Equilibrium, Acids and Bases
A. Dynamic Equilibrium
reactions are often reversible… which means that not
only are the products formed but the reactants can be
reformed
and C + D  A + B
eg) A + B  C + D
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 the rates of the forward and reverse reactions are
equal and we observe no macroscopic ( visible ) changes
(system is a closed system)
B. Classes of Reaction Equilibria
 the rate of a reaction depends on:
1. the EK (temperature) of the particles
2. orientation of collisions (shapes of particles)
3. # of collisions/second (concentration)
 there are 4 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
4. quantitative to the left (percent rxn <1% )
<1%
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!)
 when different combinations of H2(g), I2(g), and HI(g) were
mixed and the concentrations measured, it was discovered
that an equilibrium was reached in all cases:
H2(g) + I2(g) ⇌ 2 HI(g)
 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
 this law can be expressed mathematically:
For the reaction
aA + bB ⇌ cC + dD
The law is:
Keq = [C]c [D]d
[A]a [B]b
 Keq is the equilibrium constant …it is constant for the
reaction at a given temperature
 it is common to ignore the units for Keq and list it only as a
numerical value (since depends on the powers of the various
[ ] terms)
 when determining Keq use only the species that are in
gas or aqueous phase
***unless all states are the same, then use them all
 the higher the value of Keq, the greater the tendency for the
reaction to favor the forward direction (the products)
 Keq 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)
Keq =
[NO2(g)]2
[NO(g)]2[O2(g)]
Example 2
Write the equilibrium law for the following reaction:
CaCO3(s) ⇌ CaO(s) + CO2(g)
Keq =
[CO2(g)]
*** do not include solids in Keq
Example 3
Write the equilibrium law for the following reaction:
2 H2O(l) ⇌ 2 H2(g) + O2(g)
Keq =
[H2(g)]2[O2(g)]
*** do not include liquids in Keq
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)
Keq = [PCl3(g) ][Cl2(g)]
[PCl5(g)]
b) If the [PCl5(g)] = 4.3 x 10-4 mol/L, the [PCl3(g) ] = 0.014
mol/L and the [Cl2(g)] = 0.014 mol/L then calculate Keq.
Keq = [PCl3(g) ][Cl2(g)]
[PCl5(g)]
= (0.014 mol/L)(0.014 mol/L)
(4.3 x 10-4 mol/L)
= 0.46
Example 5
Find the [SO3(g)] for the following reaction if Keq = 85.0 at
25.0C.
2 SO2(g)
+
O2(g)
⇌
2 SO3(g)
0.500 mol/L
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
Keq
=
Your Assignment: 1. pgs 1-2 in workbook
2. pg 431 #7; pg 434 #10
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
2 SO2(g) + O2(g) ⇌ 2 SO3(g)
Consider this rxn:
Concentration
(mol/L)
SO3(g)
SO2(g)
75
50
25
O2(g)
0
10
20
30
Time (s)
At what time does equilibrium get reached and what is the value
for Keq?
Keq =
[SO3]2
[SO2]2[O2]
=
(75)2
(50)2(25)
= 0.090
.
Equilibrium is reached at approximately 20
seconds.
Your Assignment: pg 3 in workbook
E. ICE Tables
 we can use a table set-up to calculate the equilibrium
concentrations and/or Keq for any system
 you must be able to calculate all equilibrium [ ]
before you can use the equilibrium law
Example 1
Find the value for Keq for the following data:
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
+0.214 mol/L
0.214 mol/L
+
I2(g)
0
+0.214 mol/L x 1/1
+0.214 mol/L
***Now you can use these equilibrium [ ]’s to
calculate Keq
Keq = [H2(g)][I2(g)]
[HI(g)]2
= (0.214)(0.214)
(1.572)2
= 0.0185
Example 2
Phosphorus pentachloride gas decomposes into phosphorus
trichloride gas and chlorine gas. If the [PCl5(g)]i = 8.1 x 10-3 mol/L
and the [PCl3(g)]i = 0.298 mol/L, calculate the equilibrium [ ] of all
chemical species and the Keq. The [Cl2g]eq = 2.00 x 10-3 mol/L.
PCl5(g)
I
⇌
8.1 x 10-3 mol/L
PCl3(g)
0.298 mol/L
+
Cl2(g)
0 mol/L
C –2.00 x 10-3 mol/L x 1/1 +2.00 x 10-3 mol/L x 1/1 +2.00 x 10-3 mol/L
E
6.1 x 10-3 mol/L
0.300 mol/L
2.00 x 10-3 mol/L
Keq = [PCl3(g)][Cl2(g)]
[PCl5(g)]
= (0.300)(2.00 x 10-3)
(6.1 x 10-3)
= 9.8 x 10-2
 you will get questions where only initial [ ] are given
 use “x”
to represent the change in concentration
 once you calculate x, you can +/- from the [ ]i to get the [ ]eq
Example 1
PCl5(g) decomposes into PCl3(g) and Cl2(g) 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
⇌
2.00mol/2.00L = 1.00 mol/L
PCl3(g)
0 mol/L
C
–x mol/L x 1/1
+x mol/L
E
1.00 – x mol/L
0 + x mol/L
= x mol/L
+
Cl2(g)
0 mol/L
+x mol/L x 1/1
0 + x mol/L
= x mol/L
Keq = [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 Keq 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 2
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
⇌
1.0mol/2.0L = 0.50 mol/L
2 NO(g)
+
Cl2(g)
0 mol/L
0 mol/L
C
–x mol/L x 2/1
= –2x mol/L
+x mol/L x 2/1
= +2x mol/L
+x mol/L
E
0.50 – 2x mol/L
0 + 2x mol/L
= 2x mol/L
0 + x mol/L
= x mol/L
Keq = [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
Your Assignment: pgs 4-5 in workbook
F. 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 opposite side
 a decrease in the [ ] of the products or reactants favours
the same side
eg) N2(g) + 3 H2(g) ⇌ 2 NH3(g)
↑ [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
 this process is used in industry to produce desired products
eg) the Haber Process for making ammonia from nitrogen
and hydrogen
 also happens in our bodies in our blood
eg)
Hb + O2(g) ⇌ HbO2
***as the [O2(g)] increases in our lungs there is a shift
towards the product (HbO2)
2. Temperature Changes
 energy is treated like a reactant or product
eg)
reactants + energy ⇌ products
reactants ⇌ products + energy
 if cooled, the equilibrium shifts to the side with energy
 if heated, the equilibrium shifts to the side without energy
3. Volume (Pressure) Changes
 with gases , volume and pressure are related
 the concentration of a gas is related to volume (pressure)
http://michele.usc.edu/java/gas/gassim.html
 an increase in [ ] caused by a decrease in volume
(increase in pressure) causes a shift towards the side of the
equation with fewer moles
eg) N2(g) + 3 H2(g) ⇌ 2 NH3(g)
4 moles
2 moles
 if the number of moles are the same on both sides of the
reaction, then there is no shift in the equilibrium
Your Assignment: 1. pgs 6-7 in workbook
2. pg 440 #12-15
 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
Your Assignment: pg 8 in workbook
G. Ionization of Water
 the equilibrium of water can be written as follows:
H2O(l) ⇌ H+(aq) + OH-(aq)
 the equilibrium law is:
Kc = [H+(aq)][ OH-(aq)]
 the equilibrium constant for water is designated as Kw
Kw = [H+(aq)][ OH-(aq)]
 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 always constant and therefore can be used to
determine the [H+(aq)] or the [ OH-(aq)]
 it is unlikely that H+(aq) (which basically is a proton ) would
exist alone in water
 it is attracted to the negative end of a water molecule
forming the hydronium ion, H3O+(aq) (also called the
“hydrated proton”)
 therefore H3O+(aq) is
interchangeable with H+(aq)
H. pH and pOH
 in 1909, Soren Sorenson devised the pH scale
 it is used because the [H3O+(aq)] is very small and
cumbersome to write
 logarithmic scale based on whole numbers that are powers
of 10
***this means that a change of 1 on the pH scale is a
10-fold change in [H3O+(aq)]
HCl(aq) acid rain
0 1
strong acid
3
pure
H2O blood
7
8
NaOH(aq)
14
strong base
 the pH of a solution can be found by using the [H3O+(aq)]
(which is the same thing as the [H+(aq)] )
 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)]
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.
Kw = [H3O+(aq)] [OH-(aq)]
1.0  10-14 = [H3O+(aq)] (2.4  10-3)
[H3O+(aq)] = 4.16… 10-12 mol/L
pH =  log[H3O+(aq)]
=  log(4.16…  10-12 mol/L)
= 11.38
Example 3
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)
m = 10.3 g
M = 74.10 g/mol
n=m M
= 10.3 g  74.10 g/mol
= 0.139…mol
+
2 OH-(aq)
v = 0.500 L
n = 0.139…mol  2/1
= 0.278…mol
C=n V
= 0.278…mol  0.500L
= 0.556…mol/L
Example 3 (continued)
Kw = [H3O+(aq)] [OH-(aq)]
1.0  10-14 = [H3O+(aq)] (0.566…)
[H3O+(aq)] = 1.79… 10-14 mol/L
pH =  log[H3O+(aq)]
=  log(1.79…  10-14 mol/L)
= 13.745
 you could also be given the pH and asked to calculate the
[H3O+(aq)]
 use the number of decimal places in the pH value to
determine the sig digs in your answer
[H3O+(aq)] = 10-pH
Example 1
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
 just as pH deals with [H+(aq)], pOH
deals with [OH-(aq)]
***p just means -log
 at SATP… pH + pOH = 14
 to calculate the pOH or the [OH-(aq)], use the same formulas
as pH but substitute the OH-(aq) values:
pOH =  log[OH-(aq)]
and
[OH-(aq)] = 10-pOH
Your Assignment: pgs 9-10 in workbook
 if you dilute a solution, the pH will increase
and the
pOH will decrease
 to find how many times stronger one solution is over
another, subtract the pH’s then go 10x where x is the
difference
eg) How many times more acidic is a solution with a
pH of 3.45 over a solution with a pH of 6.20?
6.20 – 3.45 = 2.75
10x = 102.75 = 562 X more acidic
Acids
Bases
 electrolyte
 electrolyte
 taste sour
 taste bitter
 litmus turns red (pink)
 litmus turns blue
 pH<7
 pH>7
 neutralize bases
 neutralize acids
 reacts with metals (H2(g))
and carbonates (CO2(g))
 vinegar, lemon juice
 tums, ammonia
2. Arrhenius Definition(1887)
 acids dissociate into H+(aq) ions
 bases dissociate into OH-(aq) ions
 limitation: some species predicted to be acids by
Arrhenius are actually bases, some things are
unpredictable
eg) HCO3-(aq)
3. Modified Arrhenius Definition
 it is unlikely that H+(aq) (which basically is a proton) would
exist alone in water…it would form the hydronium ion,
H3O+(aq)
 the modified definition gives us the following:
acids
HA(aq) +
bases
B(aq) +
(other than hydroxides)
H2O(l) ⇌
H2O(l) ⇌
H3O+(aq) +
BH+(aq) +
A-(aq)
OH-(aq)
 limitation: when writing these acid and base formulas
you can run into the situation where two equations can
be written for the same substance and you must have
water present
4. Brønsted-Lowry Definition (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) bicarbonate ions, hydrogen sulphate ions, water
Your Assignment: pg 11 #1-4 in workbook
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
rxn 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
Example 1
What is the pH of a 0.10 mol/L acetic acid?
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO-(aq)
 CH3COOH is the acid and CH3COO- is the conjugate base
 H2O is the base and H3O+ is the conjugate acid
Your Assignment: pg 11 #1,3,4,5 in workbook
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 faster it reacts with other
substances
1. Strong Acids
 acids that ionize quantitatively in water to form H3O+(aq)
 percent rxn = 100%
 the bigger the Ka (Keq for acids) the more the products
are favoured
 top 6 acids on the table (pg 11 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. Strong Bases
 according to Arrhenius, bases are substances that increase
the hydroxide [ ] of a solution
 all ionic hydroxides are strong bases
 percent rxn = 100%
eg) NaOH(aq)  Na+(aq) + OH-(aq)
Ba(OH)2(aq)  Ba2+(aq) + 2 OH-(aq)
 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
hydroxide ions (think about the dissociation equation!)
Example
Calculate the [OH-(aq)] of a 0.600 mol/L solution of
Ca(OH)2(aq).
[OH-(aq)] = x[BH(aq)]
= x[Ca(OH)2(aq)]
= 2(0.600 mol/L)
= 1.20 mol/L
3. 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 …you cannot
use just the [WA] because it is not 100% dissociated
 the Ka law is an equilibrium law and
is devised the same
way we did Keq
eg)
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO-(aq)
Ka = [H3O+(aq)][CH3COO-(aq)]
[CH3COOH(aq)]
 you will be required to figure out the [H3O+(aq)] before you
can calculate the pH
 you have the [WA]
the [A-(aq)]
and the Ka value but you don’t have
 since the mole ratio for [H3O+](aq):[A-(aq)] is 1:1 , they
have the same [ ] (this is a dissociation !)
eg)
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO-(aq)
x
x
Ka = [H3O+(aq)][CH3COO-(aq)]
[CH3COOH(aq)]
Ka =
(x)(x)
[CH3COOH(aq)]
Ka =
x2
[CH3COOH(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)
= 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)
= 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
[H3O +(aq)]2
[H2CO3 (aq)]
= (3.31…x10-4 mol/L)2
0.25
= 4.4 x 10-7 mol/L
Ka =
 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 %
4. Weak Bases
 do not dissociate completely in water…just like weak
acids
B + HOH(l) ⇌ BH+(aq) + OH-(aq)
 Kb is the dissociation constant or equilibrium constant for
bases
 you need to use two things to calculate [OH-(aq)]:
1. use the Kb expression from the dissociation equation
2. use the fact that Ka  Kb = 1.00  10-14
[OH-(aq)] = (Kb)([WB])
Example
You have a 15.0 mol/L NH3(aq) solution.
a) Find Kb.
Ka = 5.6  10-10 mol/L
Kb = K w
Ka
= 1.00  10-14
5.6  10-10
= 1.8  10-5 mol/L
b) Find the pH.
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
Kb = [NH4+( aq)][OH-(aq)]
[NH3(aq)]
[OH-(aq)] = (Kb)([NH3(aq)])
[OH-(aq)] = (1.8  10-5)(15.0)
[OH-(aq)] = 1.63… x 10-2 mol/L
pOH = -log[OH-(aq)]
pH = 14 – pOH
= -log(1.63…x 10-2 mol/L)
= 14 – 1.78…
= 1.78…
= 12.214
Your Assignment: pgs 12-13 in workbook
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
 don’t forget to include water
 dissociate ionic compounds
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, the <50% (favours
reactants) ⇌
 if H3O+ and the base is stronger than (below or
including) F-, then stoichiometric 
 if OH- and the acid is stronger than (above or
including) H2PO4-, then stoichiometric 
Example 1
Predict the acid-base reaction that occurs when sodium
hydroxide is mixed with vinegar.
List:
OH-(aq) +
Na+(aq)
OH-(aq)
B
SB
CH3COOH(aq)
CH3COOH(aq)
A
SA
H2O(l)
H2O(l)
A/B
+ CH3COO-(aq)
*** Reaction has OH-(aq), and CH3COOH(aq) is higher than
H2PO42-(aq) rxn is stoichiometric
Example 2
Predict the acid-base reaction when ammonia is mixed with
HCl(aq).
List:
NH3(aq
B
SB
H3O+(aq)
A
SA
Cl-(aq)
B
H2O(l)
A/B
NH3(aq) +
H3O+(aq)
H2O(l)
+ NH4+ (aq)
Your Assignment: pgs 14-15 in workbook
M. Polyprotic and Polybasic Substances
 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)…the conjugate base also appears on the acid list
on pg 11
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
monobasic
 a base that can accept more than one proton is called
polybasic ( dibasic or tribasic )
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, monobasic or polybasic:
monoprotic acid; monobasic
1. HSO4-(aq)
polyprotic acid; monobasic
2. H2PO4-(aq)
monoprotic acid; polybasic
3. HPO42-(aq)
monoprotic acid; monobasic
4. HCO3-(aq)
5. H2O(l)
monoprotic acid; monobasic
 reactions involving polyprotic acids or polybasic
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)
OH-(aq)
B
SB
HOOCCOOH(aq) H2O(l) HOOCCOO-(aq) OOCCOO2-(aq)
A
SA
A/B
A/B
SA
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)
I-(aq)
A/B
B
SB
H3O+(aq)
H2O(l)
A
SA
A/B
H2PO4-(aq)
A/B
SB
H3PO4(aq)
A
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)
Your Assignment: pg 15 in workbook
O. Acid-Base Stoichiometry
 use the same rules of stoich that we always have
 write the net reaction first using the acid-base reaction
prediction rules (List, identify A, B, SA, SB etc)
 if the question says “at the second endpoint”, then you know
the reaction is dealing with something that is polyprotic or
polybasic …determine all reaction steps then use the
net reaction to do the stoich calculation
Example 1
A 25.0 mL NH3(aq) solution is titrated with 0.100 mol/L HCl(aq).
Calculate the concentration of the NH3(aq) using the following
data: final buret reading: 16.30 mL
initial buret reading: 0.60 mL
NH3(aq) H3O+(aq) Cl-(aq) H2O(l)
SB
SA
B
A/B
NH3(aq)
+
V = 0.0250 L
C=?
n = 0.001570 mol × 1/1
C = n/V
= (0.001570 mol)
0.0250 L
= 0.0628 mol/L
H3O+(aq) ⇌
H2O(l) +
v = 16.30 mL – 0.60 mL
= 15.70 mL
= 0.01570 L
C = 0.100 mol/L
n = CV
= (0.100 mol/L)/(0.01570L)
= 0.001570 mol
NH4+(aq)
Example 2
What is the volume of 0.500 mol/L KOH(aq) needed to react
with 10.0 mL of 0.500 mol/L nitric acid?
NO3-(aq)
B
H3O+(aq)
+
H3O+(aq)
SA
K+(aq)
–
OH-(aq)
OH-(aq)
SB
⇌
H2O(l)
A/B
2 H2O(l)
V = 0.0100 L
V=?
C = 0.500 mol/L
C = 0.500 mol/L
n = CV
n = 0.00500 mol X 1/1
= (0.500 mol/L)(0.0100L)V = n
= 0.00500 mol
C
= 0.00500 mol
0.500 mol/L
= 0.0100 L
Example 3
What is the pH of a mixture of 50.0 mL of 0.30 mol/L HCl(aq)
with 30.0 mL of 0.40 mol/L NaOH(aq)?
Cl-(aq)
B
H3O+(aq)
+
H3O+(aq)
SA
Na+(aq) OH-(aq)
–
SB
OH-(aq)
⇌
H2O(l)
A/B
2 H2O(l)
V = 0.0500 L
V = 0.0300 L
C = 0.30 mol/L
C = 0.40 mol/L
n = CV
n = CV
= (0.30 mol/L)(0.0500 L) = (0.40 mol/L)(0.0300 L)
= 0.015 mol
= 0.012 mol
excess (some left
limiting (used up)
over…use to find pH)
H3O+(aq)
+
OH-(aq)
nH3O+ left = 0.015 mol – 0.012 mol
= 0.0030 mol
Vtotal = 0.0500 L + 0.0300 L = 0.0800L
C = n/V
= 0.0030 mol/0.0800 L
= 0.0375 mol/L
pH = -log(0.0375 mol/L)
= 1.43
⇌
2 H2O(l)
Your Assignment: pgs 18-19
P. 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:
o the initial pH of the curve must be the pH of the
sample
o the co-ordinate of the equivalence point must be
correct in terms of pH and volume
o the “over-titration” must be asymptotic with the
pH of the titrant
o number of equivalence points must match the
number of quantitative reactions occurring
 titrant selection:
o if the sample is an acid, titrant should be a
strong base such as NaOH(aq) or KOH(aq)
o if the sample is a base, titrant should be HCl(aq)
(strong, monoprotic, minimizes secondary rxns)
 you need to be able to interpret pH curves:
1. Strong Monoprotic with Strong Monoprotic
o pH of 7
at the equivalence point
SA titrated with SB
SB titrated with SA
14
14
pH EP 7
pH EP 7
0
0
volume
volume
2. Weak Monoprotic with Strong Monoprotic
o if weak acid, then pH of >7 at equivalence point
o if weak base, then pH of <7 at equivalence point
o 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
o more than 1 equivalence point
o middle buffer region at pH 7
WA(poly) titrated with SB
WB(poly) titrated with SA
14
14
EP2
EP1
pH
7
pH
EP1
0
volume
7
EP2
0
volume
2. Indicators for Titrations
 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
L. Acid/Base 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) +
acid
eg) litmus red
H2O(l) ⇌ In-(aq) +
base
conjugate
base
blue
H3O+(aq)
conjugate
acid
 several indicators can be used to determine the approximate
pH of a solution
eg) Solution 1:
indigo carmine is blue
pH is 11.4
thymol blue is blue
pH is 9.6
thymolphthalein is blue
pH is  10.6
 Solution 1 has a pH between 10.6 and 11.4
eg) Solution 2:
methyl violet is blue
pH is  1.6
orange IV is yellow
pH is 2.8
methyl orange is red
pH is  3.2
 Solution 2 has a pH between 2.8
and 3.2
Your Assignment: pg 11 #7,8 in workbook
Your Assignment: pg 20 in workbook
3. Buffers
 buffers are chemicals that, when added to water, protect
the solution from large pH changes when acids or
bases are added to them
they are used to calibrate pH meters and control the
rate of pH sensitive reactions (eg. in the blood)
typical buffers are conjugate pairs such as a weak acid
and the salt of the conjugate base
 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 by the addition of
too much acid or base
 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 – CH3COO-
pKa = -log(1.8 × 10-5) = 4.7
3. pH of 10.0
HCO3- – CO32-
pKa = -log(4.7 × 10-11) = 10.3
Example 1
Using an acetic acid – sodium acetate buffer system, show what
happens when:
a) a small amount of HCl(aq) is added
CH3COOH(aq)
A
Na+(aq)
–
CH3COO-(aq) H3O+(aq) Cl-(aq) H2O(l)
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)
A
SA
Na+(aq)
–
CH3COO-(aq) OH-(aq)
CH3COOH(aq) + OH-(aq)
B
B
SB
H2O(l)
A/B
⇌ CH3COO-(aq) +
H2O(l)
Example 2
One important buffer is described by the following equilibrium:
H2PO4-(aq) + H2O(l) ⇌ HPO42-(aq) + H3O+(aq)
A sample of this buffer contains 3.2  10-3 mol/L H2PO4-(aq) and
9.8  10-4 mol/L HPO42-(aq) at equilibrium. Calculate the pH of
the sample.
Ka = [HPO42-(aq)][H3O+(aq)]
[H2PO4-(aq)]
6.2  10-8 = (9.8  10-4) [H3O+(aq)]
3.2  10-3
[H3O+(aq)] = 2.02… 10-7 mol/L
pH = -log[H3O+(aq)]
= -log(2.02…10-7 mol/L)
= 6.69
Your Assignment: pg 21
Q. Society and Technological Connections
 acid deposition has been an environmental concern for
many years
 using pages 488-491 in your text and your data booklet,
describe the following:
1. definition and formation of acid deposition
2. environmental impact and measures being taken by
industries to reduce emissions
3. technology of acid deposition
4. social aspects of acid deposition