Acids and Bases
Chapter 8
Acids & Bases
There are many substances that change the
hydrogen ion concentration [H+] when
dissolved in water.
 Binary
acids (HF, HCl, HBr, HI, H2S, etc)
 Oxyacids (HNO3, H2SO4, HClO4, etc)
 Organic acids (acetic acid, methanoic acid, etc)
 Hydroxy bases (NaOH, Ca(OH)2, KOH, etc)
 Amine bases (methylamine, dimethylamine, etc.)
Definition of acids & bases

Empirical definition – defines based on
observed properties in experimental
conditions.
Empirical definition of an acid
 Turns
litmus red
 Reacts with base metals to produce H2 gas
 Reacts with carbonates to produce CO2 gas
 Conducts electricity
 Tastes sour
 Feels dry or rough
 Is colourless in phenolphthalein
 Turns bromothymol blue to a yellow
Empirical definition of a base
 Turns
litmus blue
 Does not react with metals or carbonates
 Conducts electricity
 Tastes bitter
 Feels slippery
 Turns pink in phenolphthalein
 Turns blue in bromothymol blue
Acid-Base Theories

Årrhenius Definition of acids & bases
on the production of H+ or OH- when
a substance is dissolved in water
 An acid produces hydrogen ions (H+) or
hydronium ions (H3O+) when dissolved in
water.
HBr(aq) + H2O  H3O+(aq) + Br-(aq)
 Founded
H2O + H +(aq)  H3O+(aq)
Acid-Base Theories

Årrhenius Definition of acids & bases
base produces hydroxide ions (OH-) when
dissolved in water.
KOH(s) + H2O(l)  K+(aq) + OH-(aq)
A
Acid-Base Theories

Årrhenius Definition of acids & bases
 Adequately
explains the behaviour and
properties of most acids and bases, but fails
to explain how compounds such as ammonia
(NH3(aq)) can behave as bases without any
hydroxide ions (OH-) in the substance.
NH3(g) + H2O(l)  NH4+(aq) + OH-(aq)
Brønsted-Lowry Theory
The second theory of acids/bases uses
the transfer of protons (H+).
 Brønsted-Lowry Acid

A

substance that donates protons.
Brønsted-Lowry Base
A
substance that accepts protons.
Brønsted-Lowry Theory
H2O(l)
Base
+
HCl(aq)  H3O+(aq) +
Acid
Conjugate
acid
Cl-(aq)
Conjugate
base
HCl donates the H+ – Acid
H2O will accept the H+ – Base
H3O+ can donate the H+ it received – Conjugate acid
Cl- can accept available H+ ions – Conjugate base
Acid-Base Equlibria
In the Brønsted-Lowry theory of acids &
bases the concept of equilibrium is implied.
 The acid-base system appears reversable.
 Acids donate hydronium ions in the forward
reaction while the conjugate acids donate
hydronium ions in the reverse reaction.

Brønsted-Lowry equilibrium
H2O(l)
Base
+
HCl(aq)  H3O+(aq) +
Acid
Conjugate
acid
Cl-(aq)
Conjugate
base
Forward reaction proton donator – Acid
Forward reaction proton acceptor – Base
Reverse reaction proton donator – Conjugate acid
Reverse reaction proton acceptor – Conjugate base
Brønsted-Lowry equilibrium
H2O(l)
Base


+
HCl(aq)  H3O+(aq) +
Acid
Cl-(aq)
Conjugate
Conjugate
acid
base
Competition for protons occurs between the base and the
conjugate base.
The state of the equilibrium will be determined by the
capacity for the acid to produce protons in comparison
with the conjugate acid’s capacity to do the same.
The Acid-Base Equilibrium



Strong acids & bases dissociate completely
(100%) therefore a quantitative reaction and a
poor equilibrium condition.
Weak acids & bases dissociate partially and
therefore offer excellent conditions for an
equilibrium setting.
The stronger the acid, the weaker its
conjugate base and in a similar manner, the
weaker the acid the stronger its conjugate
base.
The Acid-Base Equilibrium
aHA(aq) + bH2O(l)  cH3O+(aq) +
Acid
Base
K acid

Conjugate

Conjugate
acid
c

acid Conjugate
a
Acid 
dA-(aq)
Conjugate
base
d

base
Amphoteric (amphiprotic)
 These
are substances that appear to
act as Brønsted-Lowry acids or bases
in different chemical reactions.
 An equilibrium condition.
 A substance that can donate or accept
protons.
 Like water or the bicarbonate ion in
baking soda.
Amphoteric (amphiprotic)
H2O(l) + H2O(l)  H3O+(aq) + OH-(aq)
Acid

Base
Congugate
acid
Congugate
base
Water acts both as an acid and a base in its
“autoionization”.
Amphoteric (amphiprotic)

As most acid-base systems are produced in
water this means there are possibly two
equilibria at play.
Amphoteric (amphiprotic)
H2O(l) + HCO3- (aq)  H2CO3 (aq) + OH-(aq)
Acid
Base
Congugate
Congugate
acid
base
H2O(l) + HCO3- (aq)  CO3-(aq) + H3O+(aq)
Base
Acid
Congugate
Congugate
base
acid
Polyprotic acids



A polyprotic acid is capable of donating more
than one proton (H2CO3(aq), H2SO4(aq))
There is an ionization constant for each proton
donation (Ka1, Ka2, etc.) as the ionization occurs
in steps. (Table Pg. 803)
The Ka values become smaller with each
ionization step, as the removal of a proton from a
negatively charged object becomes more difficult.
Polyprotic acids



However, the most ionization occurs in the first
step.
 Ka1>> Ka2 > Ka3 .. . .
Consequently, the [H+] is predominantly
established in the first ionization with the Ka1
value. Subsequent ionizations (Ka2 & Ka3) only
add minimal amounts of [H+].
Use Ka1 to determine the pH of the solution at
equilibrium.
Weak & Strong classifications

A strong acid/base dissociates very well
 It
breaks up into its ions when dissolved in water
 These tend to form quantitative dissociation reactions

A weak acid/base dissociates poorly
 These
do not dissociate well and appear
predominantly in the undissociated form
 These tend to form equilibria
Very strong
acids
100% ionized
in water
The strongest
proton donor that
can exist in water
Weak acids
in water
Does not react
with water as
an acid
Very weak
bases
HClO4
ClO4-
HNO3
NO3-
HCl
Cl-
H3O+
H2O
HF
F-
HNO2
NO2-
HC2H3O2
C2H3O2-
HOCl
OCl-
NH4+
NH3
H2O
OH-
NH3
NH2-
Very weak
acids
Very strong
bases
Do not react
with water to
any measurable
extent
Weak bases
in water
The strongest proton
acceptor that can
exist in water
Reacts 100 %
with water
Acid-Base properties of Salt Solutions
A salt is a solid that when dissolved in
water dissociated into ions.
 Some salts may contain ions that alter the
[H+] or pH of a solution.
 The acidic or basic properties of a salt
solution arise from the reaction of the
dissociated ions of the salt with water.

Salts that form neutral solutions

The salt of a strong acid/strong base dissolves in
water to form neutral solutions.
 Strong acid supplies
 HCl  H+ + Cl-
the anion of the salt
 Strong base supplies
 NaOH  Na+ + OH-
the cation of the salt
 When
the water is removed (evaporation) a salt
remains



HCl + NaOH  H+ + OH- + Cl- + Na+  NaCl
Includes groups 1 & 2, but not Be2+
The solution has a pH of 7
Salts that form acidic solutions
The salt of a weak base (cation) and
strong acid (anion) dissolves in water to
form acidic solutions.
 The cation reacts with water to liberate H+
 The solution has a pH less than 7
NH4Cl  dissociates  NH4+ & ClNH4+ + H2O  H3O+(aq) + NH3(aq)

Salts that form basic solutions
The salt of a strong base (cation) and
weak acid (anion) dissolves in water to
form basic solutions.
 The anion reacts with water to liberate OH The solution has a pH greater than 7
NaC2H3O2 dissociates  Na+ & C2H3O2C2H3O2- (aq)+ H2O  HC2H3O2(aq) + OH-(aq)

Weak base/weak acid salts
The salt of a weak base (cation) and weak
acid (anion) dissolves in water and both
ions react with water.
 If Ka>Kb, the solution is acidic.
If Kb>Ka, the solution is basic.
 When an ion reacts with water to produce
an acidic or basic solution (break water
into its ions) it is called hydrolysis.

Weak base/weak acid salts


NH4HCO3 is the salt of a weak base (NH3) and a
weak acid (H2CO3). Both ions react with water.
NH4+ + H2O  NH3 + H3O+
HCO3 + H2O  H2CO3 + OHKa for NH4+ = 5.6 x 10-10
Kb for HCO3- = 2.2 x 10-8
Therefore, the solution would have more OH- in
solution (Kb>Ka) and have a large pH. It is basic!
Anion from
Strong acid
(Large Ka)
Anion from
Weak acid
(Small Ka)
Cation from Neither ions reacts Anion reacts with
with
water
so
the
water
to
produce
OH
Strong base
pH is neutral.
so the pH is basic.
(Large Kb)
Very large Kb.
Cation from Cation reacts with
to produce
Weak base water
H+ so the pH is
(Small Kb)
acidic.
Very large Ka.
Both ions react with
water.
Ka = Kb neutral
Ka > Kb acidic
Kb > Ka basic
Acid-Base Reactions

Neutralization reaction
 Double
displacement reaction between an
acid and a base to produce a salt and water
(solvent).
 The acidic and basic properties are destroyed
– neutralized.
Acid-Base Reactions

Titration
Analytical
lab technique used to determine the
concentration of a solution.
Quantitative neutralization reaction
Titrant – the solution of known concentration in
buret.
Sample – the solution of unknown
concentration, but known volume (in flask).
Acid-Base Reactions

Titration
solution – a stock solution that is
of known concentration that is used to create
the titrant.
 Primary standard – a chemical that is
available in a pure and stable form that can
be used to produce an accurate
concentration.
 Standardization – a titration used to find the
concentration of the titrant using a a primary
standard.
 Standard
Acid-Base Reactions
 Titration
 Indicator – an acid base indicator that will
change colour at a known pH to signify a
specific pH in the neutralization reaction.
 End point – the point in the titration when the
indicator changes colour.
 Equivalence point – the point in the titration
when chemically equivalent amounts of
reactants have reacted. Usually an
equilibrium has been established.
34
Titration of a strong acid with a
strong base
At the equivalence point [H+]=[OH-] or
pH=7.
 Phenolphthalein is a popular indicator
because it is colourless in acidic solutions
and pink in basic.
 Remember to consider the reaction at the
molar level. (convert to moles!)

 C=n/V
and C1V1 = C2V2
Titration of a strong acid with a
strong base
Titration curve – a plot of the pH vs. Volume of
titrant added.
14
Equivalence point
(pH=7.0)
pH

7
0
Vol. of titrant
Titration of a strong base with a
strong acid
14
pH
Equivalence point
(pH=7.0)
7
0
Vol. of titrant
Titration of a weak acid with a
strong base
In the titration of a weak acid with a strong
base the neutralization occurs equilibrating
the number of H+ and OH- ions.
 In doing so, a salt ion is produced that
adjusts the hydrolysis of water due to the
ions “desire” to form a weak acid.
 The shift in the ionization of water produces
a pH other than 7 at the equivalence point.

Titration of a weak acid with a
strong base
14
pH
Equivalence point
(pH> 7.0)
7
Buffer region
0
½ way to Equivalence
point
Vol. of titrant
Mechanics for a Weak acid –
Strong base titration
The quantitative neutralization reaction
adjusts the quantities of acid, base, salt &
water produced.
 As the neutralization progresses the
quantity of acid becomes reduced while the
salt (conjugate base) increases.
 This process results in a shift in the weak
acid ionization equilibrium, toward the
reactants.

Mechanics for a Weak acid –
Strong base titration
Remember, the neutralization is consuming
weak acid while producing conjugate base.
 At the equivalence point, the weak acid has
been completely neutralized (there is none
left)
 The conjugate base has been increased to
the quantity of the neutralized weak acid.

Mechanics for a Weak acid –
Strong base titration
The equilibrium shifts to a situation where
the Conjugate Base is in sufficient quantity
to behave as a reactant in its own ionization
equilibrium.
 This results in the production of Weak Acid
and hydroxide ion as the equilibrium is
established.
 Hence, a basic pH value at equivalence
point.

Mechanics for a Weak acid –
Strong base titration
Start of Titration
HA
+
NaOH
Buffer region
(Both weak acid (HA)
and conjugate base
(A-) are present)
HA
+
NaOH
A-
+
H2O
HA
+
NaOH
A-
+
H2O
HA +
NaOH
A-
+
A-
H2O
+
H2O
Equivalence point
(H+ has reacted
with OH- completely)
Auto-ionization of H2O
A- +
H2 O ⇌
HA
+ OH-
Titration of a weak base with a
strong acid
14
pH
Buffer region
7
Equivalence point
(pH<7.0)
0
Vol. of titrant
Titration of a polyprotic acid with a
strong base
14
pH
Second
Equivalence
point
7
First Equivalence
point
0
Vol. of titrant
Buffer
During the “buffer region” one has a
combination of weak acid and its
conjugate base.
+
 A buffer solution can “absorb” either H or
–
OH ions added to the solution.
 A buffer is an equal mixture of a weak acid
and its conjugate base.

Buffer solution of H2CO3 &
HCO3
Consumes H+
H +(aq) + HCO3- (aq)  H2CO3 (aq) + H2O(l)
Conjugate
base
Weak
acid
Consumes OHOH -(aq) + H2CO3 (aq) HCO3-(aq) + H2O (l)
Weak
acid
Conjugate
base
Acid-base calculations
Determine [H+] and pH
 Percent ionization
 Use ka to determine [H+] and pH
 Neutralization reactions

Determine [H+] and pH
[ H  ]  2.50 x 10-6
mol
L
pH  - log[ H  ]
  log( 2.50 x 10-6 )
 5.602
pH  8.36
[ H  ]  10  pH
 10 8.36
 4.4 x 10 -9
mol
L
Determine [H+] and pH

The pH and pOH add up to 14 for a
conjugate acid-base pair.
14 = 𝑝𝐻 + 𝑝𝑂𝐻
𝑝𝑂𝐻 = 14 − 3.5
= 10.5
Determine pKa and pKb

The pKa and pKb add up to 14 for a
conjugate acid-base pair.
14 = 𝑝𝐾𝑎 + 𝑝𝐾𝑏
𝑝𝐾𝑏 = 14 − 3.5
= 10.5
𝐾𝑏 = 10−𝑝𝐾𝑏
= 10−10.5
= 3.162 × 10−11
Percent Ionization
[ionized acid]
% ionization 
 100%
[acid solution]

[H ]

100%
[HA]
Use ka to determine [H+] and pH

Given the concentration and equilibrium
constant for an acid one can determine the
hydronium ion concentration and the pH of
the solution.
Calculate the hydrogen ion concentration and the
pH of a 0.15 mol/L solution of hydrofluoric acid
(HF(aq)) with a ka of 6.6 x 10-4.
1.
2.
Produce a balanced chemical equation for the
hydrolysis equilibrium.
+
HF (aq) + H2O (l)  H3O (aq) + F (aq)
Create an equilibrium expression.


[ H 3O ][ F ]
ka 
[ HF ]
 6.6 x 10
-4
Calculate the hydrogen ion concentration and the
pH of a 0.15 mol/L solution of hydrofluoric acid
(HF(aq)) with a ka of 6.6 x 10-4.
3.
Generate and ICE table to account for
concentration changes.
+
HF (aq) + H2O (l)  H3O (aq) +
Initial
Change
Equilibrium
0.15 mol L-1
-x
0.15 -x
0
+x
0+x=x
F
(aq)
0
+x
0+x=x
Calculate the hydrogen ion concentration and the
pH of a 0.15 mol/L solution of hydrofluoric acid
(HF(aq)) with a ka of 6.6 x 10-4.
4.
Use the equilibrium expression to generate a
mathematical model for the equilibrium.
[ H 3O  ][ F  ]
ka 
[ HF ]
( x)( x)
-4
6.6 x 10 
(0.15  x)
6.6 x 10 -4 (0.15  x)  x 2
0  x 2  (6.6 x 10 -4  0.15)  (6.6 x 10 -4 x)
0  x 2  6.6 x 10 -4 x  9.9 x 10 -5
Calculate the hydrogen ion concentration and the
pH of a 0.15 mol/L solution of hydrofluoric acid
(HF(aq)) with a ka of 6.6 x 10-4.
5.
Resolve for “x” and relate to [H+].
0  x 2  6.6 x 10 -4 x  9.9 x 10 -5
 b  b 2  4ac
x
2a
x  9.6 x 10 -3 mol L or - 1.03 x 10 -2
x = [H+] = 9.6 x10-3 mol/L
mol
L
Calculate the hydrogen ion concentration and the
pH of a 0.15 mol/L solution of hydrofluoric acid
(HF(aq)) with a ka of 6.6 x 10-4.
6.
Calculate the pH.
[ H  ]  9.6 x 10 -3 mol L
pH  - log[ H  ]
  log(9.6 x 10 -3 )
 2.02
The [H+] = 9.6 x10-3 mol/L
and the pH= 2.02.
[H+] = 9.6 x10-3 mol/L has
two significant digits (9.6)
while the exponent is a
certain value. The pH value
must reflect the certainty of
the exponent and
significance of the
measured coefficient.
(2 – indicates the exponent,
.02 – the two SD of the
coefficient)
2.02  2SD
Neutralization of a weak/strong
combination
Stoichiometric calculation of quantities
during neutralization.
 Determine adjusted specie concentration
due to modified volumes
 Look at the common ion affect on the
hydrolysis of water equilibrium.

Calculate the pH of a sample when 25.00 mL of
0.200 mol/l HClO (aq) is titrated with 10.00 mL of
0.200 mol/L KOH (aq) .
There are two reactions to consider in this situation:
 The neutralization reaction which is a quantitative
reaction that is governed by stoichiometry.
HClO (aq) + KOH (aq) → H2O (l) + KClO (aq)
 The neutralization produces a salt that adjusts an acidbase equilibrium system for the conjugate acid-base
pair.
HClO (aq) + H2O (l) ⇌ H3O+ (aq) + ClO- (aq)
 This might be considered as a common ion effect where
the neutralization generates and ion that affects an
equilibrium.
Calculate the pH of a sample when 25.00 mL of
0.200 mol/l HClO (aq) is titrated with 10.00 mL of
0.200 mol/L KOH (aq) .
1.
Use the neutralization reaction to determine the quantity
of acid and base that react.
HClO (aq) + KOH (aq) → H2O (l) + KClO (aq)
HClO (aq)
KOH (aq)
Concentration (C)
0.200 mol/L
0.200 mol/L
Volume (V)
0.02500 L
0.01000L
0.200 mol
0.02500 L   0.200 mol 0.01000 L 
1L
1L
 5.00 x 10 -3 mol
 2.00 x 10 -3 mol

Moles (n)
Calculate the pH of a sample when 25.00 mL of
0.200 mol/l HClO (aq) is titrated with 10.00 mL of
0.200 mol/L KOH (aq) .
2.
Use the neutralization reaction to determine the quantity
of “salt” produced in the reaction.
HClO (aq) + KOH (aq) → H2O (l) + KClO (aq)
2.00 x 10-3 mol of KOH (aq) is used up completely and
produces 2.00 x 10-3 mol of KClO (aq) or ClO- (aq).
In doing so there is a quantity of HClO (aq) that is left
unreacted in the neutralization reaction.
The unreacted HClO (aq) is
5.00 x 10-3 mol – 2.00 x 10-3 mol = 3.00 x 10-3 mol
Calculate the pH of a sample when 25.00 mL of
0.200 mol/l HClO (aq) is titrated with 10.00 mL of
0.200 mol/L KOH (aq) .
2.
Alternate method to use the neutralization reaction to
determine the quantity of “salt” produced in the reaction.
HClO (aq) + KOH (aq) → H2O (l) + KClO (aq)
0.200 mol KOH 1 mol KClO
mol KClO=10.00 mL KOH
1000 mL KOH
1 mol KOH
mol KClO=2.00 x 10−3 mol KClO
The unreacted HClO (aq) is . . .
5.00 x 10-3 mol – 2.00 x 10-3 mol = 3.00 x 10-3 mol HClO
Calculate the pH of a sample when 25.00 mL of
0.200 mol/l HClO (aq) is titrated with 10.00 mL of
0.200 mol/L KOH (aq) .
3.
Using the quantities of weak acid and conjugate base
determine the affects on the hydrolysis of water.
HClO (aq) + H2O (l) ⇌ H3O+ (aq) + ClO- (aq)
H3O+ (aq)
ClO- (aq)
= 3.00 x 10-3 mol/
0.035 L
= 0.0857 mol/L
0
= 2.00 x 10-3
mol/ 0.035 L
= 0.0571 mol/L
[Change in]
-x
+x
+x
[Equilibrium]
0.0857 - x
x
0.0571 + x
HClO (aq)
[Initial]
Calculate the pH of a sample when 25.00 mL of
0.200 mol/l HClO (aq) is titrated with 10.00 mL of
0.200 mol/L KOH (aq) .
4.
Use the ka value and equilibrium expression to solve for
[H+] and pH
HClO (aq) + H2O (l) ⇌ H3O+ (aq) + ClO- (aq)
[H 3O  ] [ClO - ]
ka 
[HClO]
(x) (0.0571  x)

(0.0857 - x)
(Consider the application of the
“500 rule”)
(x) (0.0571)
 2.9 x 10 -8
0.0857
(2.9 x 10 -8 )(0.0857)
x
0.0571
 4.35 x 10 -8 mol L
Calculate the pH of a sample when 25.00 mL of
0.200 mol/l HClO (aq) is titrated with 10.00 mL of
0.200 mol/L KOH (aq) .
4.
Use the ka value and equilibrium expression to solve for
[H+] and pH
HClO (aq) + H2O (l) ⇌ H3O+ (aq) + ClO- (aq)
[H  ]  4.35 x 10 -8 mol L
pH  - log [H  ]
  log(4.35 x 10 -8 )
 7.362