Download 2nd Nine Weeks Notes

2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
I. Introduction.
A. Chemical Kinetics.
1. Chemical kinetics is the study of the changes in concentrations of
reactants or products as a function of time.
2. Examples:
2. It is the study of reaction rates.
3. Each reaction has its own characteristic rate.
* It is either slow, fast, or reversible.
B. What affects the rate?
1. Concentration: molecules must collide in order to react.
2. Physical state: molecules must mix in order to collide.
3. Temperature: molecules must collide with enough energy to react.
4. The use of a catalyst.
II. Reaction Rates.
A. Terms.
1. A reaction rate is the change in concentration of a reactant or a
product per unit of time.
2. An average rate is the change in concentration of reactants (or
products) over a finite time period.
3. An instantaneous rate is the reaction rate at a particular time, given
by the slope of a tangent to a plot of reactant concentration vs. time.
III. Rate Laws: An Introduction.
A. Rate Law.
1. A rate law is an expression in which the reaction rate will depend only
on the concentration of the reactants.
4. An initial rate is the instantaneous rate at the point at which the
reactants are mixed, that is, at t = 0.
B. Reaction Rates.
1. General form:
2. General form:
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2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
3. A reaction order is a positive or negative exponent, for a reactant,
for which the concentration is raised to in a rate law.
* Examples:
3. Because the differential and integrated rate laws for a given
reaction are related in a well-defined way, the experimental
determination of either of the rate laws is sufficient.
4. Experimental convenience usually dictates which type of rate law is
determined experimentally.
5. Knowing the rate law for a reaction is important mainly because we
can usually infer the individual steps involved in the reaction from
the specific form of the rate law.
IV. Determining the Form of the Rate Law.
A. Introduction.
1. The first step in understanding how a given chemical reaction occurs
is to determine the form of the rate law.
2. We must determine experimentally the power to which each reactant
concentration must be raised in the rate law.
a. An exponent of “1” is referred to as first order.
b. An exponent of “2” is referred to as second order.
c. I think you get the point!
B. Method of Initial Rates.
1. Again, the initial rate is the instantaneous rate at t = 0.
B. Types of Rate Laws.
1. There are two types of rate laws.
a. The differential rate law (often called simply the rate law)
shows how the rate of reaction depends on concentration.
b. The integrated rate law shows how the concentrations of
species in the reaction depend on time.
2. The overall reaction order is the sum of the orders for the various
reactants.
3. Example: For the reaction below, determine the experimental rate
law.
NO2(g) + CO(g) ----> NO(g) + CO2(g)
2. Because we typically consider reactions only under conditions where
the reverse reaction is unimportant, our rate laws will involve only
concentrations of reactants.
* The value of the exponents in a rate law can only be determined
from experimentation; it cannot be written from the balanced
equation.
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2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
4. Determining a Rate Law.
* Determine the rate law for
V. The Integrated Rate Law.
A. First-Order Rate Laws.
1. A first-order reaction means that as the concentration doubles, the
rate also doubles, etc.
H2 + I2 ----> 2HI
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3. Equation:
4. Important things to note from the above equation.
a. The equation shows how the concentration of A depends on time.
if the initial concentration of A and the rate constant k are
known, the concentration of A at any time can be calculated.
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2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
b. This equation is in the form of y = mx + b, where a plot of y vs.
x is a straight line with slope m and intercept b. In the
equation,
y = ln[A]
x = t
m = -k
6. First-Order Rate Laws II.
* Using the data in the previous example, calculate [N2O5] at 150 s
after the start of the reaction.
b = ln[A]o
Thus, for a first-order reaction, plotting the natural logarithm
of concentration vs. time always gives a straight line. This fact
is often used to test whether a reaction is first order or not.
For the reaction
aA -----> products
the reaction is first order in A if a plot of ln[A] vs. time is a
straight line. Conversely, if this plot is not a straight line, the
reaction is not first order in A.
B. Half-Life of a First-Order Reaction.
1. The time required for a reactant to reach half its original
concentration is called the half-life of a reactant.
c. This integrated rate law for a first-order reaction also can be
expressed in terms of a ratio of [A] and [A]o as follows:
2. Equation:
5. First-Order Rate Laws I.
* The decomposition of N2O5 in the gas phase was studied at
constant temperature.
2N2O5(g) -----> 4NO2(g) + O2(g)
[N2O5] (mol/L)
0.1000
0.0707
0.0500
0.0250
0.0125
0.00625
3. For a first-order reaction the half-life does not depend on
concentration.
Time (s)
0.00
50.00
100.00
200.00
300.00
400.00
4. A certain first-order reaction has a half-life of 20.0 minutes.
a. Calculate the rate constant for this reaction.
b. How much time is required for this reaction to be 75%
complete?
Using these data, verify that the rate law is first order in
[N2O5], and calculate the value of the rate constant, where the
rate = -[N2O5]/t.
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2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
C. Second-Order Rate Laws.
1. If the concentration doubles, the rate quadruples.
a. Is this reaction first order or second order?
b. What is the value of the rate constant for the reaction?
c. What is the half-life for the reaction under these conditions?
2. Equation:
3. Characteristics of the equation.
a. A plot of 1/[A] vs. t will produce a straight line with a slope equal
to k.
b. [A] depends on time and can be used to calculate [A] at any time
t, provided k and [A]o are known.
4. Half-Life.
* Equation:
5. Determining Rate Laws.
* Butadiene reacts to form its dimer (when two identical molecules
combine, the resulting molecule is called a dimer) according to
the equation
2C4H6(g) -----> C8H12(g)
The following data were collected for this reaction at a given
temperature.
D. Zero-Order Rate Laws.
1. Regardless of what the concentration does, the rate is constant (does
not change).
[C4 H6 ] (mol/L) Time (+/- 1 s )
0.01000
0
0.00625
1000
0.00476
1800
0.00370
2800
0.00313
3600
0.00270
4400
0.00241
5200
0.00208
6200
2. Equation:
3. Half-life.

Equation:
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2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
E. A Summary.
1. Rate Law.
a. Zero-order:
5. Relationship of rate constant to the slope of straight line.
a. Zero-order:
b. First-order:
b. First-order:
c. Second-order:
c. Second-order:
2. Integrated Rate Law.
a. Zero-order:
6. Half-Life.
a. Zero-order:
b. First-order:
b. First-order:
c. Second-order:
c. Second-order:
3. Slope.
a. Zero-order:
VI. Reaction Mechanisms.
A. A Reaction Mechanism.
1. A reaction mechanism is a series of simpler reactions that sum to the
overall reaction.
b. First-order:
c. Second-order:
2. Mechanisms are influenced by elementary steps, molecularity, and
rate.
B. Elementary Reactions and Molecularity.
1. Elementary reactions (steps) are simple reactions that describe a
single molecular event in a proposed reaction mechanism.
4. Plot needed to give a straight line.
a. Zero-order:
2. Elementary steps are characterized by molecularity, the number of
reactant particles involved in a proposed reaction mechanism.
a. A unimolecular reaction is an elementary reaction that involves
the decomposition or rearrangement of a single particle.
* Example:
b. First-order:
c. Second-order:
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2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
b. A bimolecular reaction is an elementary reaction involving two
reactant species.
* Examples:
c. Some termolecular reactions exist, but are rare.
* Example:
3. The rate law of an overall reaction is never determined from the
balanced equation, but through experimentation.
4. The rate law for an elementary step, however, can be deduced from
the reaction stoichiometry. The reaction order equals the
molecularity.
C. The Rate Limiting Step of a Reaction Mechanism.
1. A rate-limiting (rate-determining) step is the slowest step in a
reaction mechanism and therefore is the step that limits the overall
rate.
5. Examples:
2. A reaction intermediate is a substance that is formed and used up
during the overall reaction and therefore does not appear in the
overall reaction.
a. They are usually unstable relative to the reactants and products.
b. They are molecules with normal bonds and are sometimes stable
enough to be isolated.
D. Constructing the Mechanism from the Rate Law.
1. We can never prove, from data only, that a particular mechanism is
the way the chemical change actually occurs.
2. Regardless of the elementary steps that are proposed for the
mechanism, they obey three criteria.
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2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
a. The elementary steps must add up to the overall equation.
(1) I2(g) <=====> 2I(g)
[fast, reversible]
b. The elementary steps must be physically reasonable (uni- or
bimolecular).
(2) H2(g) + I(g) <=====> H2I(g)
[fast, reversible]
(3) H2I(g) + I(g) -----> 2HI(g)
[slow; rate limiting]
c. They mechanism must be consistent with the rate law.
Show that the mechanism is consistent with the rate law.
3. Mechanisms with a slow initial step.
a. The overall rate law includes only species up to and including
those in the rate-determining step.
b. Each step in the mechanism has its own transition state.
4. Mechanisms with a fast initial step.
a. Some mechanisms have a fast equilibrium step first that must be
consumed by the slow step.
b. In writing the rate law for the rate-determining step, keep in
mind that an overall rate law must include only those substances
in the overall equation.
* Write rate laws for both directions of the fast equilibrium
step and for the slow step.
* Show that the slow step’s rate law is equivalent to the overall
rate law by expressing the intermediate concentration in
terms reactant concentration; set the forward rate law of
the fast, reversible step equal to the reverse rate law, and
solve for the concentration of the intermediate.
* Substitute the concentration of the intermediate into the
rate law for the slow step to obtain the overall rate law.
E. Problems.
1. The kinetics of the gas-phase reaction between H2 and I2,
H2(g) + I2(g) -----> 2HI(g); rate = k [H2][I2]
has been studied extensively. The accepted mechanism is
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2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
2. The kinetics of the formation of carbon tetrachloride from
chloroform,
VII. The Effect of Temperature on the Rate Constant.
* The Arrhenius Equation.
1. The equation:
1/2
CHCl3(g) + Cl2(g) -----> CCl4(g) + HCl(g); rate = k [CHCl3][Cl2]
has been studied extensively. The accepted mechanism is
(1) Cl2(g) <=====> 2Cl(g)
[fast, reversible]
2. Activation energy (Ea) is the minimum amount of energy a molecule
(2) CHCl3(g) + Cl(g) <=====> HCl(g) + CCl3(g) [rate limiting]
(3) CCl3(g) + Cl(g) -----> CCl4(g)
must have to react.
[fast]
3. R = 8.314 J/mol K; Ea is usually expressed in kJ!
Show that the mechanism is consistent with the rate law.
4. Determining Activation Energy I.
* The reaction
2N2O5(g) -----> 4NO2(g) + O2(g)
was studied at several temperatures, and the following values for k
were obtained:
k (s -1 )
2 .0 x1 0 -5
7 .3 x 1 0 -5
2 .7 x 1 0 -4
9 .1 x 1 0 -4
2 .9 x 1 0 -3
T (C )
20
30
40
50
60
Calculate the value of Ea for this reaction.
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2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
5. Determining Activation Energy II.
* The gas-phase reaction between methane and diatomic sulfur is
given by the equation
B. Collision Theory
1. Def’n: reaction rates are the result of frequent and energetic
molecular collisions.
CH4(g) + 2S2(g) ----> CS2(g) + 2H2S(g)
2. Factors.
a. Reactant concentration
* The more moles of reactants present, the increased likelihood
of a collision.
At 550 C the rate constant for this reaction is 1.1 L/mol s, and at
625 C the rate constant is 6.4 L/mol s. Using these values,
calculate Ea for this reaction.
* Rate depends on the product of the reactants.
b. Temperature
* Increasing the temperature increases the average speed of
the reactants. This in turn increases the frequency of the
collisions.
* Although average speed increases, most collisions do not result
in a reaction.
* If the energy of collision results in a reaction, this minimum
energy is called activation energy (Ea).
* Comparing T2 > T1
VIII. Model for Chemical Kinetics.
A. Energy Diagrams
1. Def’n: a method of depicting PE of a system during a reaction as a
smooth curve.
2. Components: Hrxn, Ea (fwd), Ea (rev), and transition state.
- Higher T always results in more collisions with the
minimum Ea.
- For an exothermic reaction, Ea
then Ea(rev).
10
(fwd)
is always greater
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
YIELDS THE PRODUCT MORE QUICKLY.
c. Molecular Structure
* Collisions are labeled as effective or ineffective. Effective
collision increase the rate as particles line up in such a way
that a reaction occurs upon collision.
2. A catalyst LOWERS THE ACTIVATION ENERGY by providing a
different mechanism that requires less energy.
3. Diagram:
C. Transition State Theory
1. Def’n: describes the role of energy in reaction rates.
2. Moving particles have mostly KE with little PE.
3. When particles collide KE decreases and PE increases as IMA
increase.
4. If the PE meets or exceeds Ea, then a reaction occurs as bonds are
broken and made.
5. Transition State Theory claims that the reaction is not so “cut and
dry.” It states that species form that are neither a product nor
reactant, but something in between.
6. This unstable, momentary species is called the activated complex or
transition state.
7. Therefore, Ea is used to stretch and deform bonds in order to reach
a transition state.
8. Diagram:
XI. Catalysts
* Def’n: a substance that increases the reaction rate without being
consumed in the reaction.
1. A catalyst speeds up the forward and reverse reactions. A reaction
with a catalyst DOES NOT YIELD MORE PRODUCT, but rather
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2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
I. Equilibrium Conditions
A. Systems at equilibrium are still dynamic (changing). However, no NET
CHANGE is observed.
Forms of K (cont.)
1. Keq = equilibrium constant
2. Kc = equilibrium constant in terms of concentration.
3. Kp = equilibrium concentration in terms of pressure.
B. A system is said to be at equilibrium when the rate of the forward
reaction is equal to the rate of the reverse reaction.
* RATE DEPENDENT, CONCENTRATION INDEPENDENT.
4. Ka = acid dissociation constant
5. Kb = base dissociation constant
6. Kw = ion-product constant for water
C. When the ratio of the products to reactants is a constant value, a
system is at equilibrium. This is the law of mass action.
* CONCENTRATION DEPENDENT, RATE INDEPENDENT.
7. Ksp = solubility product constant
8. Kf = formation constant
E. Calculating K.
1. The following equilibrium concentrations were observed for the
Haber process at 127 C.
II. The Equilibrium Constant, K.
A. K is calculated form EQUILIBRIUM CONCENTRATIONS! Its values
may only be calculated experimentally.
-2
B. According to the law of mass action, for reaction
a. Forward Reaction.
N2(g) + 3H2(g) <=====> 2NH3(g)
jA + kB <=====> lC + mD
equilibrium constant = equilibrium expression
l
-1
m
K = [C] [D]
j
k
[A] [B]
C. Conditions of K.
1. Large K. The product concentration is very large with virtually
no reactant concentration (the reaction has gone to completion).
K = 2.2 x 10
b. Reverse Reaction.
2NH3(g) <=====> N2(g) + 3H2(g)
22
2. Small K. The reactant concentration is very large with virtually
no product concentration.
K = 1.0 x 10
-3
[NH3] = 3.1 x 10 M, [N2] = 8.5 x 10 M, [H2] = 3.1 x10 M
-30
c. Multiply by Factor n.
1/2N2(g) + 3/2H2(g) <=====> NH3(g)
3. Intermediate K. Significant concentration of all substances are
present.
K = 0.211
D. Forms of K.
* Equilibrium can be established for many categories of reaction
systems. A subscript after the “K” signifies the type of system.
* K is written without units!
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2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
2. Consider the following:
(1) Br2(g) <=====> 2Br(g)
4. Kp: Equilibrium of Gases.
(2) Br(g) + H2(g) <=====> HBr(g) + H(g)
PV = nRT
or P = (n/V)RT
or
P = CRT
where C represents the molar concentration of a gas.
(3) H(g) + Br2(g) <=====> HBr(g) + Br(g)
(4) 2Br(g) <=====> Br2(g)
* The reaction for the formation of nitrosyl chloride
2NO(g) + Cl2(g) <=====> 2NOCl(g)
was studied at 25 C. The pressures at equilibrium were found
to be
PNOCl = 1.2 atm
-2
PNO = 5.0 x10 atm
-1
PCl2 = 3.0 x10 atm
Calculate the value of Kp for the reaction at 25 C.
3. Summary
a. For forward reaction
5. Relating Kc and Kp.
jA + kB <=====> lC + mD,
l
m
K = [C] [D]
j
k
[A] [B]
Kp = Kc(RT)
jA + kB <=====> lC + mD,
j
k
K’ = 1 = [A] [B]
l
m
K [C] [D]
* Using the value of Kp obtained in the previous example,
calculate the value of Kc at 25 C for the reaction
2NO(g) + Cl2(g) <=====> 2NOCl(g)
njA + nkB <=====> nlC + nmD
n
nl
nm
K’’ = K = [C] [D]
nj
nk
[A] [B]
d. For an overall reaction of two or more steps,
Koverall = K1 x K2 x K3 x
gas
ngas = moles of gaseous product - moles of gaseous reactant
b. For reverse reaction
c. For reaction
n
........
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2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
6. Homogeneous Equilibria vs. Heterogeneous Equilibria
a. Homogeneous Equilibria involves all substances in the same
phase of matter.
2. Conditions.
a. Q = K. The system is at equilibrium. No shifting is required.
b. Q > K. The numerator is large in comparison to the denominator.
In other words, product conditions > reactant conditions.
The reaction shifts to the left to establish equilibrium.
b. Heterogeneous Equilibria involves substances in different
phases of matter.
* If pure solids or pure liquids are involved in the reaction,
their concentrations are not included in the equilibrium
expressions because their concentrations do not vary.
c. Q < K. The denominator is large in comparison to the numerator.
In other words, reactant conditions > product conditions.
The reaction shifts to the right to establish equilibrium.
* Only consider AQUEOUS and GASEOUS PHASES!
3. For the synthesis of ammonia at 500 C, the equilibrium constant is
-2
c. Write expressions for Kc and Kp for the following reactions:
6.0 x 10 . Predict the direction in which the system will shift to
reach equilibrium in each of the following cases:
* The decomposition of solid phosphorus pentachloride to liquid
phosphorus trichloride and chlorine gas.
* Deep blue solid copper(II) sulfate pentahydrate is heated to
drive off water vapor to form white copper(II) sulfate.
III. The Reaction Quotient, Q.
A. Q is obtained by applying the law of mass action to INITIAL
CONCENTRATIONS!
B. Q is useful in determining which direction a reaction must shift to
establish equilibrium.
C. K vs. Q.
1. Recall:
a. K is calculated from equilibrium concentrations or pressures.
b. Q is calculated from initial concentrations or pressures.
14
Concentration (M)
[NH3]o
[N2]o
Experiment 1
1.0 x 10
Experiment 2
2.0 x 10
Experiment 3
1.0 x 10
-3
-4
-4
1.0 x 10
-5
1.5 x 10
5.0
-5
[H2]o
2.0 x10
-3
3.54 x10
-1
-3
1.0 x10
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
IV. Equilibrium Problems.
A. Developing Strategies.
1. Calculating Equilibrium Pressure I.
* Dinitrogen tetroxide in its liquid state was used as one of the
fuels on the lunar lander for the NASA Apollo missions. in the
gas phase it decomposes to gaseous nitrogen dioxide:
3. Calculating Equilibrium Concentration I.
* Carbon monoxide react with steam to produce carbon dioxide and
hydrogen. At 700 K the equilibrium constant is 5.10. Calculate
the equilibrium concentrations of all species if 1.000 mol of each
component is mixed in a 1.000-L flask.
N2O4(g) <=====> 2NO2(g)
Consider an experiment in which gaseous N2O4 was placed in a
flask and allowed to reach equilibrium at a temperature where Kp
= 0.133. At equilibrium, the pressure of N2O4 was found to be
2.71 atm. Calculate the equilibrium pressure of NO2(g).
4. Calculating Equilibrium Concentration II.
* Assume that the reaction for the formation of gaseous hydrogen
fluoride from hydrogen and fluorine has an equilibrium constant
2
of 1.15 x 10 at a certain temperature. In a particular
experiment, 3.000 mol of each component was added to a 1.500L flask. Calculate the equilibrium concentrations of all species.
2. Calculating Equilibrium Pressure II.
* At a certain temperature a 1.00-L flask initially contained 0.298
-3
mol PCl3(g) and 8.70 x 10 mol PCl5(g). After the system had
-3
reached equilibrium, 2.00 x 10 mol Cl2(g) was found in the flask.
Gaseous PCl5 decomposes according to the following reaction
PCl5(g) <=====> PCl3(g) + Cl2(g)
Calculate the equilibrium concentrations of all species and the
value of K.
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2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
V. Solving Equilibrium Problems.
A. Procedure.
1. Write the balanced chemical equation.
C. Determining Equilibrium Concentrations from Kc or Kp.
* The equilibrium between nitric oxide, oxygen, and nitrogen is
described by the following equation: 2NO(g) <===> N2(g) + O2(g);
30
Kc = 2.3 x 10
2. Write the mass-action expression in terms of Q.
at 298 K. In the atmosphere, PO2 = 0.209 atm and
PN2 = 0.781 atm. What is the equilibrium partial pressure of NO?
3. Convert all amounts into the correct units.
4. When reaction direction is not given, compare Q with K.
5. Construct a reaction table.
6. Substitute the amounts into Q.
7. To simply the math, assume that x is negligible.*
8. Solve for x.**
9. Solve for the equilibrium amounts.
* The assumption is that [A]init - x = [A]init. If the assumption is
justified, then x/[A]init < 5% error.
** You may need the quadratic equation,
x=
D. Calculating Equilibrium Pressures.
* Assume that gaseous hydrogen iodide is synthesized from hydrogen
gas and iodine vapor at a temperature where the equilibrium
2
-b + b -4ac
-----------------
2
-1
constant is 1.00 x10 . Suppose HI at 5.000 x10 atm, H2 at
2a
-2
-3
1.000x10 atm, and I2 at 5.000 x10 atm are mixed in a 5.000-L
B. Calculating Kp from Pressure Data.
flask. Calculate the equilibrium pressures of all species.
* The reaction between 1.000 atm NO and 1.000 atm O2 to produce
NO2 at 184 C was studied. At equilibrium, PO2 = 0.506 atm.
Calculate Kp.
16
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
E. Determining Equilibrium Concentrations from Initial Concentrations
and Kc.
G. Calculating Equilibrium Concentrations Through the Use of Simplifying
Assumptions.
* In a study of halogen bond strengths, 0.50 mol I2 was heated in a
* The decomposition of HI at low temperature was studied by injecting
2.50 mol HI into a 10.32-L vessel at 25 C. What is [H2] at equilibrium
2.5-L vessel, and the following reaction occurred: I2(g) <===> 2I(g).
-3
for the reaction 2HI(g) <===> H2(g) + I2(s)? Kc = 1.26 x10 .
-10
a) Calculate [I2]eq and [I]eq at 600 K; Kc = 2.94 x 10 .
b) Calculate [I2]eq and [I]eq at 2000 K; Kc = 0.209.
F. Predicting Reaction Direction and Calculating Equilibrium
Concentrations.
* An inorganic chemist studying the reactions of phosphorus halides
mixes 0.1050 mol PCl5 with 0.0450 mol Cl2 and 0.0450 mol PCl3 in a
0.5000-L flask at 250 C:
PCl5(g) <=====> PCl3(g) + Cl2(g)
Kc = 4.2 x10
-2
a) In which direction will the reaction proceed?
b) If [PCl5] = 0.2065 M at equilibrium, what are the equilibrium
concentrations of the other components?
17
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
VI. Le Chatelier’s Principle.
A. Definition: Whenever a system at equilibrium is disturbed, it acts in
such a way as to relieve the stress.
c) CaC2O4(s) <=====> CaCO3(s) + CO(g)
B. The Effect of a Change in Concentration.
* In a study of the chemistry of glass etching, an inorganic chemist
examines the reaction between sand (SiO2) and hydrogen fluoride
D. The Effect of a Change in Temperature.
1. A rise in T will increase Kc for a system with a positive Hrxn.
at high temperature:
SiO2(s) + 4HF(g) <=====> SiF4(g) + 2H2O(g)
2. A rise in T will decrease Kc for a system with a negative Hrxn.
Predict the effect on [SiF4] when a) H2O(g) is removed; b) some liquid
3. How would a decrease in temperature affect the partial pressure of
the underlined substance and Kp for the following reactions?
water is added at 220 C; c) some HF is removed; d) some sand is
removed.
a) C(graphite) + 2H2(g) <=====> CH4(g)
b) N2(g) + O2(g) <=====> 2NO(g)
H = 181 kJ
c) P4(s) + 10Cl2(g) <=====> 4PCl5(g)
C. The Effect of a Change in Pressure (Volume).
1. A decrease in volume (increase in pressure) decreases n.
2. An increase in volume (decrease in pressure) increases n.
3. How would you change the pressure (via a change in volume) of the
following reaction mixtures to decrease the yield of products?
a) 2SO2(g) + O2(g) <=====> 2SO3(g)
b) 4NH3(g) + 5O2(g) <=====> 4NO(g) + 6H2O(g)
18
H = -75 kJ
H = -1528 kJ
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
I. The Nature of Acids and Bases.
A. Water is the product of all neutralization reactions between a strong
acid and a strong base.
B. The Arrhenius Definitions.
a) You smell ammonia when NH3 dissolves in water.
+
b) The odor goes away when you add an excess of HCl to the
solution in (a).
1. An acid contains H and forms H3O (hydronium).
-
2. A base contains OH and forms OH (hydroxide).
C. Proton Transfer: The Bronsted-Lowery Definitions.
+
1. An acid is a proton (H ) donor.
c) The odor returns when you add an excess of NaOH to the
solution in (b).
+
2. A base is a proton (H ) acceptor.
D. Conjugate Acid-Base Pairs.
1. Any species formed after a proton transfer is labeled as
“conjugate” to the original acid or base.
2. The direction of the reaction always proceeds from the strong acid
and strong base (always on the same side) toward the weak acid and
weak base (always on the same side).
II. Acid Strengths.
A. Strong acids dissociate completely (100%) in water.
3. Identifying Conjugate Acid-Base Pairs.
* Identify the conjugate acid-base pairs in the following reactions:
-
+
-
HA(aq) + H2O(l) ------> H3O (aq) + A (aq)
+
a) CH3COOH(aq) + H2O(l) <====> CH3COO (aq) + H3O (aq)
Kc = [H3O+] [ A-]
[HA]
At equilibrium, Qc = Kc >> 1 because [HA] is approx. 0.
-
B. Weak acids are most undissociated.
-
b) H2O(l) + F (aq) <====> OH (aq) + HF(aq)
+
-
HA(aq) + H2O(l) <====> H3O (aq) + A (aq)
Kc = [H3O+] [ A-]
[HA]
+
4. Predicting the Direction of a Neutralization Reaction.
* Explain with balanced equations, in which you show the net
direction of the reaction, each of the following observations:
-
At equilibrium, Qc = Kc << 1 because [H3O ] and [ A ] are
approx. 0.
19
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
C.
K is the acid dissociation constant.
a
1. Ka applies to WEAK ACIDS ONLY!!
+
d) The annilium ion (C6H5NH3 )
2. The smaller the Ka value, the weaker the acid.
3+
e) The hydrated aluminum(III) ion [Al(H2O)6]
D. Classifying Strengths.
1. Strong Acids.
a. Hydrohalic acids HCl, HBr, and HI.
6. Classifying Acid and Base Strength from the Chemical Formula
* Which of the following is the stronger acid or base:
a) HClO or HClO3
b. Oxyacids in which O > H by two or more
2. Weak Acids.
a. Hydrohalic acid HF.
b) HCl or CH3COOH
b. H not bonded to an O.
c) NaOH or CH3NH2
c. Oxyacids in which O > or = H by 1.
* Using a Ka table from your book, arrange the following species
d. Organic acids.
-
-
-
in order of increasing base strength: H2O, F , Cl , and CN .
3. Strong Bases.
a. M2O or MOH where M = Group 1 metals.
b. MO or M(OH)2 where M = Group 2 metals (beginning at Ca).
E. Amphoteric Substances.
1. Amphoteric substances can behave as either an acid or as a base.
* Water is the most useful amphoteric substance in acid-base
chemistry.
4. Weak Bases.
a. Ammonia.
b. Organic Amines.
2. The Autoionization of Water and the Ion-Product (Dissociation)
Constant for Water,
5. Acid Dissociation (Ionization ) Reactions.
* Write the simple dissociation (ionization) reaction
(omitting water) for each of the following acids.
a) Hydrochloric acid
K .
w
+
-
H2O(l) + H2O(l) <====> H3O (aq) + OH (aq)
+
-
Kc = Kw = [H3O ] [OH ] = 1.0 x 10
b) Acetic acid
+
-
-7
[H3O ] = [OH ] = 1.0 x 10 M
+
-
a) [H3O ] > [OH ] = Acidic.
c) The ammonium ion
20
-14
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
+
-
+
-
III. The pH Scale.
b) [H3O ] < [OH ] = Basic.
+
A. pH = -log[H3O ]
where neutral = 7.00
acidic = < 7.00
basic = 7.00 <
c) [H3O ] = [OH ] = Neutral.
+
-
B. pOH = -log[OH ]
-
3. Calculating [H3O ] and [OH ].
+
-
* Calculate [H3O ] or [OH ] as required for each of the following
C. pH + pOH = 14
solutions at 25 C, and state whether the solution is neutral,
acidic, or basic.
-5
D. pK = -log K
-
a) 1.0 x10 M OH
E. pH Problems.
1. Calculate pH and pOH for each of the following solutions at 25 C.
-
a) 1.0 x 10-3 M OH
-7
-
b) 1.0 x10 M OH
+
c) 10.0 M H3O
+
b) 1.0 M H3O
4. Autoionization of Water.
-13
* At 60 C, the value of Kw is 1.0 x10 .
2. The pH of a sample of human blood was measured to be 7.41 at 25 C.
a) Using LeChatelier’s principle, predict whether the reaction
+
+
H2O(l) + H2O(l) <====> H3O (aq) + OH (aq)
is exothermic or endothermic.
+
-
Calculate pOH, [H3O ], and [OH ] for the sample.
-
-
b) Calculate [H3O ] and [OH ] in a neutral solution at 60 C.
21
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
VI. Calculating the pH of a Strong Acid Solution.
A. Focusing on the SOLUTION components.
* Consider a 1.0 M solution of HCl. It is a strong acid. Therefore, it
dissociates nearly 100% as follows
+
V. Calculating the pH of Weak Acid Solutions and Weak Acid Equilibria.
A. Solving Weak Acid Equilibrium Problems.
1. List all the major species in the solution.
+
2. Choose the species that can produce H3O , and write balanced
-
HCl(aq, 1M) ------> H3O (aq) + Cl (aq)
+
equations for the reactions producing H3O .
+
-
We can conclude that the solution contains only H3O and Cl ions
3. Using the values for the equilibrium constants for the reactions you
have written, decide which equilibrium will dominate in producing
rather than HCl molecules.
+
B. Focusing on the MAJOR SPECIES, those solution components present
in relatively high amounts.
H3O .
4. Write the equilibrium expression for the dominant equilibrium.
+
* Consider the same 1.0 M solution of HCl. The major species are H3O ,
-
-
5. List the initial concentrations of the species participating in the
dominant equilibrium.
Cl , and H2O. OH is considered a minor species.
+
-
HCl(aq, 1M) ------> H3O (aq) + Cl (aq)
and
+
6. Write the equilibrium concentrations in terms of x.
-
7. Substitute the equilibrium concentrations into the equilibrium
expression.
H2O(l) + H2O(l) <====> H3O (aq) + OH (aq)
C. Calculating pH of Strong Acids.
1. Calculate the pH of 0.10 M HNO3
8. Solve for x the “easy” way; that is, by assuming that [HA]0 - x =
[HA]0.
9. Use the 5% rule to verify whether the approximation is valid.
+
10. Calculate [H3O ] and pH.
B. Two Key Assumptions.
+
1. The [H3O ] from the autoionization of water is so much smaller than
C. Calculating pH of Strong Acids (cont.).
2. Calculate the pH of 1.0 x 10
-10
+
[H3O ] from the HA dissociation that we can neglect it:
M HCl.
+
Therefore,
+
+
[H3O ] = [H3O ]from HA + [H3O ]from water
+
+
[H3O ] = [H3O ]from HA
22
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
+
2. A weak acid has a small Ka; therefore, it dissociates to such a small
2. The conjugate acid of ammonia, NH4 , is a weak acid. If a 0.2 M
extent that we can neglect the change in its concentration to find its
equilibrium concentrations:
+
NH4Cl solution has a pH of 5.0, what is the Ka of NH4 ?
[HA] = [HA]init - [HA]dissoc = [HA]init
C. Problems.
-
1. The hypochlorite ion (ClO ) is a strong oxidizing agent often found
in household bleaches and disinfectants. It is also the active
ingredient that forms when swimming pool water is treated with
chlorine. In addition to its oxidizing abilities, the hypochlorite ion
has a relatively high affinity for protons (it is a much stronger base
-
than Cl , for example) and forms the weakly acidic hypochlorous acid
-8
(HOCl, Ka = 3.5 x 10 ). Calculate the pH of a 0.100 M aqueous
solution of hypochlorous acid.
3. Cyanic acid (HOCN), an extremely acrid, unstable substance, is used
+
industrially to make cyanates. What is the [H3O ] and pH of 0.10 M
-4
HOCN? Ka of cyanic acid is 3.5 x10 .
23
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
D. The pH a Mixture of Weak Acids.
* Calculate the pH of a solution that contains 1.00 M HCN (Ka =
-10
E. Percent Dissociation.
1. Percent Dissociation =
-4
6.2 x10 ) and 5.00 M HNO2 (Ka = 4.0 x 10 ). Also calculate the
-
concentration of cyanide ion (CN ) in this solution at equilibrium.
amount dissociated (mol/L) x 100
initial concentration (mol/L)
-5
2. Calculate the percent dissociation of acetic acid (Ka = 1.8 x 10 ) in
each of the following solutions.
a) 1.00 M HC2H3O2
b) 0.100 M HC2H3O2
24
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
+
VI. Weak Bases and Their Relation to Weak Acids.
3. Conclusion: For solutions of any weak acid HA, [H3O ] decreases as
+
-
B: (aq) + H2O(l) <=====> BH + OH (aq)
[HA]0 decreases, but the percent dissociation increases as [HA]0
decreases.
The Base Dissociation Constant (K )
b
4. Explanation:
Kb = [BH+] [OH-]
[B:]
A. The pH of Strong Bases.
-2
* Calculate the pH of a 5.00 x 10 M NaOH solution.
5. Lactic acid (HC3H5O3) is a waste product that accumulates in muscle
tissue during exertion, leading to pain and a feeling of fatigue. In a
0.100 M aqueous solution, lactic acid is 3.7% dissociated. Calculate
the Ka for this acid.
B. The pH of Weak Bases I.
-5
* Calculate the pH for a 15.0 M solution of NH3 (Kb = 1.8 x 10 ).
25
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
C. The pH of Weak Bases II.
VII. Polyprotic Acids.
* Characteristics.
-4
* Calculate the pH of a 1.0 M solution of methylamine (Kb = 4.38 x 10 ).
+
1. A polyprotic acid has more than one ionizable proton (H ).
2. Each successive Ka value get smaller (Ka1 > Ka2 > Ka3). Therefore,
the first dissociation step makes a significant contribution to the
+
equilibrium concentration of [H3O ]. This means that a calculation of
the pH for a solution of a typical weak polyprotic acid is identical to
that for a weak monoprotic acid.
3. Sulfuric acid is unique in being a strong acid in its first dissociation
step and a weak acid in its second step. For relatively concentrated
solutions of sulfuric acid (1.0 M or higher), the large concentration of
+
H3O from the first dissociation step represses the second step,
+
which can be neglected as a contributor of H3O ions. For dilute
solutions of sulfuric acid, the second step does make a significant
contribution, and the quadratic equation must be used to obtain the
+
total H3O concentration.
4. Problems.
* Calculate the pH of a 5.0 M H3PO4 solution and the equilibrium
D. Pyridine (C5H5N), an important solvent and base in organic syntheses,
has a pKb of 8.77. What is the pH of 0.10 M pyridine?
-
2-
concentrations of the species H3PO4, H2PO4 , HPO4 , and
3-
PO4 .
26
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
-2
* Calculate the pH of a 1.00 x 10 M H2SO4 solution.
* Calculate the pH of a 1.0 M H2SO4.
27
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
VIII. Acid-Base Properties of Salt Solutions.
A. The salt of a strong acid and strong base yields a neutral solution
because the ions do not react with water.
B. The salt of a strong acid and weak base yields an acidic solution because
the cation acts as a weak acid.
+
+
-
NH4Cl(s) -------> NH4 (aq) + Cl (aq)
D. The Acidity of Hydrated Metal Ions.
n+
+
NH4 (aq) + H2O(l) -------> NH3(aq) + H3O (aq)
1. For M , a small, highly charged ion,
ACIDIC!
* Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3 is
n+
(n-1)
M(H2O)x (aq) + H2O (l) <=====> M(H2O)x-1OH
-5
1.8 x 10 .
+
(aq) + H3O (aq)
2. Calculate the pH of a 0.010 M AlCl3 solution. The Ka value for
3+
-5
Al(H2O)6 is 1.4 x 10 .
C. The salt of a strong base and weak acid yields a basic solution because
the anion acts as a weak base.
+
-
CH3COONa(s) ------> Na (aq) + CH3COO (aq)
-
-
CH3COO (aq) + H2O(l) -------> CH3COOH(aq) + OH (aq)
BASIC!
* Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is
-4
7.2 x 10 .
E. For salts from a weak acid and weak base, the Ka and Kb of the ions
must be compared.
28
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
F. Problems,
* Predict whether an aqueous solutions of each of the following salts
will
be acidic, basic, or neutral.
a) NH4C2H3O2
IX. Molecular Properties and Acid Strength.
A. Trends in Binary Nonmetal Hydride Activity.
1. Consider Groups 16 and 17 only.
2. From left to right, electronegativity increases, polarity increases.
+
H removal is easier, and acidity increases.
3. From top to bottom, bond length increases, bond strength decreases,
+
b) NH4CN
H removal is easier, and acidity increases.
B. Trends in Oxyacid Acidity.
1. With the same number of oxygens around the central nonmetal (E),
acid strength increases with the electronegativity of E.
HClO > HBrO > HIO
c) Al2(SO4)3
2. With different numbers of oxygens around E, acid strength
increases with the number of oxygen atoms.
d) CH3NH3NO3
+
* The oxygens pull the electron density away from E, and H is
more easily removed.
X. The Leveling Effect.
A. Definition: The inability of a solvent to distinguish the strengths of any
acid (or base) that is stronger than the conjugate acid (or
base) of the solvent.
e) KClO2
f) CsI
B. Water treats all strong acids (or bases) the same as if they are all
equal in strength (HCl, HBr, and HI).
C. By dissolving an acid in a weaker base than water, such a acetic acid,
different levels of “protonation” occurs and strengths are determined.
g) NH4F
XI. Lewis Acids and Bases.
A. This is the broadest definition.
h) Cu(CH3COO)2
B. A base is an electron pair acceptor and an acid is an electron pair
donor.
C. This includes: electron deficient compounds, polar double bonded
+
compounds, metal ions, and protons (H ).
29
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
I. Acid-Base Equilibria.
A. Solutions of Acids or Bases Containing a Common Ion.
1. A common ion is one that is both dissolved in an aqueous solution and
introduced to the existing equilibrium system.
2. The components of a buffer are a conjugate acid-base pair.
3. Approaching Buffer Problems.
a. Assume that the reaction goes to completion, and carry out the
stoichiometric calculations (based on MOLES).
2. The shift in equilibrium as a result of the introduction of a common
ion is called the common ion effect.
* This is an application of Le Chatelier’s Principle.
3. Acidic Solutions Containing Common Ions.
b. Carry out the equilibrium calculations.
+
-
4. Calculating the Effect of Added H3O and OH on a Buffer pH.
-
* Calculate the pH of a buffer of 0.50 M HF and 0.45 M F (a)
before and (b) the addition of 0.40 g NaOH to 1.0 L of the
+
* The equilibrium concentration of H3O in a 1.0 M HF solution is
-2
-4
2.7 x 10 M, and the percent dissociation of HF is 2.7%.
buffer. Ka of HF = 6.8 x 10 .
+
Calculate [H3O ] and the percent dissociation of HF in a solution
-4
containing 1.0 M HF (Ka = 7.2 x 10 ) and 1.0 M NaF.
II. Buffered Solutions.
A. A buffered solution is one that resists a change in its pH when either
+
hydroxide ions or protons (H3O ) are added.
1. Very little change in pH is witnessed even when a strong acid or base
is added.
30
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
5. Special Points.
a. Buffered solutions are simply solutions of weak acids or bases
containing a common ion. The pH calculations on buffered
solutions require exactly the same procedures introduced last
chapter.
OR
-
pH = pKa + log ([A ]/[HA]) = pKa + log ([BASE]/[ACID])
* For a particular buffering system (conjugate acid-base
b. When a strong acid or base is added to a buffered solution, it is
best to deal with the stoichiometric calculation first. Then,
consider the equilibrium calculation.
6. Buffering: How Does It Work?
-
pair), all solutions that have the same ratio [A ]/[HA]
will have the same pH.
7. The pH of a Buffered Solution I.
* Calculate the pH of a solution containing 0.75 M lactic acid (Ka =
+
a. The equilibrium concentration of H3O , and thus the pH, is
-4
1.4x 10 ) and 0.25 M sodium lactate. Lactic acid (HC3H5O3) is
-
determined by the ratio of [HA]/[A ].
-
a common constituent of biological systems. For example, found
in milk and is present in human muscle tissue during exertion.
+
HA + H2O <====> A + H3O
+
-
+
Ka = [H3O ] [A ]
therefore, [H3O ] = Ka [HA]
-
[HA]
[A ]
-
b. If OH- is added to the system, HA is converted to A , and the
-
ratio of [HA]/[A ] decreases.
-
-
HA + OH <====> A + H2O
c. However, if the amounts of HA and A- originally present are
-
very large compared with the amount of OH added, the change
-
in the [HA]/[A ] ratio will be very small.
+
d. Similar reasoning applies when adding H3O to a system.
e. The Henderson-Hasselbach Equation.
+
[H3O ] = Ka [HA]
-
[A ]
+
-
-log[H3O ] = -log Ka -log ([HA]/[A ])
-
pH = pKa - log ([HA]/[A ])
31
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
8. The pH of a Buffered Solution II.
III. Buffer Capacity.
A. Buffer capacity represent the amount of protons or hydroxide ions the
buffer can absorb without significantly changing the pH.
-5
* A buffered solution contains 0.25 M NH3 (Kb = 1.8 x 10 ) and
0.40 M NH4Cl. Calculate the pH of this solution.
-
B. The pH of a buffered solution is determined by the ratio of [A ]/[HA].
The capacity of a buffered solution is determined by the magnitudes of
-
[HA] and [A ].
C. Adding Strong Acid to a Buffered Solution II.
* Calculate the change in pH that occurs when 0.0100 mol gaseous HCl is
added to 1.0-L of each of the following substances:
Solution A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2
Solution B: 0.050 M HC2H3O2 and 0.050 M NaC2H3O2
For acetic acid, Ka = 1.8 x 10
9. Adding Strong Acid to a Buffered Solution I.
* Calculate the pH of the solution that results when 0.10 mol gaseous
HCl is added to 1.0-L of the buffered solution in the previous
example.
32
-5
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
D. The pKa of the weak acid to be used in the buffer should be as close as
possible to the pH.
1. Preparing a Buffer.
* A chemist needs a solution buffered at pH 4.30 and can choose
from the following acids and their salts:
-3
a) chloroacetic acid (Ka = 1.35 x 10 )
-5
b) propanoic acid (Ka = 1.3 x 10 )
-5
c) benzoic acid (Ka = 6.4 x 10 )
-8
d) hypochlorous acid (Ka = 3.5 x 10 )
-
Calculate the ratio [HA]/[A ] required to yield the pH 4.30.
Which system works best?
IV. Titration and pH Curves.
A. A titration curve is a plot of pH vs. volume of added titrant.
B. An alternative way of looking at molarity.
* Because titrations involve small concentrations, and mL are of often
used in titrations, we refer to the amount of a substance in millimoles,
or mmol. Therefore,
Molarity = mmol/mL
2. How would you prepare a benzoic acid/benzoate buffer with pH 4.25,
starting with 5.0-L of 0.050 M sodium benzoate (C6H5COONa)
+
-
C. An equivalence point exists when [H3O ] = [OH ].
solution and adding the acidic component? Ka of benzoic acid
D. All volumes in a titration are considered to be additive.
-5
(C6H5COOH) = 6.3 x 10 .
E. Always label the equivalence point and the point where pH = pKa.
F. Strong Acid-Strong Base Titration Curves.
1. Before the addition.
* pH is calculated directly from the initial concentration.
2. Additions before the equivalence point.
a. Construct a “stoichiometry” reaction table.
33
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
b. Determine MOLES of acid in excess (not neutralized).
+
c. Divide MOLES by the TOTAL VOLUME to obtain [H3O ].
d. Calculate the pH.
3. Additions at the equivalence point.
+
-
* The pH ALWAYS is equal to 7.00 when [H3O ] = [OH ].
4. Additions beyond the equivalence point.
a. Construct a “stoichiometry” reaction table.
b. Determine MOLES of base in excess (not neutralized).
-
c. Divide MOLES by the TOTAL VOLUME to obtain [OH ].
d. Calculate the pOH, then the pH.
5. Problem: 50.0 mL of 0.200 M HNO3 are titrated with 0.100 M NaOH.
Calculate the pH after the additions of 0.0, 10, 20, 50, 100,150, and
200 mL samples of NaOH. Then, construct a titration curve and label
it properly.
34
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
G. Weak Acid-Strong Base Titration Curves.
1. Before the addition.
a. Construct an “equilibrium” reaction table ONLY!
-
+
b. Ka = [A ] [H3O ]
[HA]
g. Calculate the pOH, then the pH.
h. The equivalence point is ALWAYS >7!
+
4. Additions beyond the equivalence point.
a. Construct a “stoichiometry” reaction table.
and obtain [H3O ].
b. Determine MOLES of base in excess (not neutralized) and the
MOLES of conjugate base.
c. Calculate the pH.
2. Additions before the equivalence point.
a. Construct a “stoichiometry” reaction table.
c. Divide MOLES by the TOTAL VOLUME,
-
+
-
-5
5. Problem: 50.0 mL of 0.10 M acetic acid (Ka = 1.8 x 10 ) are titrated
with 0.10 M NaOH. Calculate the pH after the additions of 0, 10, 25,
40, 50, 60, and 75 mL samples of NaOH. Then, construct a titration
curve and label it properly.
d. Construct an “equilibrium” reaction table.
-
+
[HA]
+
and obtain [H3O ].
f. Calculate the pH.
3. Additions at the equivalence point.
a. Construct a “stoichiometry” reaction table.
b. Determine MOLES of conjugate base formed.
-
c. Divide MOLES by the TOTAL VOLUME to obtain [A ].
d. Calculate Kb (Ka x Kb = Kw).
e. Construct an “equilibrium” reaction table, reacting the conjugate
base with water.
-
+
f. Kb = [OH ] [BH ]
-
calculate pOH, then the pH.
c. Divide MOLES by the TOTAL VOLUME to obtain [H3O ] and [A ].
e. Ka = [A ] [H3O ]
-
d. Because [OH ]excess >> [OH ]conj. base, use [OH ]excess to
b. Determine MOLES of acid in excess (not neutralized) and
MOLES of conjugate base formed.
-
and obtain [OH ].
[B:]
35
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
H. Weak Base-Strong Acid Titration Curves.
1. Before the addition.
a. Construct an “equilibrium” reaction table ONLY!
-
+
b. Kb = [OH ] [BH ]
-
and obtain [OH ].
[B:]
c. Calculate the pOH, then the pH.
2. Additions before the equivalence point.
a. Construct a “stoichiometry” reaction table.
b. Determine MOLES of base in excess (not neutralized) and
MOLES of conjugate acid formed.
+
c. Divide MOLES by the TOTAL VOLUME to obtain [BH ] and [B:].
d. Construct an “equilibrium” reaction table.
-
+
e. Kb = [OH ] [BH ]
-
and obtain [OH ].
[B:]
f. Calculate the pOH, then the pH.
3. Additions at the equivalence point.
a. Construct a “stoichiometry” reaction table.
b. Determine MOLES of conjugate acid formed.
+
c. Divide MOLES by the TOTAL VOLUME to obtain [BH ].
d. Calculate Ka (Ka x Kb = Kw).
e. Construct an “equilibrium” reaction table, reacting the conjugate
base with water.
-
+
f. Ka = [A ] [H3O ]
[HA]
36
+
and obtain [H3O ].
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
g. Calculate the pH.
4. Additions beyond the equivalence point.
a. Construct a “stoichiometry” reaction table.
b. Determine MOLES of acid in excess (not neutralized) and the
MOLES of conjugate acid.
c. Divide MOLES by the TOTAL VOLUME,
+
+
-
d. Because [H3O ]excess >> [H3O ]conj. acid, use [OH ]excess to
calculate the pH.
5. Problem: 20.0 mL of 0.10 M triethylamine, (CH3CH2)3N, (Kb = 5.2 x
-4
10 ) are titrated with 0.100 M HCl. Calculate the pH after the
additions of 0.0, 10, 15, 19, 19.95, 20, 20.05, and 25 mL samples of
HCl. Then, construct a titration curve and label it properly.
V. Acid-Base Indicators.
A. Acid-base indicators mark the end point of a titration by changing
color.
* Note: The equivalence point is defined by the reaction stoichiometry.
B. For strong acid-strong base titrations, color changes are sharp, allowing
more flexibility in choosing an indicator.
C. For weak acid titrations, less flexibility is exists in choosing an
indicator.
D. Choose possible indicators for the three previous titration examples.
37
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
VI. Solubility Equilibria and the Solubility Product.
A.
K is the solubility product constant is equilibrium between solid solute
sp
and dissolved ions.
n+ p
z- q
B. Ksp = [M ] [X ]
C. Writing Ksp for Slightly Soluble Ionic Compounds.
* Write the solubility product expression for:
a. Magnesium carbonate
F. Calculating Solubility from Ksp.
-7
* The Ksp for copper(II) iodate, Cu(IO3)2, is 1.4 x10 at 25 C.
b. Iron(II) hydroxide
Calculate its solubility at 25 C.
c. Calcium phosphate
d. Silver sulfide
D. Calculating Ksp from Solubility I.
-4
* Copper(I) bromide has a measured solubility of 2.0 x10 M at 25 C.
Calculate its Ksp value.
G. Relative Solubilities
* Relative solubilities can only be predicted by comparing Ksp values only
E. Calculating Ksp from Solubility II.
* Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a
solubility of 1.0 x 10
-15
for salts that produce the same total number of ions.
M at 25 C.
H. The Common Ion Effect.
-11
* Calculate the solubility of solid CaF2 (Ksp = 4.0 x 10 ) in a 0.025 M
NaF solution.
38
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
4. Predicting Whether a Precipitate Will Form.
* Phosphate in natural waters often precipitates as insoluble salts,
2+
3-
such as Ca3(PO4)2. In a certain river, [Ca ]init = [PO4 ]init =
-9
1.0 x 10 M. Will Ca3(PO4)2 precipitate? Ksp of Ca3(PO4)2 =
1.2 x 10
I. pH and Solubility.
-29
.
+
1. If a compound contains the anion of a weak acid, addition of H3O
from a strong acid will increase the compounds solubility.
2. Predicting the Effect on Solubility of Adding a Strong Acid.
* Write balanced equations to show how addition of HNO3 will
affect the solubility of a) calcium fluoride; b) zinc sulfide; c)
silver iodide.
VII. Precipitation and Qualitative Analysis.
A. Selective precipitation is a technique in which metal ions in aqueous
solutions are separated by using a reagent whose anion forms a
precipitate with only one or a few metal ions in the mixture.
-4
+
-3
2+
B. A solution contains 1.0 x10 M Cu and 2.0 x10 M Pb . If a source of
-
-8
I is added gradually to this solution, will PbI2 (Ksp = 1.4 x10 ) or CuI (Ksp
J. Predicting the Formation of a Precipitate: Qsp vs. Ksp.
-12
= 5.3 x10 ) precipitate first? Specify the concentration of I
necessary to begin precipitation of each salt.
1. If Qsp = . Ksp, then the solution is saturated and no change occurs.
2. If Qsp > Ksp, then a precipitate forms until the solution is saturated.
3. If Qsp < Ksp, then the solution is unsaturated and no precipitate
forms.
39
-
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
2+
2+
3. Ion group 3 (base-insoluble sulfides and hydroxides): Zn , Mn ,
2+
2+
2+
3+
3+
Ni , Fe , Co as sulfides, and Al , Cr as hydroxides.
As the solution is made basic, and more H2S is added, then basic
2-
solution produces a higher concentration of [S ], which leads to
the precipitation of the above ions. Also, because the solution is
-
basic and OH is present, some insoluble hydroxides precipitate.
2+
2+
2+
4. Ion group 4 (insoluble phosphate): Mg , Ca , Ba
Only Group I and II cations remain (except ammonium). By adding
3-
PO4 , the Group II ions are precipitated.
+
+
5. Ion group 5 (Alkali metal and ammonium ions): Na , K , NH4
+
At this point, either a flame or a litmus test must be done. Sodium
produces a characteristic yellow-orange color, and potassium gives
a violet color. Moist red litmus paper turns blue when being held
+
over a solution containing NH4 that has been made basic with
NaOH because the NH3 produced reacts with H2O on the paper:
C. Qualitative analysis involves the separation and identification of ions in
a solution based on ion group characteristics.
+
1. Ion group 1 (insoluble chlorides):
2+
Ag , Hg , Pb
NH4
2+
2+
+
3+
3+
2+
+
-
+ OH (turns litmus blue)
6. Some helpful hints.
a. Distinguishing insoluble hydroxides from group 1 ions
3+
-
3+
-
Al (aq) + 3OH (aq) <====> Al(OH)3 (s) (white)
3+
2. Ion group 2 (acid-insoluble sulfides): Cu , Cd , Hg , As , Sb ,
2+
-
+ OH (aq; added) -----> NH3 + H2O
NH3 + H2O (moist red litmus) -----> NH4
When dilute HCl is added to a mixture of common cations, the
above ions are precipitated out of solution as insoluble chlorides.
All other chlorides remain soluble and in solution.
2+
+
4+
Pb , Bi , Sn , Sn
Fe (aq) + 3OH (aq) <====> Fe(OH)3 (s) (brown)
After the insoluble chlorides are removed, the solution is still
acidic, since HCl was added. If H2S is added to solution, only the
2-
most insoluble sulfides (above) will precipitate, since [S ] is
+
relatively low because of the high concentration of H3O . All other
sulfides remain dissolved in solution.
40
2nd 9 Weeks Notes
Kinetics, Equilibrium, Acids and Bases, Aqueous Equilibria
6. Helpful hints (continued).
b. The effects of adding NH3.
Step 5: Added HCl (aq) to solution from Step 2 ===> white precipitate.
Step 6: Flame test on original mixture ===> pale blue color of ordinary flame.
+
+
2+
2+
Ag (aq) + 2NH3(aq) <====> Ag(NH3)2 (aq) (colorless)
Cu (aq) + 4NH3(aq) <====> Cu(NH3)4 (aq) (blue)
+
2+
c. Separating Ag and Cu by adding HCl.
+
+
+
+
Ag(NH3)2 (aq) + 2H3O (aq; added) <===> Ag (aq) + NH4 (aq) + H2O (l)
2+
Cu(NH3)4
+
(aq) + 4H3O (aq; added) <===> Cu
+
2+
+
(aq) + 4NH4 (aq) + 4H2O (l)
-
Ag (aq) + Cl (aq; added) <====> AgCl (s) (white)
2+
-
2-
Cu (aq) + 4Cl (aq; added) <====> CuCl4 (aq) (green)
d. Separating aluminum and iron by adding solid NaOH.
-
-
Al(OH)3 (s) + OH (aq; added) -----> Al(OH)4 (aq)
-
Fe(OH)3 (s) + OH (aq; added) -----> no reaction
e. Confirming the presence of aluminum with HCl and Na2HPO4.
-
+
+
3+
Al(OH)4 (aq) + H3O (aq) -----> Al(OH)3 (s) + H3O (aq) -----> Al (aq)
Al
3+
2-
+
(aq) + HPO4 (aq; added) + H2O (l) -----> H3O (aq) + AlPO4 (s) (white)
8. Another student in the class obtained the following test results on
a different mixture of the same possible ions:
f. Confirming the presence of iron with HCl and KSCN.
+
3+
Fe(OH)3 (s) + H3O (aq) -----> Fe (aq) + 6H2O (l)
3+
-
Step 1: Added NH3 (aq) to ion mixture ===> white precipitate in colored
solution.
Step 2: Centrifuged ===> deep-blue solution.
Step 3: Treated precipitate with NaOH (aq) ===> dissolve completely.
Acidified and obtained white precipitate with Na 2HPO4.
2+
Fe (aq) + SCN (aq; added) -----> FeSCN (aq) (red)
7. A student was given a solution containing salts of one or more of the
+
+
3+
2+
Step 4: Added HCl (aq) to solution from Step 2 ===> turned green, and white
precipitate formed.
Step 5: Centrifuged precipitate from Step 4 and washed it to remove
excess HCl. Treated it with NH3 ===> dissolved to form colorless
3+
the following cations: K , Ag , Al , Cu , and Fe . From the
following laboratory results, determine the cations present:
Step 1: Added NH3 (aq) to ion mixture ===> brown precipitate.
Step 2: Centrifuged ===> colorless solution.
Step 3: Treated precipitate with NaOH (aq) ===> some solid remained, but
some may have dissolved.
Step 4: Neutralized solution in Step 3 with NH 4Cl (aq) ===> no precipitate.
solution.
Step 6: Flame test on original mixture ===> green flame with no red visible.
41