Equilibrium Basic Concepts
• Reversible reactions do not go to completion.
– They can occur in either direction

a Ag  + b Bg   c Cg  + d Dg 
• Chemical equilibrium exists when two opposing reactions occur
simultaneously at the same rate.
– A chemical equilibrium is a reversible reaction that the forward
reaction rate is equal to the reverse reaction rate.
• Chemical equilibria are dynamic equilibria.
– Molecules are continually reacting, even though the overall
composition of the reaction mixture does not change.
– Double arrow (⇋) indicates that both the forward and reverse reactions
are occurring simultaneously and is read “is in equilibrium with”
– At equilibrium, the composition of the system no longer changes with
time
– Composition of an equilibrium mixture is independent of the direction
from which equilibrium is approached
Equilibrium Basic Concepts
• One of the fundamental ideas of chemical equilibrium is that equilibrium can be
established from either the forward or reverse direction.
• Equilibrium state is achieved when the rate of the forward reaction equals the rate
of the reverse reaction: Ratef = Rater
• Under a given set of conditions, there must be a relationship between the
composition of the system at equilibrium and the kinetics of a reaction
represented by rate constants. The rates of the forward and reverse reactions can
be represented as:
• The ratio of the rate constants yields a new constant, the equilibrium constant
(K), a unitless quantity and is defined as K = kf /kr.
•
Equilibrium constants are uniless because they actually involve a thermodynamic quantity
called activity. Activities are directly related to molarity
• This implies that equilibrium mixture is determined by the magnitudes of the rate
constants for the forward and reverse reactions.
Equilibrium Basic Concepts
Product Favored, ∆G˚ negative
Systems can reach
equilibrium when reactants
have NOT converted
completely to products.
In the case
∆Grxn is < ∆Gorxn
state with both reactants and
products present is MORE
STABLE than complete
conversion.
Developing an Equilibrium Constant
Expression
•
The ratio of the product of the equilibrium concentrations of the
products (raised to their coefficients in the balanced equation) to the
product of the equilibrium concentrations of the reactants (raised to
their coefficients in the balanced equation) is always a constant under
a given set of conditions:
[C]c [D]d
Kc =
[A]a [B]b
is called the equilibrium equation.
• This relationship is known as the law of mass action where
– Kc is the equilibrium constant for the reaction
– Right side of the equation is called the equilibrium constant
expression
• Relationship is true for any pair of opposing reactions regardless of
the mechanism of the reaction or of the number of steps in the
mechanism.
The Equilibrium Constant
• Write equilibrium constant expressions for the following reactions
at 500oC. All reactants and products are gases at 500oC.
The Equilibrium Constant
• One liter of equilibrium mixture from the following system at a high
temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086
mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate Kc for
the reaction.
• Equil []’s
0.028 M
0.172 M
0.086 M
The Equilibrium Constant
Products favored 103 >
K > 10-3 Reactants favored
– Values of K greater than 103 indicate a strong tendency for reactants to form products, so
equilibrium lies to the right, favoring the formation of products (kf >>kr)
– Values of K less than 10–3 indicate that the ratio of products to reactants at equilibrium is
very small; reactants do not tend to form products readily, and equilibrium lies to the left,
favoring the formation of reactants (kf<<kr)
– Values of K between 103 and 10–3 are not very large or small, so there is no strong
tendency to form either products or reactants; at equilibrium, there are significant
amounts of both products and reactants (kf  kr)
The Equilibrium Constant
• Equilibrium can be approached from either direction in a chemical reaction,
so the equilibrium constant expression and the magnitude of the
equilibrium constant depend on the form in which the chemical reaction is
written.
• When a reaction is written in the reverse direction,
cC + dD ⇋ aA + bB
K and the equilibrium constant expression are inverted:
K´= [A]a [B]b
[C]c [D]d
so K´ = 1/K
Variation of Kc with the
Form of the Balanced Equation
• The value of Kc depends upon how the balanced equation is
written. From the prior examples we have:
PCl5
Equil []’s 0.028 M
PCl3 + Cl2
0.172 M
0.086 M
Kc = [PCl3][Cl2]/[PCl5] = 0.53
PCl3 + Cl2
Equil. []’s 0.172 M 0.086 M
PCl5
0.028 M
Kc’ = [PCl5]/[PCl3][Cl2] = 1.89
2PCl5
Equil []’s 0.028 M
2PCl3 + 2Cl2
0.172 M
0.086 M
Kc ’’ = [PCl3] 2[Cl2] 2/[PCl5] 2 = 0.28
2PCl3 + 2Cl2
Equil. []’s 0.172 M
0.086 M
2PCl5
0.028 M
Kc ‘“ = [PCl5]2/[PCl3]2[Cl2]2 = 3.56
The Equilibrium Constant
• At a given temperature 0.80 mole of N2 and 0.90 mole of H2 were introduced
in an evacuated 1.00-liter container. After equilibrium was established the
equilibrium concentrations were determined to be 0.20 mole of NH3, 0.70
moles N2, and 0.60 moles H2. Calculate Kc for the reaction.
Partial Pressure and Kc
• For gas phase reactions the equilibrium constants can be
expressed in partial pressures rather than concentrations.
• For gases, the pressure is proportional to the concentration.
• We can see this by looking at the ideal gas law.
–
–
–
–
PV = nRT
P = nRT/V
n/V = M
P= MRT and M = P/RT
(PC)c (PD)d {[C](RT)}c{[D](RT)}d
Kp =
=
a
b
(PA) (PB) {[A](RT)}a{[B](RT)}b
[C]c(RT)c[D]d(RT)d [C]c[D]d (RT)c(RT)d
= [A]a(RT)a[B]b(RT)b = [A]a[B]b x(RT)a(RT)b
=
Kc{(RT)(c+d)-(a+b)}
(nprod-nreact )
= Kc(RT)
Dn
= Kc(RT)
Partial Pressures and the
Equilibrium Constant
• Consider this system at equilibrium at 5000C.
2SO2(g) + O2(g)
2SO3(g)
• Kp is unitless.
• Partial pressures are expressed in atmospheres or mmHg, so the molar
concentration of a gas and its partial pressure do not have the same numerical
value but are related by the ideal gas constant R and the temperature
Kp = Kc(RT)Dn
Kc = Kp(RT)-Dn
where K is the equilibrium constant expressed in units of concentration and Dn is
the difference between the number of moles of gaseous products and gaseous
reactants; temperature is expressed in Kelvins.
• If Dn = 0, Kp = Kc
Relationship Between Kp and Kc
• Kc is 49 for the following reaction at 450oC. If 1.0 mole of H2 and
1.0 mole of I2 are allowed to reach equilibrium in a 3.0-liter
vessel,

H 2g   I 2g   2 HI g 
(a) How many moles of I2 remain unreacted at equilibrium?
(b) What are the equilibrium partial pressures of H2, I2 and HI?
(c) What is the total pressure in the reaction vessel?
Relationship Between Kp and Kc
•
Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25oC, in a vessel
in which the total pressure is 0.25 atmosphere. What is the value of Kp?
2 NOBrg  
 2 NOg  + Br2g 
Homogeneous and Heterogeneous
Equilibria
• Homogeneous equilibrium
– When the products and reactants of an equilibrium reaction form a single
phase, whether gas or liquid
– Concentrations of the reactants and products can vary over a wide range
• Heterogeneous equilibrium
– A system whose reactants, products, or both are in more than one phase
– An example is the reaction of a gas with a solid or liquid
• Molar concentrations of pure liquids and solids do not vary with
temperature, so they are treated as constants, this simplifies their
equilibrium constant expressions
Heterogeneous Equlibria
• Heterogeneous equilibria have more than one phase present.
– For example, a gas and a solid or a liquid and a gas.
•
How does the equilibrium constant differ for heterogeneous equilibria?
– Pure solids and liquids have activities of unity.
– Solvents in very dilute solutions have activities that are essentially unity.
– The Kc and Kp for the reaction shown above are:
Heterogeneous Equlibria
Heterogeneous Equlibria
• What are Kc and Kp for this reaction?
Heterogeneous Equlibria
• What are Kc and Kp for this reaction?
Equilibrium Constant Expressions
for the Sums of Reactions
An aside: Similar to Hess’ Law we can
add several reactions together to get to an
overall reaction not listed in a table.
To determine K for the sum of the
reactions simply multiply all the values
for each intermediate step to get the K for
the overall reaction.
Solving Equilibrium Problems
• Two fundamental kinds of equilibrium problems
1. Those in which the concentrations of the reactants
and products at equilibrium are given and the
equilibrium constant for the reaction needs to be
calculated
2. Those in which the equilibrium constant and the
initial concentrations of reactants are known and the
concentration of one or more substances at
equilibrium needs to be calculated
Calculating an Equilibrium Constant
from Equilibrium Concentrations
• can be calculated when equilibrium concentrations, or partial
pressures are substituted into the equilibrium constant expression
for the reaction.
• When equilibrium concentrations are not given the equilibrium
concentrations can be obtained from the initial concentrations of
the reactants and the balanced equation for the reaction, as long as
the equilibrium concentration of one of the substances is known.
• Equilibrium constants can be used to calculate the equilibrium
concentrations of reactants and products by using the quantities or
concentrations of the reactants, the stoichiometry of the balanced
equation for the reaction, and a tabular format to obtain the final
concentrations of all species at equilibrium.
Uses of the Equilibrium Constant, Kc
• The equilibrium constant, Kc, is 3.00 for the following reaction at a given
temperature. If 1.00 mole of SO2 and 1.00 mole of NO2 are put into an evacuated
2.00 L container and allowed to reach equilibrium, what will be the concentration of
each compound at equilibrium?
Uses of the Equilibrium Constant, Kc
•
The equilibrium constant is 49 for the following reaction at 450oC. If 1.00 mole of HI is put
into an evacuated 1.00-liter container and allowed to reach equilibrium, what will be the
equilibrium concentration of each substance?
The Reaction Quotient
• The mass action expression or reaction quotient has the
symbol Q.
– Q has the same form as Kc
• The major difference between Q and Kc is that the
concentrations used in Q are not necessarily equilibrium
values.
• Why do we need another “equilibrium constant” that does not
use equilibrium concentrations?
• Q will help us predict how the equilibrium will respond to an
applied stress.
• Q may be derived from a set of values measured at any time
during the reaction of any mixture of the reactants and
products, regardless of whether the system is at equilibrium
• To make this prediction we compare Q with Kc.
The Reaction Quotient (Q)
• Comparing the magnitudes of Q and K allows the
determination of whether a reaction mixture is
already at equilibrium and, if it is not, how to
predict whether its composition will change with
time (whether the reaction will proceed to the right
or to the left)
1. If Q = K, the system is at equilibrium, no further
change in the composition of the system will occur
unless the conditions are changed
2. If Q < K, then the ratio of the concentrations of
products to the concentration of reactants is less
than the ratio at equilibrium; reaction will proceed
to the right, forming products at the expense of
reactants
3. If Q > K, then the ratio of the concentrations of
products to the concentrations of reactants is greater
than at equilibrium; reaction will proceed to the left,
forming reactants at the expense of products
The Reaction Quotient
• The equilibrium constant for the following reaction is 49 at 450oC.
If 0.22 mole of I2, 0.22 mole of H2, and 0.66 mole of HI were put
into an evacuated 1.00-liter container, would the system be at
equilibrium? If not, what must occur to establish equilibrium?
Le Châtelier’s Principle
•
•
•
•
When a system at equilibrium is perturbed in some way, the
effects of the perturbation can be predicted qualitatively using
Le Châtelier’s principle.
This principle states that if a stress is applied to a system at
equilibrium, the composition of the system will change to relieve
the applied stress.
Stress occurs when any change in the system affects the
magnitude of Q or K.
Three types of stresses can change the composition of an
equilibrium mixture:
1. A change in the concentrations (or partial pressures) of the
components by the addition or removal of reactants or
products
2. A change in the total pressure or volume
3. A change in the temperature of the system
Disturbing a System at Equlibrium:
Predictions
1
Changes in Concentration of Reactants and/or Products
• Also true for changes in pressure for reactions involving
gases.
• An equilibrium is disturbed by adding or removing a reactant
or product
1. Stress of an added reactant or product is relieved by reaction in the
direction that consumes the added substance
a. Add reactant—reaction shifts right toward product
b. Add product—reaction shifts left toward reactant
2. Stress of removing reactant or product is relieved by reaction in the
direction that replenishes the removed substance
a. Remove reactant—reaction shifts left
b. Remove product—reaction shifts right
Disturbing a System at Equlibrium:
Predictions
2 Changes in Volume (and pressure for reactions involving gases)
• Increase in pressure (due to decrease in volume) results in a reaction in the
direction of a fewer number of moles of gas
• Decrease in pressure (due to increase in volume) results in a reaction in the
direction of a greater number of moles of gas
1. Decrease volume—molarity increases
2. If reactant side has more moles of gas
a. Increase in denominator is greater than increase in numerator and Qc < Kc
b. To return to equilibrium, Qc must increase; the numerator of the Qc
expression must increase and denominator must decrease—it shifts toward
fewer moles of gas (reactants to products)
3. If product side has more moles of gas
a. Increase in numerator is greater than increase in denominator and Qc > Kc
b. b. To return to equilibrium, Qc must decrease; the denominator of the Qc
expression must decrease and the numerator must increase—it shifts
toward fewer moles of gas (products to reactants)
– Predict what will happen if the volume of this system at equilibrium is
changed by changing the pressure at constant temperature:
Disturbing a System at Equlibrium:
Predictions
3 Changing the Reaction Temperature
Changes in temperature can change the value of K without affecting Q (Q  K)
• Predictions on the response of a system to a change in requires knowledge
regarding DHrxn. Remember:
1. Exothermic (heat is lost by the system, DH<0): reactants ⇋ products + heat
2. Endothermic (heat is gained by the system, DH>0): reactants + heat ⇋ products
• Le Châtelier’s principle predicts
1. that an exothermic reaction will shift to the left (toward reactants) if the
temperature of the system is increased (heat is added);
2. that an endothermic reaction will shift to the right (toward the products) if the
temperature of the system is increased;
3. that if DHrxn = 0, then a change in temperature has no affect on composition.
• Increasing the temperature increases the magnitude of the equilibrium constant for
an endothermic reaction
• Increasing the temperature decreases the equilibrium constant for an exothermic
reaction
Disturbing a System at Equlibrium:
Predictions
• Introduction of a Catalyst
– Catalysts decrease the activation energy of both the forward and reverse
reaction equally.
• Catalysts do not affect the position of equilibrium.
– The concentrations of the products and reactants will be the same
whether a catalyst is introduced or not.
– Equilibrium will be established faster with a catalyst.
Disturbing a System at Equlibrium:
Predictions
• How will an increase in pressure (caused by decreasing the
volume) affect the equilibrium in each of the following reactions?
Disturbing a System at Equlibrium:
Predictions
• How will an increase in temperature affect each of the
following reactions?
Disturbing a System at Equlibrium:
Predictions
• A 2.00 liter vessel in which the following system is in equilibrium contains
1.20 moles of COCl2, 0.60 moles of CO and 0.20 mole of Cl2. Calculate the
equilibrium constant.
Disturbing a System at Equlibrium:
Predictions
• An additional 0.80 mole of Cl2 is added to the vessel at the same temperature.
Calculate the molar concentrations of CO, Cl2, and COCl2 when the new
equilibrium is established.
Controlling the Products of
Reactions
• One of the primary goals of modern chemistry is to control the
identity and quantity of the products of chemical reactions.
• Two approaches
1. To get a high yield of a desired compound, make the rate of the
desired reaction much faster than the rate of any of the other possible
reactions that might occur in the system; altering reaction conditions to
control reaction rates, thereby obtaining a single product or set of
products is called kinetic control.
2. Thermodynamic control—consists of adjusting conditions so that at
equilibrium only the desired products are present in significant
quantities.
Disturbing a System at Equlibrium:
Predictions
• Given the reaction below at equilibrium in a closed container at 500oC. How would
the equilibrium be influenced by the following?
•
An Application of Equilibrium:
The Haber Process
The Haber process is used for the commercial production of ammonia.
– This is an enormous industrial process in the US and many other countries.
– Ammonia is the starting material for fertilizer production.
Fritz Haber
1868-1934
Nobel Prize, 1918
Carl Bosch
1874-1940
Nobel Prize, 1931
o
DG
Relationship Between
rxn
and the Equilibrium Constant
• The relationship for K at conditions other than thermodynamic
standard state conditions is derived from this equation.
DG  DG  RT lnQ
o
or
DG  DG o  2.303 RT log Q
When Q < K or Q > K, reaction is spontaneous.
When Q = K reaction is at equilibrium
When ∆G = 0 reaction is at equilibrium
Therefore, ∆G˚ = - RT ln K
Acids and bases
1.
2.
3.
Autoionization reaction of liquid water
pH, pOH, and pKw
conjugate acid-base pairs
4. acid or base strength and the magnitude of Ka, Kb, pKa, and pKb
5.
leveling effect
6. To be able to predict whether reactants or products are favored in an
acid-base equilibrium
7. use molecular structure and acid and base strengths
8. use Ka and Kb values to calculate the percent ionization and pH of a
solution of an acid or a base
9. calculate the pH at any point in an acid-base titration
10.common ion effects and the position of an acid-base equilibrium
11.
how a buffer works and how to use the HendersonHasselbalch equation to calculate the pH of a buffer
Acids and bases
• There are three classes of strong electrolytes.
1 Strong Water Soluble Acids
Remember the list of strong acids from Chapter 4.
2 Strong Water Soluble Bases
The entire list of these bases was also introduced in Chapter 4.
3 Most Water Soluble Salts
The solubility guidelines from Chapter 4 will help you remember these
salts.
• Weak acids and bases ionize or dissociate partially,
much less than 100%, and is often less than 10%.
Table of Common Ions
Common Negative Ions (Anions)
Monovalent
Hydride
Fluoride
Chloride
Bromide
Iodide
Hydroxide
Permangante
Cyanide
Thiocynate
Acetate
Nitrate
Bisulfite
Bisulfate
Bicarbonate
Dihydrogen phosphate
Nitrite
Amide
Hypochlorite
Chlorite
Chlorate
Perchlorate
HFlClBrIOHMnO4CNSCNCH3COONO3HSO3HSO4HCO3H2PO4NO2NH2ClOClO2ClO3ClO4-
Divalent
Oxide
Peroxide
Sulfide
Selenide
Oxalate
Chromate
Dichromate
Tungstate
Molybdate
Tetrathionate
Thiosulfate
Sulfite
Sulfate
Carbonate
Hydrogen phosphate
O2O22S2Se2C2O42CrO42Cr2O72WO42MoO42S4O62S2O32SO32SO42CO32HPO42-
Trivalent
Nitride
N3-
Phosphate
PO43-
Table of Common Ions
Common Positive Ions (Cations)
Monovalent
Hydronium (aqueous)
Hydrogen (proton)
Lithium
Sodium
Potassium
Rubidium
Cesium
Francium
Silver
Ammonium
Thalium
Copper I
O+
H3
H+
Li+
Na+
K+
Rb+
Cs+
Fr+
Ag+
NH4+
Tl+
Cu+
Divalent
Magnesium
Calcium
Strontium
Beryllium
Manganese II
Barium
Zinc
Cadmium
Nickel II
Palladium II
Platinum II
Copper II
Mercury II
Mercury I
Iron II
Cobalt II
Chromium II
Lead II
Tin II
Mg2+
Ca2+
Sr2+
Be2+
Mn2+
Ba2+
Zn2+
Cd2+
Ni2+
Pd2+
Pt2+
Cu2+
Hg2+
Hg22+
Fe2+
Co2+
Cr2+
Pb2+
Sn2+
Trivalent
Aluminium
Antimony III
Bismuth III
Al3+
Sb3+
Bi3+
Iron III
Cobalt III
Chromium III
Fe3+
Co3+
Cr3+
Water Solubility of Ionic Compounds
If one ion from the “Soluble
Compound” list is present in
a compound, the compound
is water soluble.
Acids and bases
• Most salts of strong or weak electrolytes can dissolve in water to produce a neutral,
basic, or acidic solution, depending on whether it contains the conjugate base of a
weak acid as the anion (A–) or the conjugate acid of a weak base as the cation
(BH+), or possibly both.
• Salts that contain small, highly charged metal ions produce acidic solutions in H2O.
• The most important parameter for predicting the effect of a metal ion on the acidity
of coordinated water molecules is the charge-to-radius ratio of the metal ion.
• The reaction of a salt with water to produce an acidic or basic solution is called a
hydrolysis reaction, which is just an acid-base reaction in which the acid is a
cation or the base is an anion.
Acids and bases
Acid (HCl)
Base (NaOH)
Arrhenius
Brönsted-Lowery
Lewis
Two species that differ by only a proton constitute a conjugate acid-base pair.
1. Conjugate base has one less proton than its acid; A– is the conjugate base of HA
2. Conjugate acid has one more proton than its base; BH+ is the conjugate acid of B
3. Conjugates are weaker than strong parents and stronger than weak parents.
4. All acid-base reactions involve two conjugate acid-base pairs.
• HCl (aq) + H2O (l)  H3O+ (aq) + Cl– (aq)
parent acid
parent base
conjugate acid conjugate base
Water is amphiprotic: it can act as an acid by donating a proton to a
base to form the hydroxide ion, or as a base by accepting a proton from
an acid to form the hydronium ion, H3O+. Substances that can behave
as both an acid and a base are said to be amphoteric.
Acids and bases can be defined in different ways:
1. Arrhenius definition: An acid is a substance that dissociates in water to
produce H+ ions (protons), and a base is a substance that dissociates in water
to produce OH– ions (hydroxide); an acid-base reaction involves the reaction
of a proton with the hydroxide ion to form water.
– Three limitations
1. Definition applied only to substances in aqueous solutions.
2. Definition restricted to substances that produce H+ and OH– ions
3. Definition does not explain why some compounds containing hydrogen
such as CH4 dissolve in water and do not give acidic solutions
2. Brønsted–Lowry definition: An acid is any substance that can donate a
proton, and a base is any substance that can accept a proton; acid-base
reactions involve two conjugate acid-base pairs and the transfer of a proton
from one substance (the acid) to another (the base). Not restricted to aqueous
solutions, expanding to include other solvent systems and acid-base reactions
for gases and solids. Not restricted to bases that only produce OH– ions .
Acids still restricted to substances that produce H+ ions. Limitation 3 not
dealt with.
3. Lewis definition: A Lewis acid is an electron-pair acceptor, and a Lewis base
is an electron-pair donor .
Acids and bases
CO2 + H2O
Baking Soda NaHCO3
Soda Pop
CO2
Atmosphere
.8317
CO2+3H2O
KH
2.5×10−4
H2CO3+2H2O
K1
H2CO3 + H2O
HCO3- + H+
HCO3- + H2O
CO32- + H+
5.61×10−11
K2
HCO3-+H3O++H2O
CO32-+2H3O+
CO2 + CaCO3 + H2O
2HCO3- + Ca2Biological Calcification
(not a reversible reaction)
unneeded critter's shells
fall to ocean floor
H2CO3
You should know the
strong acids & bases
pH, a Concentration Scale
pH: a way to express acidity -- the concentration of H+ in solution.
Low pH: high [H+]
High pH: low [H+]
pH = log (1/ [H+]) = - log [H+]
Acid
Formula
pH at half
equivalence
point
Acetic
CH3COOH
4.7
Nitrous
HNO2
3.3
Hydrofluoric
HF
3.1
Hypochlorous
HClO
7.4
Hydrocyanic
HCN
9
Acidic solution
Neutral
Basic solution
pH < 7
pH = 7
pH > 7
pH, a Concentration Scale
• A convenient way to express the acidity and basicity of a solution is the pH
and pOH scales.
• The pH of an aqueous solution is defined as:
Acid
Formula
pH
Acetic
CH3COOH
4.7
Nitrous
HNO2
3.3
Hydrofluoric
HF
3.1
Hypochlorous
HClO
7.4
Hydrocyanic
HCN
9
Titration
Titrant
Equivalence Point
Primary Standard
End Point
Secondary Standard
Titration
1. Add solution from the buret.
2. Reagent (base) reacts with
compound (acid) in solution
in the flask.
3. Indicator shows when exact
stoichiometric reaction has
occurred.
4. Net ionic equation
H+ + OH- --> H2O
5. At equivalence point
moles H+ = moles OH-
http://www.chem1.com/acad/webtext/abcon/abcon-2.html
Autoionization of Water
• Because water is amphiprotic, one water molecule can react with another to form an
OH– ion and an H3O+ ion in an autoionization process:
2H2O(l)⇋H3O+ (aq) + OH– (aq)
• Equilibrium constant K for this reaction can be written as
Kc = [H3O+] [OH–]
[H2O]2
• 1 L of water contains 55.5 moles of water. In dilute aqueous solutions:
• The water concentration is many orders of magnitude greater than the ion
concentrations. The concentration is essentially that of pure water. Recall that the
activity of pure water is 1.
• When pure liquid water is in equilibrium with hydronium and hydroxide ions at
25ºC, the concentrations of hydronium ion and hydroxide ion are equal:
[H3O+]=[OH–] = 1.0 x 10–7 M
[H3O+][OH–] = 1.0 x 10–14 M = Kw
pH = pOH = 7 pH + pOH = pKw = 14
Kc [H2O]2 = Kw = [H3O+][OH–] = 1.0 x 10–14
Leveling Effect
It’s all because of Gibbs Free Energy
No acid stronger than H3O+ and no base
stronger than OH– can exist in aqueous
solution, leading to the phenomenon
known as the leveling effect.
• Any species that is a stronger acid than
the conjugate acid of water (H3O+) is
leveled to the strength of H3O+ in aqueous
solution because H3O+ is the strongest
acid that can exist in equilibrium with
water.
• In aqueous solution, any base stronger
than OH– is leveled to the strength of
OH– because OH– is the strongest base
that can exist in equilibrium with water
• Any substance whose anion is the
conjugate base of a compound that is a
weaker acid than water is a strong base
that reacts quantitatively with water to
form hydroxide ion
http://www.chem1.com/acad/webtext/abcon/abcon-2.html
Acid-Base Equilibrium Constants:
Ka, Kb, pKa, and pKb
• The magnitude of the equilibrium constant for an ionization
reaction can determine the relative strengths of acids and bases
• The general equation for the ionization of a weak acid in water,
where HA is the parent acid and A– is its conjugate base, is
HA(aq) + H2O(l) ⇋ H3O+(aq) + A–(aq)
• The equilibrium constant for this dissociation is
K = [H3O+] [A–]
[H2O] [HA]
• The concentration of water is constant for all reactions in
aqueous solution, so [H2O] can be incorporated into a new
quantity, the acid ionization constant (Ka):
Ka = K[H2O] = [H3O+] [A–]
[HA]
Ionization Constants for Weak
Monoprotic Acids and Bases
• Strong acids and bases ionize essentially completely in water;
the percent ionization is always approximately 100%,
regardless of the concentration
• The percent ionization in solutions of weak acids and bases is
small and depends on the analytical concentration of the weak
acid or base; percent ionization of a
weak acid or a weak base actually
increases as its analytical
concentration decreases and
percent ionization increases as the
magnitude of the ionization
constants Ka and Kb increases
Ionization Constants for Weak Acids and Bases
When is a 1M solution not a 1 M solution?
 A 1 M solution is prepared by dissolving 1 mol of acid or base in water and
adding enough water to give a final volume of exactly 1 L.
 If the actual concentrations of all species present in the solution were listed, it
would be determined that none of the values is exactly 1 M because a weak
acid or a weak base always reacts with water to some extent.
 Only the total concentration of both the ionized and unionized species is equal
to 1 M.
 The analytical concentration (C) is defined as the total concentration of all
forms of an acid or base that are present in solution, regardless of their state of
protonation.
 Thus; a 1 M solution has an analytical concentration of 1 M, which is the sum
of the actual concentrations of unionized acid or base and the ionized form.
Ionization Constants for Weak
Monoprotic Acids and Bases
• Let’s look at the dissolution of acetic acid, a weak acid, in water as an
example.
• The equation for the ionization of acetic acid is:
• The equilibrium constant for this ionization is expressed as:
Ionization Constants for Weak
Monoprotic Acids and Bases
• The water concentration in dilute aqueous solutions is very high.
• 1 L of water contains 55.5 moles of water.
• Thus in dilute aqueous solutions:
• The water concentration is many orders of magnitude greater than the ion
concentrations.
• Thus the water concentration is essentially that of pure water.
– Recall that the activity of pure water is 1.
Ionization Constants for Weak
Monoprotic Acids and Bases
• We can define a new equilibrium constant for weak acid equilibria that uses
the previous definition.
– This equilibrium constant is called the acid ionization constant.
– The symbol for the ionization constant is Ka.
Ionization Constants for Weak
Monoprotic Acids and Bases
• The ionization constant values for several acids are given below.
– Which acid is the strongest?
– Are all of these acids weak acids?
– What is the relationship between Ka and strength?
– What is the relationship between pKa and strength?
– What is the relationship between pH and strength?
Acid
Formula
Ka value
pKa value
-log Ka
pH of 1M
analytical [HA]
Acetic
CH3COOH
1.8 x 10-5
4.7
2.4
Nitrous
HNO2
4.5 x 10-4
3.3
1.7
Hydrofluoric
HF
7.2 x 10-4
3.1
1.6
Hypochlorous
HClO
3.5 x 10-8
7.5
3.7
Hydrocyanic
HCN
4.0 x 10-10
9.4
4.7
Determining Ka and Kb
• The ionization constants Ka and Kb are equilibrium constants that
are calculated from experimentally measured concentrations.
• What does the concentration of an aqueous solution of a weak acid
or base exactly mean?
– A 1 M solution is prepared by dissolving 1 mol of acid or base in water and adding
enough water to give a final volume of exactly 1 L.
– If the actual concentrations of all species present in the solution were listed, it would
be determined that none of the values is exactly 1 M
– Only the total concentration of both the ionized and unionized species is equal to 1
M.
– The analytical concentration (C) is defined as the total concentration of all forms of
an acid or base that are present in solution, regardless of their state of protonation.
– 1 M solution has an analytical concentration of 1 M, the sum of the actual
concentrations of unionized acid or base and the ionized form.
Two common ways to obtain the concentrations
1. By measuring the electrical conductivity of the solution, which is related to the
total concentration of ions present
2. By measuring the pH of the solution, which gives [H+] or [OH–]
Determining Ka and Kb
• RICE Procedure for determining Ka for a weak acid
and Kb for a weak base
1. The analytical concentration of the acid or base ionization Reaction is
the Initial concentration
2. The stoichiometry of the reaction with water determines the Change in
concentrations
3. The final (Equilibrium) concentrations of all species are calculated
from the initial concentrations and the changes in the concentrations
4. Inserting the final concentration into the equilibrium constant
expression enables the value of Ka or Kb to be calculated
Ionization Constants for Weak
Monoprotic Acids: the math
In a 0.12 M solution of a weak monoprotic acid, HY, the acid is 5.0% ionized.
Calculate the ionization constant for the weak acid.
R
HY
H3O+ +
Y-
I
C
E
• Use the concentrations that were just determined in the ionization constant
expression to get the value of Ka.
Ionization Constants for Weak Monoprotic
Acids the MATH
• The pH of a 0.10 M solution of a weak monoprotic acid, HA, is found to be 2.97.
What is the value for its ionization constant?
• Use the [H3O+] and the stoichiometry
of the ionization reaction to determine
concentrations of all species.
R
Simplifying Assumption: Is the change
significant? Later we will find that in
general, if the Ka/[] is < 1x10-3 you
can apply the simplifying assumption.
C
HA
I
E
• Calculate the ionization constant from this information.
H3 O + +
A-
Ionization Constants for Weak Monoprotic Acids
the MATH
Calculate the concentrations of the various species in 0.15 M acetic acid, CH3COOH,
solution.
1.
It is always a good idea to write down the ionization reaction and the ionization
constant expression.
2.
Next we combine the basic chemical concepts with some algebra to solve the
problem
+
-
R CH3COOH
I
C
E
H3O + CH3COO
Ionization Constants for Weak
Monoprotic Acids and Bases
• Substitute these algebraic
quantities into the ionization
expression.
• Solve the algebraic equation,
using a simplifying assumption
that is appropriate for all weak
acid and base ionizations.
• Solve the algebraic equation, using a
simplifying assumption that is
appropriate for all weak acid and base
ionizations.
Ionization Constants for Weak
Monoprotic Acids and Bases
• Complete the algebra and solve for the concentrations of the species.
•
Note that the properly applied simplifying assumption gives the same result as
solving the quadratic equation does.
Ionization Constants for Weak Monoprotic
Acids: the MATH
•
Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution.
Ka= 4.0 x 10-10 for HCN
R
HCN
H3O+ + CNI
C
E
• Substitute these algebraic quantities into the ionization expression.
• Solve the algebraic equation, using the simplifying assumption that is appropriate
for all weak acid and base ionizations.
Ionization Constants for Weak
Monoprotic Acids and Bases
• Let’s look at the percent ionization of two weak acids as a function of their
ionization constants.
Solution
Ka
[H+]
pH
% ionization
0.15 M
acetic acid
0.15 M
HCN
1.8 x 10-5
1.6 x 10-3
2.80
1.1
4.0 x 10-10
7.7 x 10-6
5.11
0.0051
• Note that the [H+] in 0.15 M acetic acid is 200 times greater than for 0.15
M HCN.
[ionized HY]
% ionization =
x 100%
[unionized HY]
Ionization Constants for Weak
Monoprotic Acids and Bases
• All of the calculations and understanding we have at present can be applied
to weak acids and weak bases. Calculate the concentrations of the various
species in 0.15 M aqueous ammonia.
Ionization Constants for Weak Monoprotic
Bases: the MATH
• All of the calculations and understanding we have at present can be applied to
weak acids and weak bases. Calculate the concentrations of the various species
in 0.15 M aqueous ammonia. Kb = 1.8E-5
R
I
C
E
NH3
NH4+ +
OH-
Ionization Constants for Weak Monoprotic
Bases: the MATH
• The pH of an aqueous ammonia solution is 11.37. Calculate the molarity (original
concentration) of the aqueous ammonia solution
R
NH3
NH4+
+
OH-
I
C
E
• Examination of the last equation suggests that our simplifying assumption can be
applied. In other words (x-2.3x10-3)  x.
– Making this assumption simplifies the calculation.
Acid-Base Properties of
Solutions of Salts
• A salt can dissolve in water to produce a neutral, basic, or acidic solution,
depending on whether it contains the conjugate base of a weak acid as the
anion (A–) or the conjugate acid of a weak base as the cation (BH+), or
both.
• Salts that contain small, highly charged metal ions produce acidic solutions
in water.
• The most important parameter for predicting the effect of a metal ion on the
acidity of coordinated water molecules is the charge-to-radius ratio of the
metal ion.
• The reaction of a salt with water to produce an acidic or basic solution is
called a hydrolysis reaction, which is just an acid-base reaction in which
the acid is a cation or the base is an anion.
Acids, Bases, and ionization constants
• Acid and Base strengths can be compared using Ka and Kb values. The larger the Ka or Kb
value the more product favored the dissociation.
• An acid-base equilibrium always favors the side with the weaker acid and base.
stronger acid + stronger base
weaker acid + weaker base
• In an acid-base reaction the proton always reacts with the strongest base until totally
consumed before reacting with any weaker bases.
• Any substance whose anion is the conjugate base of a weak acid weaker than OH- reacts
quantitatively with water to form more hydroxide ions.
H2O
Step 1. NaCH3COO → Na+ + CH3COOAcetate ion is the conjugate base of acetic acid, a weak acid.
Step 2. CH3COO- + H2O
CH3COOH + OH• Hydrolysis: Aqueous solutions of salts that dissociate into both:
1. A strong conjugate acid and a strong conjugate base are neutral (KNO3).
2. A strong conjugate acid and a weak conjugate base are acidic (HCl).
3. A strong conjugate base and a weak conjugate acid are basic (NaOH).
4. A weak conjugate base and a weak conjugate acid can be neutral, basic or acidic:
• The comparison of the values of Ka and Kb determine the pH of these solutions.
a. Kbase = Kacid make neutral solutions (NH4CH3OO)
b. Kbase > Kacid make basic solutions (NH4ClO)
c. Kbase < Kacid make acidic solutions (CH3)3NHF
Polyprotic Acids and Bases
• Polyprotic acids contain more than one
ionizable proton, and the protons are
lost in a stepwise manner.
• The fully protonated species is always
the strongest acid because it is easier to
remove a proton from a neutral
molecule than from a negatively
charged ion; the fully deprotonated
species is the strongest base.
• Acid strength decreases with the loss of
subsequent protons, and the pKa
increases.
• The strengths of the conjugate acids and
bases are related by pKa + pKb = pKw,
and equilibrium favors formation of the
weaker acid-base pair.
Polyprotic Acids
• Many weak acids contain two or more acidic hydrogens.
– Examples include H3PO4 and H3AsO4.
• The calculation of equilibria for polyprotic acids is done in a
stepwise fashion.
– There is an ionization constant for each step.
• Consider arsenic acid, H3AsO4, which has three ionization
constants.
1 Ka1 = 2.5 x 10-4
2 Ka2 = 5.6 x 10-8
3 Ka3 = 3.0 x 10-13
• Notice that the ionization constants vary in the following fashion:
K a1  K a2  K a3
• This is a general relationship.
– For weak polyprotic acids the Ka1 is always > Ka2, etc.
Polyprotic Acids
• The first ionization step for arsenic acid is:
• The second ionization step for arsenic acid is:
• The third ionization step for arsenic acid is:
Polyprotic Acids The MATH
• Calculate the concentration of all species in 0.100 M arsenic acid, H3AsO4,
solution.
1 Write the first ionization step and represent the concentrations.
Approach this problem exactly as previously done.
R
I
C
E
The simplifying assumption cannot be used.
Using the quadratic equation x =
H3AsO4
H3O+ + H2AsO4-
Polyprotic Acids The MATH
2.
Next, write the equation for the second step ionization and represent the
concentrations and work as before.
R
I
C
E
The simplifying assumption can be used.
H2AsO4 -
H3O+ + HAsO4-2
Polyprotic Acids The MATH
3.
Finally, repeat the entire procedure for the third ionization step.
R
HAsO4 -2
H3O+
+
AsO4-3
I
C
E
The simplifying assumption can be used. 3.0x10-13/5.6x10-8 = 5.4x10-6<1.0x10-3
Polyprotic Acids
• A comparison of the various species in 0.100 M H3AsO4 solution follows.
Species
Concentration
H3AsO4
0.095 M
H+
4.9 x 10-3 M
H2AsO4-
4.9 x 10-3 M
HAsO42-
5.6 x 10-8 M
AsO43-
3.4 x 10-18 M
OH-
2.0 x 10-12 M
When a strong base is added to a solution of a polyprotic acid, the
neutralization reaction occurs in stages.
1. The most acidic group is titrated first, followed by the next most acidic, and so
forth
2. If the pKa values are separated by at least three pKa units, then the overall
titration curve shows well-resolved “steps” corresponding to the titration of
each proton
Polyprotic Acids
The Common Ion Effect and Buffers
The ionization equilibrium of a weak acid (HA) is affected by the addition of
either the conjugate base of the acid (A–) or a strong acid (a source of H+);
LeChâtelier’s principle is used to predict the effect on the equilibrium
position of the solution
• Common-ion effect—the shift in the position of an equilibrium on addition
of a substance that provides an ion in common with one of the ions already
involved in the equilibrium; equilibrium is shifted in the direction that
reduces the concentration of the common ion \
Buffers are characterized by the following:
1. the pH range over which they can maintain a constant pH—depends
strongly on the chemical properties of the weak acid or base used to
prepare the buffer (on K)
2. their buffer capacity, the amount of strong acid or base that can be
absorbed before the pH changes significantly—depends solely on the
concentration of the species in the buffered solution (the more
concentrated the buffer solution, the greater its buffer capacity)
3. observed change in the pH of the buffer is inversely proportional to
the concentration of the buffer
Buffers are characterized by the following:
1. the pH range over which they can maintain a constant pH—depends strongly on
the chemical properties of the weak acid or base used to prepare the buffer (on K)
2. buffer capacity, is the number of moles of strong acid or strong base needed to
change the pH of 1 Liter of buffer solution by 1 pH unit.
a. depends solely on the concentration of the species in the buffered solution
(the more concentrated the buffer solution, the greater its buffer capacity)
b. A general estimate of the buffer capacity is 40% of the sum of the molarities
of the conjugate acid and conjugate base
3. observed change in the pH of the buffer is inversely proportional to the
concentration of the buffer
The Common Ion Effect and Buffers
• There are two common kinds of buffer solutions:
1 Solutions made from a weak acid plus a soluble ionic salt of the weak acid.
2 Solutions made from a weak base plus a soluble ionic salt of the weak base
1.
•
Solutions made of weak acids plus a soluble ionic salt of the weak acid
One example of this type of buffer system is:
– The weak acid - acetic acid CH3COOH
– The soluble ionic salt - sodium acetate NaCH3COO
Buffer Solutions Weak Acids Plus
Salts of Their Conjugate Bases
One example of the type I of buffer system is:
The weak acid - acetic acid CH3COOH
CH3COOH
H3O+ + CH3COO~100%
NaCH3COO →Na+ + CH3COOThe soluble ionic salt - sodium acetate NaCH3COO
• This is an equilibrium problem with a starting concentration for both the
cation and anion. After calculating the concentration of H+ and the pH of a
solution that is 0.15 M in both acetic acid sodium acetate yields:
R CH3COOH
I
C
E
H3O+ + CH3COO-
Buffer Solutions Weak Acids Plus
Salts of Their Conjugate Bases
Alternatively you might have noticed:
0
[base]/[acid] = 1 = 10
log 100 = 0
pH = pKa + 0 = pKa = 4.74
[H+] = Ka = 1.8E-5
Solution
[H+]
pH
0.15 M CH3COOH
1.6 x 10-3
2.80
0.15 M CH3COOH &
0.15 M NaCH3COO buffer
1.8 x 10-5
4.74
 [H+] is ~90 times greater in pure acetic acid than in buffer solution.
Note that the pH of the buffer equals the pKa of the buffering acid.
The Common Ion Effect and Buffers
• The general expression for the ionization of a weak monoprotic acid is:
• The generalized ionization constant expression for a weak acid is:
The Common Ion Effect and Buffers
• If we solve the expression for [H+], this relationship results:
• By making the assumption that the concentrations of the weak acid and the
salt are reasonable, the expression reduces to:
The Common Ion Effect and Buffers
• The relationship developed in the previous slide is valid for buffers
containing a weak monoprotic acid and a soluble, ionic salt.
• If the salt’s cation is not univalent the relationship changes to:
The Common Ion Effect and Buffers
• Simple rearrangement of this equation and application of algebra
yields the
Henderson-Hasselbach equation.
The Henderson-Hasselbach equation is one method to calculate the pH
of a buffer given the concentrations of the salt and acid.
Henderson-Hasselbalch - Caveats and Advantages
• Henderson-Hasselbalch equation is valid for solutions whose
concentrations are at least 100 times greater than the value of
their Ka’s
• The Henderson-Hasselbach equation is one method to calculate the
pH of a buffer given the concentrations of the salt and acid.
• A special case exists for the Henderson-Hasselbalch equation when
[base]/[acid] = some power of 10, regardless of the actual
concentrations of the acid and base, where the HendersonHasselbalch equation can be interpreted without the need for
calculations:
[base]/[acid] = 10x
log 10x = x
in general pH = pKa + x
Examples:
1. when [base] = [acid], [base]/[acid] = 1 or 100, log 1 = 0, pH = pKa,
(corresponds to the midpoint in the titration of a weak acid or base)
2. when [base]/[acid] = 10 or 101, log 10 = 1 then pH = pKa + 1
3. when [base]/[acid] = .001 or 10-2, log 10 = -2 then pH = pKa -2
Buffer Solutions
There are two common kinds of buffer solutions:
I.
II.
Commonly, solutions made from a weak acid plus a soluble ionic salt of the
conjugate base of the weak acid.
Less common, solutions made from a weak base plus a soluble ionic salt of
the conjugate acid of the weak base.
Both of the above may also be prepared by starting with a weak acid (or weak
base) and add half as many moles of strong base (acid)
Buffer Solutions: Weak Bases Plus
Salts of Their Conjugate Acids
One example of the type II of buffer system is:
The weak base – ammonia NH3
NH3
The soluble ionic salt – ammonium nitrate NH4NO3
NH4 ++ OH -
→ NH
~100%
NH4NO3
4
++ NO 3
• This is an equilibrium problem with a starting concentration for both the cation
and anion. After calculating the concentration of OH- and the pOH of the
solution that is 0.15 M in aqueous ammonia, NH3, and 0.30 M in ammonium
nitrate, NH4NO3 yeilds:
R
NH3
NH4+
+
OH-
I
C
E
• Substitute the quantities determined in the previous relationship into
the ionization expression for ammonia.
The Common Ion Effect and Buffers
• We can derive a general relationship for buffer solutions that contain a
weak base plus a salt of a weak base similar to the acid buffer relationship.
– The general ionization equation for weak bases is:
• Simple rearrangement of this equation and application of algebra
yields the
Henderson-Hasselbach equation.
The Common Ion Effect and Buffers
• A comparison of the aqueous ammonia concentration to that of the buffer
described above shows the buffering effect.
 The
Solution
[OH-]
pH
0.15 M NH3
1.6 x 10-3 M
11.20
0.15 M NH3 &
0.15 M NH4NO3 buffer
9.0 x 10-6 M
8.95
[OH-] in aqueous ammonia is 180 times greater than in the buffer.
Buffering Action
• If 0.020 mole of gaseous HCl is added to 1.00 liter of a buffer solution that is
0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much
does the pH change? Assume no volume change due to addition of the HCl.
1 Calculate the pH of the original buffer solution.
NH4NO3
NH3
R
NH3
→ NH
~100%
4
++ NO 3
NH4 ++ OH NH4+
+
OH-
I
C
E
• Substitute the quantities determined in the previous relationship into
the ionization expression for ammonia.
Buffering Action
2 Next, calculate the concentration of all species after the addition of the
gaseous HCl.
– The HCl will react with some of the ammonia and change the
concentrations of the species.
– This is another limiting reactant problem.
HCl
→ H + Cl
NH3 + H+
NH4 +
~100%
R
I
C
E
NH3
+
NH4+
-
+
OH-
Buffering Action
3 Using the concentrations of the salt and base and the HendersonHassselbach equation, the pH can be calculated.
4 Finally, calculate the change in pH.
Buffering Action
1. If 0.020 mole of NaOH is added to 1.00 liter of solution that is
0.100 M in aqueous ammonia and 0.200 M in ammonium
chloride, how much does the pH change? Assume no volume
change due to addition of the solid NaOH.
→ Na + OH
~100%
NaOH
NH4 + + OH-
R
I
C
E
NH3
+
-
NH3
NH4+
+
OH-
Buffering Action
2.
3.
Using the concentrations of the salt and base and the HendersonHassselbach equation, the pH can be calculated.
Finally, calculate the change in pH.
Buffering Action
Original Solution
1.00 L of solution
containing
0.100 M NH3 and
0.200 M NH4Cl
Original
pH
Acid or base
added
New
pH
DpH
0.020 mol NaOH 9.08
+0.13
0.020 mol HCl
-0.14
8.95
8.81
• Notice that the pH changes only slightly in each case.
Preparation of Buffer Solutions
• Calculate the concentration of H+ and the pH of the solution prepared by
mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium
hydroxide solutions.
• Determine the amounts of acetic acid and sodium hydroxide prior to the
acid-base reaction.
• NaOH and CH3COOH
react in a 1:1 mole ratio.
• After the two solutions
are mixed, Calculate
total volume.
• The concentrations of
the acid and base are:
• Substitution of these
values into the ionization
constant expression (or the
Henderson-Hasselbach
equation) permits
calculation of the pH.
Preparation of Buffer Solutions
• For biochemical situations, it is sometimes important to prepare a buffer
solution of a given pH. Starting with a solution that is 0.100M in aqueous
ammonia prepare 1.00L of a buffer solution that has a pH of 9.15 using
ammonium chloride as the source of the soluble ionic salt of the conjugate
weak acid.
NH4Cl
~100%
+
NH4 +
Cl-
NH3
H2O
NH4++ OH-
• The Henderson-Hasselbalch equation is used to determine the ratio of the
conjugate acid base pair
•
•
•
•
pOH can be determined from the pH:
pKb can be looked up in a table:
[base] concentration is provided:
Solve for [acid]:
• Does this result make sense?
Titration Curves
Strong Acid/Strong Base Titration Curves
• These graphs are a plot of pH vs. volume of acid or base added
in a titration.
• As an example, consider the titration of 100.0 mL of 0.100 M
perchloric acid with 0.100 M potassium hydroxide.
– In this case, we plot pH of the mixture vs. mL of KOH
added.
– Note that the reaction is a 1:1 mole ratio.
HClO 4  KOH  KClO 4  H2O
Strong Acid/Strong Base Titration Curves
(An M1V1M2V2 problem)
• Before any KOH is added the pH of the HClO4 solution is 1.00. Remember perchloric
acid is a strong acid that ionizes essentially 100%.
HClO 4  KOH  KClO 4  H2O
mL KOH
0
20
mMol KOH†
0
2
50
90
100
5
9
10
†mMol
KOH = ml KOH∙NaOH M
*mMol HClO4 = ml HClO4∙ HClO4 M
1:1 mole ratio
‡mMol
HClO4 = ml HClO4- mMol KOH
mMol HClO4 mL soln. Ұ
10*
100
8‡
120
5‡
1‡
0‡
150
190
200
ҰmL
pH
1.00
1.18
1.48
2.28
7.00
soln = mL HClO4 + mL KOH
= 100 mL + total base added
mMol H+ = mMol HClO4
[H+] = mMol HClO4 / mL soln.
pH = -log[H+]
Strong Acid/Strong Base
Titration Curves
• We have calculated only a few points on the titration curve. Similar
calculations for remainder of titration show clearly the shape of the titration
curve.
Weak Acid/Strong Base Titration Curves
• As an example, consider the titration of 100.0 mL of 0.100 M
acetic acid, CH3 COOH, (a weak acid) with 0.100 M KOH (a
strong base).
– The acid and base react in a 1:1 mole ratio.
• Before the equivalence point is reached, both CH3COOH and
KCH3COO are present in solution forming a buffer.
– The KOH reacts with CH3COOH to form KCH3COO.
• A weak acid plus the salt of a weak acid’s conjugate base form a
buffer.
• Hypothesize how the buffer production will effect the titration
curve.
Weak Acid/Strong Base
Titration Curves
• The pH changes much more gradually around the equivalence point in the
titration of a weak acid or a weak base.
• [H+] of a solution of a weak acid (HA) is not equal to the concentration of the
acid (HA)
• [H+] depends on both its Ka and the analytical concentration of the acid (HA).
• Only a fraction of a weak acid dissociates, so [H+] is less than [HA];
therefore, the pH of a solution of a weak acid is higher than the pH of a
solution of a strong acid of the same concentration.
Weak Acid/Strong Base Titration Curves
(a RICE problem)
1.
Determine the pH of the acetic acid solution before the titration is begun.
2. Solve the algebraic equation for each addition of strong base, using a simplifying
assumption that is appropriate for all weak acid and base ionizations.
• At the equivalence point, the solution is 0.500 M in KCH3COO, the salt of a
strong base and a weak acid which hydrolyzes to give a basic solution.
3. The solution cannot have a pH=7.00 at equivalence point.
– Both processes make the solution basic. Concentrations must now be
calculated using the equation for Kb. Remember that Kw = KaKb and
that pH + pOH = 14
Weak Acid/Strong Base
Titration Curves
mL KOH mMol OH† [CH3COOH] [CH3COO- ]
•
[H3O+]
mL soln. Ұ
pH
0
0
0.100
1.34E-02
1.34E-02
100
1.87
20
2.0
0.0800
1.67E-02
8.64E-05
120
4.06
50
5.0
0.0500
3.33E-02
2.70E-05
150
4.57
90
9.0
0.0100
4.74E-02
3.80E-06
190
5.42
100
10
5.30E-06
5.00E-02
1.89E-09
200
8.72
110
10
0.000
5.00E-02
2.10E-13
210
12.68
After the equivalence point is reached, the pH is determined by the excess
KOH just as in the strong acid/strong base example.
Weak Acid/Strong Base Titration Curves
• We have calculated only a few points on the titration curve. Similar
calculations for remainder of titration show clearly the shape of the titration
curve.
•
Weak Acid/Strong Base
Titration
Curves
We have calculated only a few points on the titration curve. Similar calculations
for remainder of titration show clearly the shape of the titration curve.
Identity of the weak acid or base being titrated strongly affects the shape of the titration
curve.
The shape of titration curves as a function of the pKa or pKb shows that as the acid or base
being titrated becomes weaker (its pKa or pKb becomes larger), the pH change around the
equivalence point decreases significantly.
Weak Acid/Weak Base
Titration Curves
• Weak Acid/Weak Base Titration curves have very short
vertical sections.
• Visual indicators cannot be used.
• The solution is buffered both before and after the equivalence
point.
• Comparison of the respective Ka and Kb values can be used to
determine the pH of the equivalence points of these titrations.
a. Kbase = Kacid
b. Kbase > Kacid
c. Kbase < Kacid
neutral solutions
basic solutions
acidic solutions
Indicators
A good indicator must have the following properties:
1. Color change must be easily detected
2. Color change must be rapid
3. Indicator molecule must not react with the substance being titrated
4. The indicator should have a pKin that is within one pH unit of the
expected pH at the equivalence point of the titration
• Choosing an indicator for an acid-base titration
1. For titrations of strong acids and strong bases (and vice versa), any
indicator with a pKin between 4 and 10 will do
2. For the titration of a weak acid, the pH at the equivalence point is
greater than 7, and an indicator such as phenolphthalein or thymol blue,
with pKin > 7, should be used
3. For the titration of a weak base, where the pH at the equivalence point
is less than 7, an indicator such as methyl red or bromcresol blue, with
pKin < 7, should be used
Indicators
• The chemistry of indicators are described by the general
equation
• The ionization constant for the deprotonation of indicator Hn
is:
[H+] [n–]
Kin =
[Hn]
• The value of pKin determines the pH at which the indicator
changes color
Indicators
• If the preceding expression is rearranged the range over which the indicator
changes color can be discerned.
Indicators
The Solubility Product, Ksp
• Silver chloride, AgCl,is rather insoluble in water.
• Careful experiments show that if solid AgCl is placed in pure water and
vigorously stirred, a small amount of the AgCl dissolves in the water.
• The equilibrium constant expression for this dissolution is called a
solubility product constant.
– Ksp = solubility product constant
The Solubility Product, Ksp
• In general, the dissolution of a slightly soluble compound and its solubility
product expression are represented as:
The Solubility Product, Ksp
• The same rules apply for compounds that have more than two kinds of ions.
• One example of a compound that has more than two kinds of ions is calcium
ammonium phosphate.
Determination of Solubility Product
Constants
• One liter of saturated silver chloride solution contains 0.00192 g of dissolved
AgCl at 25oC. Calculate the molar solubility of, and Ksp for, AgCl.
• The molar solubility can be easily calculated from the data:
• The equation for the dissociation of silver chloride, the appropriate molar
concentrations, and the solubility product expression are:
• Substitution of the molar concentrations into the solubility product
expression gives:
Determination of Solubility Product
Constants
•
1.
One liter of saturated calcium fluoride solution contains 0.0167 gram of
CaF2 at 25oC. Calculate the molar solubility of, and Ksp for, CaF2.
Calculate the molar solubility of CaF2.
• From the molar solubility, we can find the ion concentrations in saturated
CaF2. Then use those values to calculate the Ksp.
– Note: You are most likely to leave out the factor of 2 for the
concentration of the fluoride ion!
Uses of Solubility Product Constants
•
•
The solubility product constant can be used to calculate the solubility of a
compound at 25oC.
Calculate the molar solubility of barium sulfate, BaSO4, in pure water and the
concentration of barium and sulfate ions in saturated barium sulfate at 25oC.
For barium sulfate, Ksp= 1.1 x 10-10.
Uses of Solubility Product Constants
•
The solubility product constant for magnesium hydroxide, Mg(OH)2, is 1.5 x 10-11.
Calculate the molar solubility of magnesium hydroxide and the pH of a saturated
magnesium hydroxide solution at 25oC.
• Be careful, do not forget the stoichiometric coefficient of 2!
• Substitute the algebraic expressions into the solubility product expression.
• Solve for the pOH and pH.
The Common Ion Effect and
Solubility
• Solubility product expression
– Equilibrium concentrations of cation and anion are inversely related
– As the concentration of the anion increases, the maximum
concentration of the cation needed for precipitation to occur decreases,
and vice versa
– Ksp is constant
• Common ion effect
– The solubility of an ionic compound depends on the concentrations of
other salts that contain the same ions.
– This dependency is an example of the common ion effect; adding a
common cation or anion shifts a solubility equilibrium in the direction
predicted by LeChâtelier’s principle.
– The solubility of any sparingly soluble salt is almost always decreased by
the presence of a soluble salt that contains a common ion.
The Common Ion Effect in Solubility
Calculations
•
Calculate the molar solubility of barium sulfate, BaSO4, in 0.010 M sodium
sulfate, Na2SO4, solution at 25oC. Compare this to the solubility of BaSO4
in pure water. (Example 20-3). (What is the common ion? How was a
common ion problem solved in Chapter 19?)
1.
Write equations to represent the equilibria.
2.
Substitute the algebraic representations of the concentrations into the Ksp
expression and solve for x.
• The molar solubility of BaSO4 in 0.010 M Na2SO4 solution is 1.1 x 10-8 M.
• The molar solubility of BaSO4 in pure water is 1.0 x 10-5 M.
– BaSO4 is 900 times more soluble in pure water than in 0.010 M sodium
sulfate!
– Adding sodium sulfate to a solution is a fantastic method to remove
Ba2+ ions from solution!
• If your drinking water were suspected to have lead ions in it, suggest a
method to prove or disprove this suspicion.
Qualitative Analysis Using Selective
Precipitation
The Ion Product
• The ion product (Q) of a salt is the product of the
concentrations of the ions in solution raised to the same
powers as in the solubility product expression.
• The ion product describes concentrations that are not
necessarily equilibrium concentrations, whereas Ksp describes
equilibrium concentrations.
• The process of calculating the value of the ion product and
comparing it with the magnitude of the solubility product is a
way to determine if a solution is unsaturated, saturated, or
supersaturated and whether a precipitate will form when
solutions of two soluble salts are mixed.
The Ion Product
• Three possible conditions for an aqueous solution
of an ionic solid
1. Q < Ksp: the solution is unsaturated, and more of the ionic solid will
dissolve
2. Q = Ksp: the solution is saturated and at equilibrium
3. Q > Ksp: the solution is supersaturated, and ionic solid will
precipitate
The Reaction Quotient in
Precipitation Reactions
•
•
The reaction quotient, Q, and the Ksp of a compound are used to calculate the concentration of
ions in a solution and whether or not a precipitate will form.
We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate,
Pb(NO3)2 solutions. Will a precipitate form?
1.
Write out the solubility expressions.
2. Calculate the Qsp for PbSO4.
– Assume that the solution volumes are additive.
– Concentrations of the important ions are:
3. Finally, calculate Qsp for PbSO4 and compare it to the Ksp.
The Reaction Quotient in
Precipitation Reactions
•
Suppose we wish to remove mercury from an aqueous solution that contains a soluble
mercury compound such as Hg(NO3)2. We can do this by precipitating mercury (II) ions as
the insoluble compound HgS. What concentration of sulfide ions, from a soluble compound
such as Na2S, is required to reduce the Hg2+ concentration to 1.0 x 10-8 M? For HgS, Ksp=3.0
x 10-53.
• What volume of the solution (1.0 x 10-8 M Hg2+ ) contains 1.0 g of
mercury?
• Equlibria that simultaneously involve two or more different equilibrium
constant expressions are simultaneous equilibria.
Simultaneous Equilibria Involving
Slightly Soluble Compounds
•
If 0.10 mole of ammonia and 0.010 mole of magnesium nitrate, Mg(NO3)2, are
added to enough water to make one liter of solution, will magnesium hydroxide
precipitate from the solution?
–
1.
For Mg(OH)2, Ksp = 1.5 x 10-11. Kb for NH3 = 1.8 x 10-5.
Calculate Qsp for Mg(OH)2 and compare it to Ksp.
–
Mg(NO3)2 is a soluble ionic compound so [Mg2+] = 0.010 M.
–
Aqueous ammonia is a weak base that we can calculate [OH-].
2. Once the concentrations of both the magnesium and hydroxide
ions are determined, the Qsp can be calculated and compared
to the Ksp.
3. Equlibria that simultaneously involve two or more different equilibrium
constant expressions are simultaneous equilibria.
Simultaneous Equilibria Involving
Slightly Soluble Compounds
• How many moles of solid ammonium chloride, NH4Cl, must be used to prevent
precipitation of Mg(OH)2 in one liter of solution that is 0.10 M in aqueous
ammonia and 0.010 M in magnesium nitrate, Mg(NO3)2 ? (Note the similarity
between this problem and Example 20-12.)
• Calculate the maximum [OH-] that can exist in a solution that is 0.010 M in Mg2+.
• Using the maximum [OH-] that can exist in solution, determine the number of
moles of NH4Cl required to buffer 0.10 M aqueous ammonia so that the [OH-]
does not exceed 3.9 x 10-5 M.
• Check these values by calculating Qsp for Mg(OH)2.
• Use the ion product for water to calculate the [H+] and the pH of the
solution.
In General How to Solve and Predict Reversible Reactions
1. Are the compounds involved strong or weak electrolytes
a. If strong then there is a 100% dissociation and no way back use a single
headed arrow and when adding the simultaneous equations together only and
add the products.
i.e. NaOH → Na+ + OHb. if a weak electrolyte use a double headed arrow and when adding
simultaneous equations together add both reactants and products
i.e. CH3COOH ↔ CH3COO- + H+
2. Determine if reactions are neutralization, dissociation, or equilibrium reactions
I. Neutralization
a. Class I – strong electrolyte/strong electrolyte: produces salt and
water. Use ICE table with units of amount (mmol or moles), cancel
terms, determine amounts of products. Stop.
b. Class II – weak electrolyte/weak electrolyte: Compare Kas or Kbs to
determine which predominates products or reactants. Use the equlibrium
equation and ICE table with units of concentration (mmol/mL or mol/L) to
determine equilibrium concentrations. Stop.
c. Class III – strong electrolyte/weak electrolyte: produces salt of the
conjugate and water and further actions, Dissociation or Equilibrium
reactions must be determined continue to 2 II
II. Dissociate any products and begin again at repeat step 1 then continue to 2 III.
Products of the dissociation though by name appear to be the same compounds
they are not identical in that they have a different source and are treated separately.
III. Equilibrium – write new stoichiometrically balanced chemical equilibrium
equations.
3. Write the equilibrium equation and solve for x and determine
[H+] or [OH-]
4. Find pH or pOH
5. Determine the pOH from pH or pH from pOH
6. Determine the [H+] or [OH-] not found in step 3
Ksp
Write a stoichiometrically balanced equation for the limited dissociation of the solid
If provided with the Ksp and looking for molar solubility of products
Write the stoichiometric ratio, i.e. 1:2
Multiply the ratio through by x, i.e. x:2x
Write the equilibrium equation, i.e. Ksp = [M2+][Y-]2
Substitute in the stoichiometrically determined x values for the [ ]s, Ksp = (x)(2x)2
Solve for x
If provided a mass determined to be in solution and looking for Ksp
convert mass to moles
convert moles to molarity
Write the equilibrium equation, i.e. Ksp = [M2+][Y-]2
Solve for Ksp