Acid and
Base
Equilibrium
Arrhenius Definition
 Acids produce hydrogen ions in aqueous
solution. (Arhennius definition)
 Bases produce hydroxide ions when dissolved
in water. (Arhennius definition)
 Limits to aqueous solutions.
 Only one kind of base.
 NH3 ammonia could not be an Arrhenius base.
Brønsted-Lowry Definitions
 And acid is an proton (H+1) donor and a base
is a proton acceptor.
 Acids and bases always come in pairs.
 HCl is an acid.
 When it dissolves in water it gives its proton
to water.
 HCl(g) + H2O(l)  H3O+1 + Cl-1
OR
 HCl  H+1 + Cl-1
 Water is a base makes hydronium ion.
General Acid/Base Equation
 The general equation is:
 HA(aq) + H2O(l)  H3O+1(aq) + A-1(aq)
 Acid + Base  Conjugate acid +
Conjugate base
 This is an equilibrium situation.
 Competition for H+1 between H2O and A-1
 The stronger base controls direction.
 If H2O is a stronger base it takes the H+1
 Equilibrium moves to right.
Acid dissociation constant Ka
 The equilibrium constant for the general
equation.
 HA(aq) + H2O(l)  H3O+1(aq) + A-1(aq)
 Ka = [H3O+1][A-1]
[HA]
 H3O+1 is often written H+1 ignoring the
water in equation (it is implied). Either
way of expressing it is fine.
Acid dissociation constant Ka
 HA(aq)  H+1(aq) + A-1(aq)
 Ka = [H+1][A-1]
[HA]
 We can write the expression for any acid.
 Strong acids dissociate completely and
their equilibrium is far to the right.
 Conjugate base must be weak.
 Don’t forget stoichiometry if you have an
acid with 2 H+1 ions like H2SO4
A Way to Describe Acids
 Strong acids
 Weak acids
 Ka is large
 Ka is small
[HA]
 A- is a weaker
base than water
 A- is a stronger
 [H+] is equal to
 [H+] <<< [HA]
base than water
More Ways to Describe Acids
 Polyprotic Acids- more than 1 acidic hydrogen
(diprotic, triprotic). Sulfuric Acid – H2SO4
 Oxyacids - Proton is attached to the oxygen of
an ion. An example is H3PO4 Phosphoric Acid
 Organic acids contain the Carboxyl group COOH with the H attached to O. An example of
this is Acetic Acid which we call vinegar.
 Organic Acids are generally very weak.
Amphoteric Nature of Water
 Behaves as both an acid and a base.
 Water autoionizes as shown below.
 2 H2O(l)  H3O+1(aq) + OH-1(aq)
 KW= [H3O+1][OH-1] = [H+1][OH-1]
 At 25ºC KW = 1.0 x 10-14
 In EVERY aqueous solution.
 Neutral solution [H+1] = [OH-1] = 1.0 x 10-7
 Acidic solution [H+1] > [OH-1]
 Basic solution [H+1] < [OH-1]
pH
 pH= -log[H+1]
 Used because [H+1] is usually very small.
 As pH decreases, [H+1] increases
exponentially.
 Log is based on a factor of 10.
 Only the digits after the decimal place
of a pH are significant.
 [H+1] = 1.0 x 10-8 pH= 8.0
 pOH= -log[OH-1] AND pKa = -log Ka
Relationships
 KW = [H+1][OH-1] = 1.0 x10-14
 -log KW = -log([H+][OH-])
 -log KW = -log[H+]+ -log[OH-]
 pKW = pH + pOH
 pH + pOH = 14.00
 [H+],[OH-], pH and pOH
OR
 Given any one of these we can find the
other three variables.
[H+1]
100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14
pH
0
1
Acidic
14 13
10-14 10-13
3
11
5
7 9
Neutral
9
7 5
11
3
13
14
Basic
1
0
pOH
10-11 10-9Basic
10-7 10-5 10-3 10-1 100
[OH-1]
Calculating pH of Solutions
 Always write down the major species
in solution.
 Strong Acids &/or Bases.
 Water
 Known Conjugates.
 Remember these are equilibria.
 Remember the stoichiometry.
 Don’t try to memorize there is no one
way to do this.
Strong Acids
 HCl, HNO3, H2SO4, HClO4 (HBr, HI)
 Completely dissociated.
 Ka is said to be infinite, since there is
no reactant left.
 [H+1] = [HA]
 [OH-1] is going to be small because of
equilibrium and is only present due to
the autoionization of water.
Weak Acids
 Ka will be small, often, very small.
 Determine whether most of the H+1 will
come from the acid or the water. You do
this by comparing Ka to Kw and see which
one has a higher dissociation.
 The rest of the problem is just like last
previous problems.
 SET UP THE ICE BOX
 DETERMINE THE SHIFT
 If [HA]< 10-7 water contributes more H+1
Example of a Ka Problem
 Calculate the pH of 2.0 M acetic acid HC2H3O2
with a Ka of 1.8 x 10-5
 Calculate pOH, [OH-1], [H+1]
HC2H3O2  H+1 + C2H3O2-1
Initial
Change
Equilibrium
1.8 x 10-5 = x2 / 2
2.0
0
0
-x
+x
+x
2.0 – x
x
x
x = .006
5% rule works
The Complete Solution
 We found that x = .006 and that this was the
amount of [H+1]. Solving for pH gives 2.22
 Solving for pOH  14 – 2.22 = 11.78
 Solving for [OH-1]  [H+1] [OH-1] = 10-14
so [OH-1] = 1.66 x 10-12
 It wasn’t asked in this problem, but .006 is
also the amount of the acetate ion in
solution.
A mixture of Weak Acids
 Determine the major species.
 The stronger will dominate.
 Bigger Ka if concentrations are comparable
 Calculate the pH of a mixture of
1.20 M HF (Ka = 7.2 x 10-4)
and
3.4 M HOC6H5 (Ka = 1.6 x 10-10)
The HF is clearly the dominant species as its Ka
is 6 factors of 10 higher. Note: Remember
that the Ka is a negative exponent.
Upon Further Review
 We can apply some LeChatelier’s Principal Ideas
to this problem.
1.20 M HF (Ka = 7.2 x 10-4)
and
3.4 M HOC6H5 (Ka = 1.6 x 10-10)
 From the problem, we see that HOC6H5 is a
minor contributor, but if LeChatelier’s Principal is
taken into account we see that it is even less
than originally thought. The addition of H+1 ions
from the HF (the major species) further shift the
HOC6H5 dissociation to the left restricting its
H+1 contribution.
Percent dissociation
=
amount dissociated x 100 %
initial concentration
 For a weak acid percent dissociation
increases as acid becomes more dilute.
 Calculate the % dissociation of 1.00 M and
.00100 M Acetic acid (Ka = 1.8 x 10-5)
 As [HA]0 decreases [H+1] decreases but %
dissociation increases.
If you know the dissociation
 What is the Ka of a weak acid that is
8.1 % dissociated as 0.100 M
solution?
 Still set up the ICE BOX and you
know that 8.1 % of the .1 M solution
is going to dissociate (x) and that
91.9 % of the solution will remain as
a molecule.
Bases
 The OH-1 is a strong base. (Arrhenius)
 Hydroxides of the alkali metals are
strong bases because they dissociate
completely when dissolved.
 The hydroxides of alkaline earths
Ca(OH)2 etc. are strong dibasic bases,
but they don’t dissolve well in water.
 Used as antacids because [OH-1] will
combine with [H+1] to form water.
Bases without OH Bases are proton acceptors. (Brønsted-
Lowry)
 NH3 + H2O  NH4+1 + OH-1
 It is the lone pair on nitrogen that accepts
the proton.
 Many weak bases contain a Nitrogen atom.
 B(aq) + H2O(l)  BH+1(aq) + OH-1(aq)
 Kb = [BH+][OH- ]
[B]
 The Law of Mass Action doesn’t change.
Strength of Bases
 Group 1 and 2 hydroxides are strong.
 Eg. NaOH, KOH, Ca(OH)2
 Most others are weak.
 The smaller the Kb  the weaker the base.
 Calculate the pH of a solution of 4.0 M
pyridine (Kb = 1.7 x 10-9)
N:
Polyprotic acids
 Always dissociate stepwise, one H+1 at a
time.
 The first H+1 comes off much easier than
the second. It’s Ka will always be greater.
 Ka1 for the first step is much bigger than
Ka2 for the second.
 Denoted Ka1, Ka2, Ka3 to clarify which step
is being referred to.
Polyprotic acid # 2
 H2CO3  H+1 + HCO3-1 Ka1= 4.3 x 10-7
 HCO3-1  H+1 + CO3-2 Ka2= 4.3 x 10-10
 The base in the first step is often the acid
in second. Your Ka values will tell you this.
 In calculations we can normally ignore the
second dissociation when determining pH.
 However, we need to solve the second
when we are looking for the anion; CO3-2
in this instance.
Calculate the Concentration
 …of all the ions in a solution of 1.00 M
Arsenic acid H3AsO4
 Ka1 = 5.0 x 10-3
Ka2 = 8.0 x 10-8
Ka3 = 6.0 x 10-10
 Each
 Set
one is done individually.
up an ICE Box. The math is easy,
the initial set up is the tricky part.
Sulfuric acid is special
 For starters, remember that it is a strong acid.
 (??? What does that tell you about Ka1 ???)
 Ka2 = 1.2 x 10-2
 Calculate the concentrations of H2SO4,
HSO4-1 and SO4-2 in a 2.0 M solution of
H2SO4
 Do the same calculations in a solution of
H2SO4 at 2.0 x 10-3 M
Salts as acids/bases - Strong
 Salts are ionic compounds.
 Solutions of salts that are made up of the
conjugates of strong acids and strong bases are
neutral. For example: NaCl & KNO3
 NaOH
 KOH
+ HCl (strong acid) NaCl
+ HNO3 (strong acid)  KNO3
(strong base)
(strong base)
 There is no equilibrium for strong acids and
bases. For teaching purposes, you can think of
these as strong salts. (This is not general chemistry
terminology, but it fits.)
 We ignore the reverse reaction because K is so
large in the forward reaction.
Salts as acids/bases - Weak
 Some salts do have a pH when in solution:
 NaNO2 will be basic in aqueous solution.
WHY, YOU ASK ? OK – I will tell you.
Here is what’s going on, the major species in solution
are Na+1, NO2-1, and from water H+1, and OH-1.
When the Na+1 and the OH-1 merge, they are soluble,
dissociating and releasing OH-1 ions into solution.
However, when H+1 and NO2-1 merge, the HNO2
molecule is favored & the H+1 ions are not in
solution.
Therefore the solution has more OH-1 ions than H+1
ions and is therefore basic.
Conjugate Acids and Bases
 When an acid or base dissociates, it leaves
behind an ion. For example.
 HNO2  H+1 + NO2-1
 Fe(OH)3  OH-1 + Fe(OH)2+1
 The H+1 or the OH-1 have already been
discussed thoroughly (see Arhennius definition of acids
and bases) but in each equation there is also an
ion that is formed.
 These are called conjugates.
 In the example above the NO2-1 is called the
conjugate base of HNO2
 The Fe(OH)2+1 is the conjugate acid of Fe(OH)3
More on Basic Salts
 If the anion of a salt is the conjugate base of a
weak acid - basic solution.
 In an aqueous solution of NaF
 The major species are Na+1, F-1, and H2O
 F-1 + H2O  HF + OH-1
 Kb =[HF][OH-1]
[F-1 ]
 but Ka = [H+1][F-1]
[HF]
 So there is a relationship between Ka and Kb
Ka tells us Kb
 The anion of a weak acid is a weak base.
 Calculate the pH of a solution of 1.00 M NaCN.
Ka of HCN is 6.2 x 10-10
 The CN-1 ion competes with OH- for the H+1
 How do we know that this is going to be a
basic solution ?
 Because the Ka of HCN is low so the molecule
is preferred and it doesn’t dissociate and
liberate any H+1 ions.
 If any NaOH forms, it will dissociate
immediately releasing OH-1 ions.
Acidic salts
 A salt with the cation of a weak base and the
anion of a strong acid will be acidic.
 Calculate the pH of a solution of 0.40 M NH4Cl
(the Kb of NH3 = 1.8 x 10-5).
 H+1 (from water) + Cl-1  forms HCl which will
immediately dissociate to form and acid soln.
 Some acidic salts are those of highly charged
metal ions.
 More on this later.
Anion of weak acid,
cation of weak base
 IF…
 Ka > Kb acidic
 Ka < Kb basic
 Ka = Kb Neutral
Structure and
Acid/Base Properties
 Any molecule with an H in it is a potential
acid.
 The stronger the X-H bond the less acidic
(compare bond dissociation energies) because the H+1
ion isn’t liberated as easily.
 The more polar the X-H bond the stronger the
acid (use electronegativities).
 The more polar H-O-X bond - stronger acid.
Strength of oxyacids
 The more oxygen hooked to the central atom,
the more acidic the hydrogen.
 HClO4 > HClO3 > HClO2 > HClO
 Remember that the H+1 is attached to an
oxygen atom.
 The oxygen atoms are electronegative and so
they pull electrons away from hydrogen atom
making it an Hydrogen ion (H+1).
Hydrated metals
 Highly charged
metal ions pull the
electrons of
surrounding water
molecules toward
them.
 Make it easier for
H+1 to come off.
Al+3 O
H
H
Acid-Base Properties
of Oxides
 Non-metal oxides dissolved in water can make
acids.
 SO3 + H2O
H2SO4
 Metal oxides dissolve in water to produce bases.
 CaO + H2O
Ca(OH)2
Lewis Acids and Bases
 Most general definition of all.
 Acids are electron pair acceptors.
 Bases are electron pair donors.
F
H
B
F
F
:N
H
H
Lewis Acids and Bases
 Boron triflouride wants more electrons.
F
H
B
F
F
:N
H
H
Lewis Acids and Bases
 Boron triflouride wants more electrons.
 BF3 is Lewis base NH3 is a Lewis Acid.
H
F
F
F
B
N
H
H
Lewis Acids and Bases
Al+3+ 6
Al
( )
( )
H
O
H
H
O
H
6
+3
Summary of Definitions of
Acids and Bases
 Arrhenius Acid: Hydrogen Ion donor
 Arrhenius Base: Hydroxide Ion donor
 Brønsted-Lowry Acid: Proton donor
 Brønsted-Lowry Base: Proton acceptor
 Lewis Acid: Electron Pair acceptor
 Lewis Base: Electron Pair donor
Titration - Introduction
 A titration is a chemistry lab procedure where an
acid is added to a base (or vice-versa) to
determine its exact concentration.
 Exact volumes are needed so a buret is usually
the tool of choice.
 Accurate measurement is needed to get a good
result.
 The formula M1V1 = M2V2 comes in handy.
 The central idea is that the number of moles of
acid will = the number of moles of base.
Titration - Procedure
 An exact amount of acid with the unknown
concentration is put in a flask.
 A small amount of an indicator is added to the
base.
 A color change might take place depending on the indicator.
 A Standard solution of base with known
concentration is put in a buret.
 The acid is carefully added until the solution
shows the desired color change which depends
on the indicator used.
Indicators
Indicators are certain chemicals that have a peculiar
property of having one color when they are in an acid
solution and another when they are in a basic
solution.
The most common indicator is phenolphthalein,
because it happens to change close to pH = 7, which
is neutral. Phenolphthalein is clear when it is in acidic
solution and a light pink when it just turns basic. If
more base is accidently added, the solution turns a
dark purple.