or basic at 25° C
Acids & Bases
a. [ H3O+
] @ 7.5 x 10– 5
M
b. [ OH
–
] @ 1.2 x 10– 3
M
c. [ H
+
] @ 5.4 x 10– 5
M
d. [ OH
–
] @ 3.6 x 10– 15
M

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
The PDP is quite straightforward and cannot be broken into any
meaningful component subparts, or otherwise more clearly restated
①
A PSP will need to be charted that addresses the burden of the question
while avoiding any unnecessary reckoning of superfluous steps
Determine whether a solution containing the species below is
neutral, acidic or basic at 25° C
The presence of any characteristic dissociation Products ( i.e. Ions ) will reliably
indicate whether or not a solution is Acidic, Basic, or Neutral
SC Solution Characterization ( Canonical Form )
SC4[DP] “for” (i.e. ‘given’ ) a concentration of
disassociation product

© Art Traynor 2011
Chemistry
Problems
Acids & Bases
Problem Solution Path (PSP) – SC4[DP]
1. Write the analytic disassociations for both Acidic & Basic solutions
2. Determine ( by comparison with the analytic expressions ) if
any of the subpart constituents ( assumed to be forward
dissociation Products ) correspond with ( i.e. indicate/signal )
a characteristic dissociation “ reaction ” .
Step Three
3. Use/Presume that the auto-ionization of water demarcates a threshold
AIPC for a neutral solution
Step Two
Step One
i.e. : BDE’s
i.e. : Constant
The PSP for SC4[DP] might profitably follow something like the following②
i.e. : PDS-E Qualitative
Comparison

© Art Traynor 2011
Chemistry
Problem Solution Path (PSP) – SC4[DP] Step One
Problems
Acids & Bases
We now implement the first step of our PSP and state the analytic BDE’s for
both the Acidic and the Basic dissociation “reactions”
③
PSP Step 1 – Analytic Balanced Dissociation Equations ( ABDE’s )
Section 14.6, (Pg. 661)
Zumdahl
If [ H3O
+
] = 7.5 x 10– 5
M , is present in solution then it would have to
have evolved as the result of a disassociation of the type described by ADE
Problem Data Set – Explicit (PDS-E)
∴ the solution is Acidic
Problem Determination Objective (PDO) – SC4[DP]
⇌HA (aq ) +H (aq)
+
A (aq)
–Acid Dissociation
Equation ( ADE )
⇌B (aq ) +BH (aq)
+Base Dissociation
Equation ( BDE )
H2O(l )+
H2O(l )+ (OH) (aq)
–

Chemistry-Chem-01_150916_01

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