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Equilibrium, Acids and Bases
A. Dynamic Equilibrium
 equilibrium theories and principles apply to a variety
of phenomena in our world
eg)
blood gases in scuba diving
CO2 in carbonated beverages
buffers in our blood
 reactions are often reversible… which means that
not only are the products formed but the
reactants can be reformed
we use the double arrow to show this relationship
eg)
A + B⇌C + D
the forward and reverse reaction will proceed at
different rates…it depends on the concentration of the
reactants and products



if we start with only the reactants A and B, the
forward reaction will initially be the fastest
as it is the only reaction possible
as the products C and D are formed, the forward
reaction will slow down and the reverse
reaction will speed up
at some point, the rates of forward and reverse
reactions become equal
Dynamic Equilibrium
forward
reaction
equilibrium
Rate
reverse
reaction
0
Time
 a system is said to be in a state of dynamic
equilibrium when:
1. the rates of the forward and reverse reactions are
equal
2. the observable (macroscopic) properties of the
system, such as temperature, pressure, concentration,
pH are constant
3. the system is a closed system at
constant temperature
B. Classes of Reaction Equilibria
 there are three classes of chemical equilibria:
1. reactants favoured (percent rxn <50%)
<50%
A + B ⇌ C + D
2. products favoured (percent rxn >50%)
>50%
A + B ⇌ C + D
3. quantitative
to the right (percent rxn >99%)
>99%
A + B ⇌ C + D
or
A + B  C + D
C. The Equilibrium Constant
 experiments have shown that under a given set of
conditions (P and T) a specific quantitative
relationship exists between the equilibrium
concentrations of the reactants and products
 one reaction that has been studied intensively is that
between H2(g) and I2(g) (simple molecules and takes
place in gas phase no solvent necessary!)
H2(g) + I2(g) ⇌ 2 HI(g)
 when different combinations of H2(g), I2(g), and HI(g)
were mixed and the concentrations measured, it was
discovered that equilibrium was reached in
all cases
 even though the equilibrium [ ] are different , the
end quotient was the same each time (within
experimental error)
 this led to the empirical generalization known as the
Law of Equilibrium which says that there is a
constant ratio between the concentrations of the
products and the concentrations of the reactants at
equilibrium
 this law can be expressed mathematically:
For the reaction
aA + bB ⇌ cC + dD
The law is:
Kc = [C]c [D]d
[A]a [B]b
Kc = equilibrium constant
A, B = reactants
C, D = products
a, b, c, d = balancing coefficients
where:
 Kc is constant for a reaction at a given
temperature …if you change the temperature, Kc
also changes
 it is common to ignore the units for Kc and list it only
as a numerical value (since depends on the powers of
the various [ ] terms)
 when determining Kc use only the species that are in
gas or aqueous
***unless all states are the same, then use them all
 the higher the value of Kc, the greater the tendency
for the reaction to favor the forward direction
(the products)
 if Kc is greater than 1,
then the reaction is
products favoured
 if Kc is less than 1,
then the reaction is reactants
favoured
 Kc indicates the
percent reaction and not the
rate of the reaction
 catalysts will not affect the [ ] at equilibrium…
they only increase the rate of the rxn
Example 1
Write the equilibrium law for the reaction of nitrogen
monoxide gas with oxygen to form nitrogen dioxide gas.
2 NO(g) + O2(g) ⇌ 2 NO2(g)
Kc =
[NO2(g)]2
[NO(g)]2[O2(g)]
Example 2
Write the equilibrium law for the following reaction:
CaCO3(s) ⇌ CaO(s) + CO2(g)
Kc =
[CO2(g)]
*** do not include solids in Kc
Example 3
Write the equilibrium law for the following reaction:
2 H2O(l) ⇌ 2 H2(g) + O2(g)
Kc =
[H2(g)]2[O2(g)]
*** do not include liquids in Kc
Example 4
Phosphorus pentachloride gas can be decomposed into
phosphorus trichloride gas and chlorine gas.
a) Write the equilibrium law for this reaction.
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Kc = [PCl3(g) ][Cl2(g)]
[PCl5(g)]
b) If the [PCl5(g)]eq = 4.3 x 10-4 mol/L, the [PCl3(g) ]eq =
0.014 mol/L and the [Cl2(g)]eq = 0.014 mol/L then
calculate Kc.
Kc = [PCl3(g) ][Cl2(g)]
[PCl5(g)]
= (0.014)(0.014)
(4.3 x 10-4)
= 0.46
Example 5
Find the [SO3(g)] for the following reaction if Kc = 85.0
at 25.0C.
2 SO2(g)
+
0.500 mol/L
Kc
O2(g)
⇌
0.500 mol/L
=
[SO3(g) ]2____
[SO2(g)]2[O2(g)]
85.0 = [SO3(g) ]2
(0.500)2(0.500)
[SO3(g) ]2 = 10.625
[SO3(g) ] = 3.26 mol/L
2 SO3(g)
???
D. Graphical Analysis
 a graph of concentration vs. time
can be used
to see when equilibrium has been reached…as soon as
the concentrations don’t change any more , you
can read this time off the graph
Example 1
Consider this rxn: 2 SO2(g) + O2(g) ⇌ 2 SO3(g)
SO3(g)
SO2(g)
O2(g)
75
Concentration 50
(mol/L)
25
0
10
20
30
Time (s)
At what time does equilibrium get reached and what is the
value for Kc?
Kc =
[SO3]2
[SO2]2[O2]
=
(75)2
(50)2(25)
= 0.090
Equilibrium is reached at approximately
20 seconds.
E. Le Châtelier’s Principle
 Le Châtelier’s principle states that when a
chemical system at equilibrium is disturbed by a
change in property of the system, the system
adjusts in a way that opposes the change
 this takes place in a three-stage process
1. initial equilibrium state
2. shifting non-equilibrium state
3. new equilibrium state
 a system can be affected by a change in concentration,
temperature and or volume (pressure)
1. Concentration Changes
 an increase
in the [ ] of the products or reactants
favours the other side of the equation
 a decrease
favours
in the [ ] of the products or reactants
the same side of the equation
eg) N2(g) + 3 H2(g) ⇌ 2 NH3(g)
***Haber-Bosch process
↑ [N2(g)] will shift the equilibrium to the products
↑ [NH3(g)] will shift the equilibrium to the reactants
 [NH3(g)] will shift the equilibrium to the products
 changes in concentration have no effect on the value
of Kc
2. Temperature Changes
 energy is treated like a reactant or product
eg)
reactants + energy ⇌ products
reactants ⇌ products + energy
 if cooled, the equilibrium shifts so more heat is
produced (same side)
 if heated, the equilibrium shifts away from the
heat so it cools down (opposite side)
 a change in temperature is the only stress that
will change
the value of Kc!!!!!!!
 if the shift is towards the product side, Kc will
increase
 if the shift is towards the reactant side, Kc will
decrease
3. Volume and Pressure Changes
 with gases, volume and pressure are related (volume ,
pressure ↑)
 the concentration of a gas is related to volume
(pressure)…volume ↓ , concentration ↑
http://michele.usc.edu/java/gas/gassim.html
 an increase in [ ]
caused by a drop in volume
causes a shift towards the side of the equation with
fewer moles of gas
eg) N2(g) + 3 H2(g) ⇌ 2 NH3(g)
4 moles
2 moles
will shift to NH3(g)
 if the number of moles are the same on both sides of
the reaction, a change in volume (pressure) has
no effect
 changes in volume and pressure have no effect on
the value of Kc
4. Colour Changes
 in many equilibrium systems, the reactants will have a
different colour than the products
 predictions can be made about the equilibrium shift
and the resulting change in colour
Example
Use the following reaction to predict the equilibrium shift
and resulting colour change when the stresses are applied
2 CrO42(aq) + 2 H3O+(aq) ⇌ 2 Cr2O72(aq) + 3 H2O(l)
yellow
orange
a) a crystal of Na2CrO4(s) is added
products, orange
b) a crystal of K2Cr2O7(s) is added reactants, yellow
c) a few drops of concentrated acid is added
products, orange
d) water is removed products, orange
e) a few crystals of NaOH(s) are added
reactants, yellow
 all of the changes that can happen to systems in
equilibrium can be shown graphically:
Example
State what change to the equilibrium takes place at
each of the labelled parts of the graph:
Manipulations of An Equilibrium System
N2(g) + 3 H2(g) ⇌ 2 NH3(g) + energy
NH3(g)
N2(g)
Concentration
(mol/L)
H2(g)
A
B
Time
(min)
C
D
Stress
Equilibrium Time
A
addition of H2(g)
B
addition of inert gas, addition of catalyst
C
decrease in volume
D
increase in energy
F. ICE Tables
 we can use a table set-up to calculate the equilibrium
concentrations and/or Kc for any system
 you must be able to calculate all equilibrium [ ]
before you can use the equilibrium law
Example 1
Hydrogen iodide gas decomposes into hydrogen gas and
iodine gas. If 2.00 mol of HI(g) is place in a 1.00 L
container and allowed to come to equilibrium at 35C, the
final concentration of H2(g) is 0.214 mol/L. Find the
value for Kc.
2 HI(g)
Initial
⇌
2.00 mol/L
Change
–0.214 mol/L x 2/1
= –0.428
Equil.
1.572 mol/L
H2(g)
0
+
I2(g)
0
+0.214 mol/L +0.214 mol/L x 1/1
0.214 mol/L
+0.214 mol/L
Kc = [H2(g)][I2(g)]
[HI(g)]2
= (0.214)(0.214)
(1.572)2
= 0.0185
Example 2
In a 500 mL stainless steel reaction vessel at 900C, carbon
monoxide and water vapour react to produce carbon dioxide and
hydrogen. Evidence indicates that this reaction establishes an
equilibrium with only partial conversion of reactants to products.
Initially, 2.00 mol of each reactant is placed in the vessel. Kc for
this reaction is 4.20 at 900C. Calculate the concentration of each
substance at equilibrium.
CO(g)
+
H2O(g)
I
2.00/0.5L
= 4.00 mol/L
2.00/0.5L
= 4.00 mol/L
C
–x mol/L
–x mol/L
E
4.00  x mol/L
4.00  x mol/L
⇌ CO2(g)
+ H2(g)
0
0
+x mol/L x 1/1
x mol/L
+x mol/L x 1/1
x mol/L
Kc = [CO2(g)][H2(g)]
[CO(g)][H2O(g)]
4.20 =
(x)(x)
(4.00  x)( 4.00  x)
4.20
=
x2
(4.00  x)2
2.05
=
x
(4.00  x)
2.05(4.00  x) = x
8.05  2.05x = x
8.05 = 3.05x
x = 2.69 mol/L
***Note, this is a perfect
square so to solve for x
simply square root both
sides of the equation,
then solve
[CO(g)] = 4.00 – x = 4.00 – 2.69 = 1.31 mol/L
[H2O (g)] = 4.00 – x = 4.00 – 2.69 = 1.31 mol/L
[CO2(g)] = x = 2.69 mol/L
[H2(g)] = x = 2.69 mol/L
Example 3
Gaseous phosphorus pentachloride decomposes into gaseous
phosphorus trichloride and chlorine gas at a temperature
where Keq = 1.00  103. Suppose 2.00 mol of PCl5(g) in a
2.00 L vessel is allowed to come to equilibrium. Calculate
the equilibrium [ ] of each species.
PCl5(g)
I
C
E
⇌
2.00mol/2.00L
= 1.00 mol/L
–x mol/L x 1/1
1.00 – x mol/L
PCl3(g)
0 mol/L
+x mol/L
x mol/L
+
Cl2(g)
0 mol/L
+x mol/L x 1/1
x mol/L
Kc = [PCl3(g)][Cl2(g)]
[PCl5(g)]
1.00  10-3 = (x)(x)
(1.00 - x)
***at this point, you would have to use the
quadratic formula to solve for x
 when the concentrations are greater than 1000 X the
equilibrium constant, we can make an
approximation that greatly simplifies our
calculations
 if Kc is very small, the equilibrium doesn’t lie very far
to the right and  x is a very small number
***in this example 1.00 – x can be assumed to be 1.00
since x is really small, so…
1.00  10-3 =
(x)(x)
(1.00 )
x2 = 1.00  10-3 x 1.00
x = 0.0316
***now you can calculate the [ ]eq for each species
…substitute x into the equilibrium values in the ICE
table
[PCl5(g)]eq = 1.00 mol/L – 0.0316 mol/L = 0.967 mol/L
[PCl3(g)]eq = 0 + 0.0316 mol/L = 0.0316 mol/L
[Cl2(g)]eq = 0 + 0.0316 mol/L = 0.0316 mol/L
Example 4
Gaseous NOCl decomposes to form gaseous NO and Cl2. At
35C the equilibrium constant is 1.6  10-5. Calculate the
equilibrium [ ] of each species when 1.0 mol of NOCl is
placed in a 2.0 L covered flask.
2 NOCl(g)
I
C
E
⇌
1.0mol/2.0L = 0.50 mol/L
–x mol/L x 2/1
= –2x mol/L
0.50 – 2x mol/L
2 NO(g)
+
Cl2(g)
0 mol/L
0 mol/L
+x mol/L x 2/1
= +2x mol/L
+x mol/L
2x mol/L
x mol/L
Kc = [NO(g)]2[Cl2(g)]
[NOCl(g)]2
1.6  10-5 = (2x)2(x)
(0.50 - 2x)2 ***using
approximation,
0.50 – 2x = 0.50
1.6  10-5 = (4x 2)(x)
(0.50 )2
4x3 = 1.6  10-5 x 0.502
x3 = 4.0  10-6 / 4
x = 0.010 mol/L
[NOCl(g)]eq = 0.50 mol/L – (2)(0.010) mol/L = 0.48 mol/L
[NO(g)]eq = 0 + (2)(0.010 mol/L) = 0.020 mol/L
[Cl2(g)]eq = 0 + 0.010 mol/L = 0.010 mol/L
G. Ionization of Water
 the equilibrium of water can be written as follows:
H2O(l) + H2O(l)  H3O+(aq) + OH-(aq)
 the equilibrium law is:
Kc = [H3O+(aq) ][ OH-(aq)]
 the equilibrium constant for water is designated as
Kw
 at 25C, neutral water has
[H+(aq)] = [OH-(aq)] = 1.0  10-7 mol/L
 Kw = (1.0  10-7) (1.0  10-7)
= 1.0  10-14 (on pg 3 of Data Booklet)
 Kw is constant and therefore can be used to determine
the [H3O+(aq) ] or the [ OH-(aq)]
eg) if [H3O+(aq)] = 1.0  104 mol/L
then [OH(aq)] = 1.0 1014 = 1.0  1010 mol/L
1.0  104
if [H3O+(aq)] = [OH(aq)], then solution is neutral
if [H3O+(aq)] > [OH(aq)], then solution is
acidic
if [H3O+(aq)] < [OH(aq)], then solution is
basic
Try These:
1. [H3O+(aq)] = 1.0  109 mol/L basic
[OH(aq)] = 1.0  105 mol/L
2. [H3O+(aq)] = 1.0  101 mol/L acidic
[OH(aq)] = 1.0  1013 mol/L
3. [H3O+(aq)] = 1.0  1012 mol/L
[OH(aq)] = 1.0  102 mol/L
basic
4. [H3O+(aq)] = 6.3  109 mol/L
[OH(aq)] = 1.6  106 mol/L
basic
5. [H3O+(aq)] = 8.1  103 mol/L
[OH(aq)] = 1.2  1012 mol/L
acidic
6. [H3O+(aq)] = 3.6  108 mol/L
[OH(aq)] = 2.8  107 mol/L
basic
H. Review of pH and pOH
 the number of digits following the decimal place
in the pH value is equal to the number of sig digs
in the [H3O+(aq)]
pH =  log[H3O+(aq)]
pOH =  log[OH-(aq)]
[H3O+(aq)] = 10-pH
[OH-(aq)] = 10-pOH
pH + pOH = 14
Example 1
Find the pH of a solution where the
[H3O+(aq)] = 4.7  10-11 mol/L.
pH =  log[H3O+(aq)]
=  log(4.7  10-11)
= 10.33
Example 2
Find the pH of a solution where the [OH-(aq)] = 2.4  10-3
mol/L.
pOH =  log[OH(aq)]
=  log(2.4  103 )
= 2.619…
pH = 14 – pOH
= 14 – 2.619…
= 11.38
Example 3
Calculate the [H3O+(aq)] if the pH of the solution is 5.25.
[H3O+(aq)] = 10-pH
= 10-5.25
= 5.6  10-6 mol/L
Example 4
Calculate the pH of a solution where 10.3 g of Ca(OH)2(s)
is dissolved in 500 mL of water.
Ca(OH)2(s) 
Ca2+(aq)
+
2 OH-(aq)
m = 10.3 g
v = 0.500 L
M = 74.10 g/mol
n = 0.139…mol  2/1
n=mM
= 0.278…mol
= 10.3 g  74.10 g/mol
C=n V
= 0.139…mol
= 0.278…mol  0.500L
= 0.556…mol/L
Example 4 (continued)
pOH =  log[OH(aq)]
=  log(0.556… )
= 0.254…
pH = 14 – pOH
= 14 – 0.254…
= 13.745
I. Brønsted-Lowry Definition of Acids & Bases
(1923)
 this theory looks at the role of the acid or base
 an acid is a chemical species (anion, cation or
molecule) that loses a proton
 a base is a chemical species that gains a proton
 like in electrochemistry where e are transferred…now
we transfer H+
H+
HCl(aq) + H2O(l) ⇌
Cl-(aq) + H3O+(aq)
H+
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
 water does not have to be involved!
H+
HCl(g) + NH3(g) ⇌ NH4Cl(s)
 a Brønsted-Lowry acid doesn’t necessarily have to
produce an acidic solution…it depends on what accepts
the proton
 an acid/base reaction is a chemical reaction in which a
proton (H+) is transferred from an acid to a
base forming a new acid and a new base
 this theory explains how some chemical species can be
used to neutralize both acids and bases
eg)
HCO3-(aq) + H3O+(aq) ⇌ H2O(l) + H2CO3(aq)
HCO3-(aq) + OH-(aq) ⇌ H2O(l) + CO32-(aq)
 a substance that appears to act as a Brønsted-Lowry
acid in some rxns and a Brønsted-Lowry base in other
rxns is said to be amphiprotic or amphoteric
eg) H2O, HCO3, HSO4, HOOCCOO etc
J. Conjugate Acids and Bases
 a pair of substances that differ only by a proton is
called a conjugate acid-base pair…the acid is on
one side of the reaction and the base is on the other
 in general, the reaction can be shown as follows:
HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)
acid
base
conjugate acid conjugate base
 the stronger an acid, the weaker its conjugate base
 the weaker an acid, the stronger its conjugate base
K. Strengths of Acids and Bases
 two different acids (or bases) can have the same [ ] but
have different strengths
eg) 1 M CH3COOH(aq) and 1 M HCl(aq) will react in the
same way but not to the same degree
 the stronger the acid, the more electricity it conducts,
the lower the pH and the
other substances
faster it reacts with
1. Strong Acids
 acids that ionize quantitatively in water to form
H3O+(aq)
 percent rxn =100%
 the bigger the Ka (Kc for acids) the more the
products are favoured
 top 6 acids on the table (pg 8-9 in Data Book) have a
very large Ka
…note the H3O+ is the strongest acid on the chart
(leveling effect)…all strong acids react to form
H3O+(aq) so it is the strongest
 when calculating pH, the [SA] = [H3O+(aq)]
so use
pH = -log[H3O+(aq)]
Example
What is the pH of a 0.500 mol/L solution of HNO3(aq)?
[H3O+(aq)] = [HNO3(aq)] = 0.500 mol/L
pH = -log[H3O+(aq)]
= -log(0.500 mol/L)
= 0.301
2. Weak Acids
 a weak acid is one that only partially ionizes in
water to form H+(aq) ions
 most ionize <50%
 Ka value is small (<1)
 to calculate pH, you need to use the Ka value
cannot use just the [WA]
100% dissociated
because it is not
…you
 the Ka law is an
equilibrium law and is devised
the same way we did Kc
eg)
HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)
Ka = [H3O+(aq)][A-(aq)]
[HA(aq)]
 you will be required to figure out the [H3O+(aq)]
before you can calculate the pH
 you have the [WA] and the Ka value but you don’t
have the [A-(aq)]
 since the mole ratio for [H3O+(aq)]:[A-(aq)] is 1:1 ,
they have the same [ ] (this is a dissociation !)
Ka = [H3O+(aq)]2
[HA(aq)]
 now you can solve for x to get the [H3O+(aq)]
[H3O+(aq)] = (Ka)([WA])
Example 1
What is the pH of a 0.10 mol/L acetic acid solution?
***check in DB…weak acid!!!!!
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO-(aq)
[H3O+(aq)] = (Ka)([WA])
= (1.8 x 10-5 mol/L )(0.10 mol/L)
= 1.34… 10-3 mol/L
pH = -log[H3O+(aq)]
= -log(1.34… 10-3 mol/L)
= 2.87
Example 2
What is the pH of a 1.0 mol/L acetic acid solution?
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO-(aq)
[H3O+(aq)] = (Ka)([WA])
= (1.8 x 10-5 mol/L )(1.0 mol/L)
= 4.24… 10-3 mol/L
pH = -log[H3O+(aq)]
= -log(4.24… 10-3 mol/L)
= 2.37
Example 3
A 0.25 mol/L solution of carbonic acid has a pH of 3.48.
Calculate Ka.
H2CO3(aq) + H2O(l) ⇌ H3O+(aq) + HCO3-(aq)
[H3O+(aq)] = 10-pH
= 10-3.48
= 3.31…  10-4 mol/L
Ka =
[H3O +(aq)]2
[H2CO3 (aq)]
= (3.31…x10-4 mol/L)2
0.25 mol/L
= 4.4 x 10-7 mol/L
 the % reaction (% ionization) can be written as a
% above the
eg) CH3COOH(aq)
⇌ in a chemical reaction:
1.3%
+ H2O(l) ⇌ H3O+(aq) + CH3COO-(aq) Ka = 1.8 x 10-5
 the % reaction be calculated using [H3O+] and [WA]
% ionization = [H3O+(aq)]  100
[WA(aq)]
Example 1
Calculate the % ionization for a 0.500 mol/L solution of
hydrosulphuric acid if the [H+(aq)] is 5.0  10-4 mol/L.
% ionization = [H3O +(aq)]  100
[WA(aq)]
= 5.0  10-4 mol/L  100
0.500 mol/L
= 0.10 %
Example 2
The pH of a 0.10 mol/L solution of methanoic acid is
2.38. Calculate the % ionization.
[H3O+] = 10-pH
= 10-2.38
= 0.00416… mol/L
% ionization = [H3O +(aq)]  100
[WA(aq)]
= (0.00416…mol/L) x (100)
0.10 mol/L
= 4.2 %
3. Strong Bases
 according to Arrhenius, bases are substances that
increase the hydroxide [ ] of a solution
 all ionic hydroxides are strong bases
 percent rxn = 100%
 strength depends on # of hydroxide ions
…
Ba(OH)2(aq) is a stronger base than NaOH (aq) at
the same [ ] because it produces 2 OH-(aq)
 [OH-(aq)] = x[BH(aq)]
where x is the number of
ions (think about the dissociation
hydroxide
equation!)
Example
Calculate the pH of a 0.0600 mol/L solution of
Ca(OH)2(aq).
[OH-(aq)] = x[BH(aq)]
= x[Ca(OH)2(aq)]
= 2(0.0600 mol/L)
= 0.120 mol/L
pOH = log(0.120)
= 0.9208…
pH = 14 – 0.9208…
= 13.079
4. Weak Bases
 do not dissociate completely
weak acids
in water…just like
 Kb is the dissociation constant or equilibrium
constant for bases
B + H2O(l) ⇌ BH+(aq) + OH-(aq)
Kb = [BH+(aq)][OH-(aq)]
[WB(aq)]
 you can calculate Kb using Ka
for the base
 Ka  Kb = KW = 1.00  1014
eg) Calculate Kb for SO42(aq).
K b = KW
Ka
= 1.00  1014
1.0  102
= 1.0  1012 mol/L
 once you have Kb, you can then solve for [OH(aq)]
using the equilibrium law (just like with weak acids!)
Kb = [BH+(aq)][OH-(aq)]
[WB(aq)]
Kb = [OH-(aq)]2
[WB(aq)]
 now you can solve for [OH(aq)]
[OH-(aq)] = (Kb)([WB])
Example
Find the pH of a 15.0 mol/L NH3(aq) solution.
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
Ka = 5.6  10-10 mol/L
Kb = K w
Ka
= 1.00  10-14
5.6  10-10
= 1.78…  10-5 mol/L
[OH-(aq)] = (Kb)([NH3(aq)])
[OH-(aq)] = (1.78…  10-5)(15.0)
[OH-(aq)] = 1.63… x 10-2 mol/L
pOH = -log[OH-(aq)]
= -log(1.63…x 10-2 mol/L)
= 1.78…
pH = 14 – pOH
= 14 – 1.78…
= 12.214
L. Predicting Acid-Base Equilibria
 acids are listed in order of decreasing strength on
the left side and bases are listed in order of
increasing strength on the right side
 when predicting reactions, the substance with the
greatest attraction for protons (the strongest
base) will react with the substance that gives up
its proton most easily (strongest acid)
 we will assume that only one proton is transferred
per reaction
 to predict the acid-base reaction, follow the following
steps:
Steps
1. List all species (ions, atoms, molecules) initially
present.
Note:  strong acids ionize into H3O+ and the
anion
 weak acids are NOT dissociated
 dissociate ionic compounds
 don’t forget to include water
2. Identify all possible acids and bases.
3. Identify the strongest acid (SA)
and
strongest base (SB)…like redox rxns the
SA is top left and the SB is bottom right.
4. To write the reaction, transfer one proton from
the acid to the base to predict the conjugate acid
and conjugate base.
5. Predict the position of the equilibrium.
Note:  if acid is above base, then >50%
(favours products) ⇌
 if base is above acid, then <50%
(favours reactants) ⇌
Example 1
Predict the acid-base reaction that occurs when sodium
hydroxide is mixed with vinegar.
List:
OH-(aq) +
Na+(aq)
OH-(aq) CH3COOH(aq)
A
B
SB
SA
CH3COOH(aq) ⇌
H2O(l)
H2O(l)
A/B
+ CH3COO-(aq)
Example 2
Predict the acid-base reaction when ammonia is mixed
with HCl(aq).
List:
NH3(aq
B
SB
NH3(aq +
)
H3O+(aq)
A
SA
Cl-(aq)
B
H2O(l)
A/B
H3O+(aq) ⇌
H2O(l)
+ NH4+ (aq)
M. Monoprotic vs. Polyprotic Acids and Bases
 an acid capable of donating only one proton is called
monoprotic
eg) HCl(aq), HNO3(aq), HOCl(aq) etc.
 if an acid can transfer more than one proton, it is
called polyprotic ( diprotic if 2 protons,
triprotic if 3 protons)
eg) Label each of the following acids as monoprotic
or polyprotic:
1. H2SO4(aq)
polyprotic
2. HOOCCOOH(aq) polyprotic
3. HCOOH(aq)
monoprotic
4. CH3COOH(aq)
monoprotic
5. H2PO4-(aq)
polyprotic
6. NH4+(aq)
monoprotic
 a base capable of accepting only one proton is called
a monoprotic base
 a base that can accept more than one proton is called
a polyprotic base ( diprotic or triprotic )
eg) PO43-(aq) can accept up to 3 H+ to form
HPO42-(aq), H2PO4-(aq), and H3PO4(aq)
respectively
eg) Label each of the following as monoprotic or
polyprotic acids, monoprotic or polyprotic bases:
1. HSO4-(aq)
2. H2PO4-(aq)
3. HPO42-(aq)
4. HCO3-(aq)
5. H2O(l)
monoprotic acid; monoprotic base
polyprotic acid; monoprotic base
monoprotic acid; polyprotic base
monoprotic acid; monoprotic base
monoprotic acid; monoprotic base
 reactions involving polyprotic acids or polyprotic
bases substances involve the same principles of
reaction prediction
 only one proton
is transferred at a time and
always from strongest acid to strongest base
Example 1
Potassium hydroxide is continuously added to oxalic acid
until no more reaction occurs.
List: K+(aq
)
2OH-(aq) HOOCCOOH(aq) H2O(l) HOOCCOO-(aq) OOCCOO
B
SB
A
SA
A/B
A/B
SA
(aq)
B
OH-(aq) + HOOCCOOH(aq)⇌
H2O(l) + HOOCCOO-(aq)
OH-(aq) + HOOCCOO-(aq) ⇌
H2O(l) + OOCCOO2-(aq)
 Net Reaction: Add all reactions together (only if all
quantitative), cancelling out any species that occur
in the same quantity on both the reactant and
product sides and summing any species that occur
more than once on the same side
OH-(aq) + HOOCCOOH(aq)
H2O(l) + HOOCCOO-(aq)
OH-(aq) + HOOCCOO-(aq)
H2O(l) + OOCCOO2-(aq)
2 OH-(aq)+ HOOCCOOH(aq)
2 H2O(l) + OOCCOO2-(aq)
Example 2
Sodium hydrogen phosphate is titrated with hydroiodic
acid. If we assume all steps are quantitative, give the net
reaction.
List:
Na+(aq) HPO42-(aq)
A/B
I-(aq)
B
SB
H3O+(aq) H2O(l)
A
SA
A/B
H2PO4-(aq) H3PO4(aq)
A/B
A
SB
H3O+(aq) + HPO42-(aq)
H2O(l) + H2PO4-(aq)
H3O+(aq) + H2PO4-(aq)
H2O(l) + H3PO4(aq)
2 H3O+(aq) + HPO42-(aq)
2 H2O(l)+ H3PO4(aq)
O. Titrations
 titrations are used to determine the pH of the
endpoint
of acid-base reactions
 the information from the titration can be plotted on a
graph, buffer regions can be analyzed and
stoichiometric calculations can be performed
1. pH Curves
 a pH curve is a graph showing the
continuous
change of pH during an acid-base reaction
 graph of pH (y-axis) vs volume of titrant
added to sample (x-axis)
 the endpoint
is the point (usually shown by a
change in indicator colour) when the reaction has
gone to completion
 the equivalence point
is the volume of titrant
required for the reaction to go to completion
 they contain a relatively flat region called the
buffer region
 all pH curves have 4 major features:
 the initial pH of the curve must be the
pH of the sample
 the co-ordinate of the equivalence
point must be correct in terms of pH
and volume
 the “over-titration” must be asymptotic
with the pH of the titrant
 number of equivalence points must
match the number of quantitative
reactions occurring
 titrant selection:
 if the sample is an acid, titrant should be a
strong base such as NaOH(aq) or KOH(aq)
 if the sample is a base, titrant should be
HCl(aq)
 you need to be able to interpret pH curves:
1. Strong Monoprotic with Strong Monoprotic
 pH of 7
at the equivalence point
SA titrated with SB
SB titrated with SA
14
14
pH EP 7
pH EP7
0
0
volume
volume
2. Weak Monoprotic with Strong Monoprotic
 if weak acid, then pH of >7 at equivalence point
 if weak base, then pH of
<7 at equivalence point
 bottom “flat” region is not as flat as with strong
acid/strong base
WA titrated with SB
WB titrated with SA
14
pH
14
EP7
0
pH
volume
EP
7
0
volume
3. Polyprotic with Strong Monoprotic
 more than 1 equivalence point
WA(poly) titrated with SB
WB(poly) titrated with SA
14
14
EP2
EP1
7
pH
EP2
pHEP1
0
7
volume
0
volume
2. Indicators
 an indicator is a substance that
changes colour
when it reacts with an acid or base and are usually
weak acids themselves
 they exist in one of two conjugate forms that are
reversible and distinctly different in color
HIn(aq) + H2O(l) ⇌
acid
eg) litmus red
base
In-(aq) + H3O+(aq)
conjugate
base
blue
conjugate
acid
 recall that to show the equivalence point of an acid-
base titration, choose an indicator:
1. whose colour change range includes the
equivalence point of the titration
2. that will react right after the sample reacts
…this means the indicator is a weaker acid or
base than the sample
3. Buffers
are chemicals that, when added to water,
protect the solution from large pH changes
when acids or bases are added to them
 buffers
they are used to calibrate pH meters
and
control the rate of pH sensitive reactions (eg.
in the blood)
 typical buffers are solutions containing relatively
large amounts of conjugate pairs such as a
weak acid and the salt of the conjugate base
eg) H2CO3 and NaHCO3
 can be selected by using pKa = –logKa
…this tells
you the pH at which the buffer is most useful
eg) Choose a buffer that would be useful for each of the
following solutions:
1. pH of 7.0
H2S – HS-
pKa = -log(8.9 × 10-8) = 7.1
2. pH of 4.5 CH3COOH – CH3COOpKa = -log(1.8 × 10-5) =
4.7
3. pH of 10.0 HCO3- – CO32pKa = -log(4.7 × 10-11) = 10.3
 the acid in the conjugate pair of the buffer protects
against any base added
 the base in the conjugate pair of the buffer protects
against any acid added
 buffers can be overwhelmed
too much acid or base
by the addition of
Example
Using an acetic acid – sodium acetate buffer system, show
what happens when:
a) a small amount of HCl(aq) is added
CH3COOH(aq) Na+(aq) CH3COO-(aq) H3O+(aq) Cl-(aq) H2O(l)
A
–
B
SB
CH3COO-(aq) + H3O+(aq) ⇌
A
SA
B
A/B
CH3COOH(aq) + H2O(l)
b) a small amount of NaOH(aq) is added
CH3COOH(aq) Na+(aq) CH3COO-(aq) OH-(aq)
A
SA
–
B
B
SB
H2O(l)
A/B
CH3COOH(aq)+ OH-(aq) ⇌ CH3COO-(aq) +
H2O(l)