Acids, Bases, and Salts
Chapters 4.3, 16 and 17
Classification of Acids and Bases
is one of the oldest in History.
The medieval Alchemists first
used the terms
“acid”, “alkali”, and “salt”.
Acids were probably the most easily
recognized chemical because of their
sour taste.
Properties of Acids
1.
2.
3.
4.
5.
6.
Have a sour taste
Change the color of some plant pigments
Dissolve or corrode certain metals and
minerals; damages skin -“Corrosive”
Produce gas bubbles when combined
with minerals.
Dissolve in water to form aqueous
solutions.
Ionize in water to produce H+ ions
Examples of Acids
HCl hydrochloric acid (stomach acid)
 H2SO4 sulfuric acid (battery acid)
 HNO3 nitric acid (used to make fertilizer)
 HBr hydrobromic acid
There are
 HI hydroiodic acid
7 strong
 HClO3 chloric acid
Acids.
 HClO4 perchloric acid
(all other acids are weak. . .)
HC2H3O2 acetic acid
H2CO3 carbonic acid

Properties of Bases
1.
2.
3.
4.
5.
6.
Have a bitter, chalky taste
Have a soapy, slippery feeling
Change the color of some plant pigments
Can burn and irritate skin - “caustic”
Bases can destroy the properties of acids
when mixed in proper proportions.
Dissolve in water to form aqueous
solution; accept H+ ions and produce OHions
•
•
•
•
•
•
•
•
Examples of Bases
NaOH sodium hydroxide (lye)
LiOH
Group IA metal
KOH
hydroxides
RbOH
CsOH
Ca(OH)2
Heavy Group IIA
Sr(OH)2
metal hydroxides
Ba(OH)2
ALL other bases are weak: NH3
Electrolytes and
Nonelectrolytes:
Strong acids and bases completely ionize
in solution making them strong
electrolytes.
 Weak acids and bases only partly ionize
making them weak electrolytes in
solution.
 Acids and Bases are the only molecular
compounds that can act as electrolytes.

Arrhenius Definition
• Acids produce hydrogen ions in aqueous
solution.
HCl + H2O ---> H+(aq) + Cl- (aq)
• Bases produce hydroxide ions when
dissolved in water.
NaOH + H2O ---> Na+ (aq) + OH- (aq)
• Limited to aqueous solutions.
• Arrhenius limits us to one kind of base.
• NH3 ammonia could not be an Arrhenius
base.
Bronsted-Lowry Definitions
• An acid is a H+ (proton) donor and a base is a H+
(proton) acceptor.
• Acids and bases always come in pairs.
• Acid + Base
Conjugate + Conjugate
acid
base
• HCl is an acid.
• When it dissolves in water it gives its proton to
water.
• HCl(g) + H2O(l)
H+ donor
H+ acceptor
H3O+ + Clconjugate
acid
conjugate
base
Bronsted-Lowry expands the list
• When ammonia is dissolved in water NH3 + H2O
H+ acceptor
BASE
H+ donor
ACID
NH4+ + OHconjugate
acid
conjugate
base
The conjugate acid is what results when the base accepts
the H+
The conjugate base is what remains after the acid has
donated the H+
Competing Pairs
NH3 + H2O
H+ acceptor
base
•
•
•
•
H+ donor
acid
NH4+ + OHconjugate
acid
conjugate
base
This is an equilibrium.
Competition for H+ between NH3 and OH
The stronger base controls direction.
If OH is a stronger base it takes the H+, forms
water.
• Equilibrium moves to left.
• If NH3 is stronger, it takes the H+, forms NH4+
• Equilibrium moves to right.
Relative strength of Acids
and Bases:

1.
2.
3.
The stronger the acid or base, the weaker its
conjugate pair.
Strong acids completely transfer their protons to
water; therefore, the conjugate bases have a
tendency to be protonated.
Weak acids only partly dissociate; therefore, the
conjugate bases show a slight ability to remove
protons from water.
Substances with negligible acidity, contain
hydrogen, but do not demonstrate any acidic
behavior; their conjugate bases are strong bases.
Lewis Acids and Bases
• Most general definition. Gives us
many things which act as acid-bases.
• Acids are electron pair acceptors.
• Bases are electron pair donors.
F
H
B
F
F
:N
H
H
Lewis Acids involve molecules which have an incomplete octet.
Lewis Bases have an unshared pair of electrons.
Lewis Acids and Bases
• Boron triflouride wants more electrons.
• It acts as a Lewis Acid.
F
H
B
F
F
:N
H
H
Ammonia, NH3 donates a pair of electrons.
It acts as a Lewis base.
Lewis Acids and Bases
• BF3 is Lewis acid
• NH3 is a Lewis base.
H
F
F
B
F
Lewis Acid
N
H
H
Lewis Base
+
H
Lewis
Acid
H+
+
Water
OH
--->
Lewis
Base
••
•
•
••
[ O - H ]Donates a pair of electrons
H2O
A New View of Water:
Autoionization
 Water
is amphoteric: it behaves as
both an acid and a base
 In any sample of water, to a very small
degree, water self-ionizes:
H2O ---> H+ + OH
actually,
2 H2O
H3O+ + OH
Kw

Since there is an equilibrium established
H3O+
2 H2O
+
OH
we can write an equilibrium expression:
[H3O+] [OH]
Keq =
[H2O] 2
We don’t include water in the equilibrium expression
because the amount is huge and constant
Kw = [H3O+] [OH]
Kw - “Ion Product Constant for water”
Kw
• Kw = [H3O+] [OH]
For any aqueous solution at 25C,
Kw = 1.0 x 10 14 (a very small amt. of ionic activity!)
This means:
Kw = [H3O+] [OH]
1.0 x 10 14 = [H3O+] [OH]
[H3O+] = [OH] = 1.0x 10 7
[H3O+] is also [H+]
So, any Acidic Solution: [H+] > [OH]
And any Basic Solution: [H+] < [OH]
Acid vs. Basic Solutions
• Remember water behaves as both an acid
and a base.
• 2H2O(l)
H3O+(aq) + OH-(aq)
• KW= [H3O+][OH-] = [H+][OH-]
• At 25ºC, KW = 1.0 x10-14
• In EVERY aqueous solution.
Therefore,
• Neutral solution [H+] = [OH-]= 1.0 x10-7
• Acidic solution [H+] > [OH-]
• Basic solution [H+] < [OH-]
pH
• pH= -log[H+]
• Used because [H+] is usually very
small,
always less than 1 M
• As pH decreases, [H+] increases
exponentially
• Helpful to see a graph of y =  log [x]
pH =  log [H+]
y =  log [x]
This is where
pH is concerned.
pH
(1,0)
[H+]
pH
• [H+] = 1.0 x 10-8 M
[H+] = 1.0 x 103 M
pH= 8.00
pH = 3.00
 Sig figs - only the digits after the
decimal place of a pH are significant only 2 sig. figs in pH = 8.00
p stands for “ log”
Also. . .
pOH = -log[OH-]
pKw = -log Kw
[H+]
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-]
Relationships
KW = [H+][OH-]
-log KW = -log([H+][OH-])
-log KW = -log[H+]+ -log[OH-]
 So, pKW = pH + pOH
and since, KW = 1.0 x10-14
 Then, 14.00 = pH + pOH
 [H+], [OH-], pH and pOH;
given any one of these we can find the other
three.
Calculations with pH
 Remember
every solution has both a
pH and a pOH.
 Remember the chemistry, but don’t try
to memorize there is no one way to do
this.
Example #1


A solution has [H+] = 1.5 x106 M
Determine the pH, pOH, and [OH].
pH = log[H+] = log [1.5 x 106 M] = 5.82
14.00 = pH + pOH
14.00 = 5.82 = pOH
pOH = 8.18
Or, take the antilog
of 8.18.
Kw = [H+] [OH]
1.0 x 1014 = 1.5 x 106 M [OH]
[OH] = 6.6 x 10 9 M
Example #2
A solution has a pOH = 4.92
 Calculate the pH, [H+], and [OH].

pH = 14.00  pOH
pH = 14.00  4.92
pH = 9.08
Take the anti-log of 9.08
[H+] = 8.3 x 1010 M
Kw = [H+] [OH]
1 x 10 14 = [8.3 x 1010 M] [OH]
[OH] = 1.2 x 105 M
Acid and Base
dissociation constants
Ka
and
Kb
Ka - “Acid dissociation
constant”
General equation for an acid:
H3O+(aq) + A-(aq)

HA(aq) + H2O(l)

Ka = [H3O+][A-]
[HA]

H3O+ is often written H+ ignoring the water
in equation (it is implied).
Ka -Acid dissociation constant
 Ignoring
the water:
 HA(aq)
H+(aq) + A-(aq)

Ka = [H+][A-]
[HA]
 We can write the expression for
any acid.
Strong Acids
HCl, HNO3, H2SO4, HBr, HI, HClO4, HClO3
 Strong acids dissociate completely.
 HCl ---> H+ (aq) +
Cl (aq)
 Ka = [H+][Cl-]
[HCl]
 Strong acids favor products.
 Equilibrium far to right. Ka is very large.
 Conjugate base must be weak.

Weak Acids








Weak acids only partly dissociate.
Weak acids favors reactants.
HF (aq)
H+ (aq) + F (aq)
Ka = [H+][F-]
[HF]
Ka will be small.
ALWAYS WRITE THE MAJOR SPECIES.
It will be an equilibrium problem from the start.
Rest is just like last chapter - an equilibrium
problem!
Summary

Strong acids

Weak acids

Ka is very large

Ka is small
[H+] is equal to [HA]
 A- is a weaker base
than water

[H+] <<< [HA]
 A- is a stronger
base than water

Types of Acids
Oxyacids - Proton is attached to the oxygen
of an ion.
Example: nitrous acid, HNO2
 Organic acids contain the Carboxyl group COOH with the H attached to O
example: Benzoic acid, C6H5COOH
 Polyprotic Acids- more than 1 acidic
hydrogen (diprotic, triprotic).

Polyprotic acids
• Always dissociate stepwise.
• The first H+ comes off much easier than the
second.
• Ka for the first step is much bigger than Ka for
the second.
• Denoted Ka1, Ka2, Ka3
• Example: Carbonic Acid
• H2CO3
H+ + HCO3-
• HCO3-
Ka1= 4.3 x 10-7
H+ + CO3-2
Ka2= 4.3 x 10-10
Almost all of the H+ comes from the first step, so successive steps
are often ignored except for sulfuric acid.
Sulfuric acid is special
H2SO4 (aq) ----> H+ (aq) + HSO4 (aq)
 In first step it is a strong acid.
 HSO4 (aq)
H+ (aq) + SO42 (aq)
 2nd step it’s weak: Ka2 = 1.2 x 10-2
 However, 2nd step too large Ka to ignore.

Ka Problems
• Problem: What is the pH of a 0.0658M HCl
solution?
• Always write the major species: HCl --> H+ + Cl
• Strong acids completely dissociate!
• [H+] = [HCl]
HCl ---> H+ (aq) + Cl (aq)
[H+] = [HCl]
[H+] = 0.0658 M
pH = log [0.0658 M] = 1.18
Try this.
• Calculate the pH of 2.0 M acetic acid
HC2H3O2. Then find the pOH and[OH-].
Weak acids only partly dissociate.
HC2H3O2
H+ + C2H3O2Ka = [H+][C2H3O2-]
Look up
[HC2H3O2]
Ka = 1.8 x 105
1.8 x 105 = [ x ][ x ]
[ 2.0 M ]
X2
10-5
= 1.8 x
x 2.0
[H+] = X = 0.0060 M
pH = 2.2
pOH = 11.8
[OH] = 1.6 x 1012 M
Bases
• Strong Bases: LiOH, NaOH, KOH, RbOH,
CsOH, Ca(OH)2, Sr(OH)2 and Ba(OH)2.
• The OH- is a strong base.
• 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.
• Again, strong bases favor products. Kb is very
large, [OH-] = [B]
• Calculations just like strong acids.
Bases without
OH
Bases are proton acceptors.
 NH3 + H2O
NH4+ + OH It is the lone pair on nitrogen that accepts
the proton.
 Many weak bases contain N

Weak Bases
NH3 + H2O
NH4+ + OH General Equation:
 B(aq) + H2O(l)
BH+(aq) + OH- (aq)
Kb = [BH+][OH- ]
[B]

Kb “base dissociation constant”
Strength of Bases
 Hydroxides
are strong.
 Others are weak.
 Smaller the Kb weaker the base.
Kb problem
• Calculate the pH of a 0.25 M NH3.
• Always write the equation and the species!
NH3 + H2O
NH4+ + OH-
Kb = [NH4+][OH- ]
[NH3]
1.8 x 105 = [ x ][ x ]
[NH3]
(1.8 x 105 ) (0.25 M) = x2
x = 0.0021 M = [OH-]
pOH = 2.67
Look up
Kb = 1.8 x 105
14.00 = pH + pOH
pH = 11.33
Percent Dissociation
The amount of the acid (or base) [HA] that
has dissociated [x] divided by the acid’s
initial conentration, [HA0], then multiplied by
100%.
 Percent Dissociation = (x/[HA0]) x 100%
 When making assumptions in equilibrium
concentrations, it is best to test the
assumption by making sure that the percent
dissociation is less than or equal to 5%.

Example
 The
percent dissociation of an acid, HA,
which is 0.100M is 2.5%. Calculate the
Ka of the acid.
 (x/.100) x 100% = 2.5%
 x = 2.5 x 10-3M
 Ka = (2.5x10-3)2 / (.100 – 2.5x10-3)
 Ka = 6.4 x 10-5
A mixture of Weak
Acids/Bases






The process is the same.
Determine the major species.
The stronger will predominate.
If one acid has a relatively highier Ka value, it will
be the focus of the solution.
Calculate the pH of a mixture:
1.20 M HF (Ka = 7.2 x 10-4) and
3.4 M HOC6H5 (Ka = 1.6 x 10-10)
HF dominates so the pH of the mixture will be based
on it.
Relationship of Ka and Kb
NH3 + H2O
NH4+ + OH- NH4+ + H2O
Kb = [NH4+] [OH-]
[NH3]
if added together,
+
=

NH3 + H3O+
Ka = [NH3] [H+]
[NH4+]
NH3 + H2O
NH4+ + OH
NH4+ + H2O
NH3 + H3O+
2 H2O
H3O+ + OH-
we get the autoionization of water equation
Relationship of Ka and Kb

Ka x Kb = ( [NH3] [H+] ) ( [NH4+] [OH-] )
[NH4+]


[NH3]
Ka x Kb = [H+] [OH-] = Kw = 1.0 x 10-14
So, Ka x Kb = Kw
Relationship of Ka and Kb

Ka x Kb = Kw = 1.0 x 1014
As the strength of an acid increases (larger Ka),
the strength of it’s conjugate base must decrease
(smaller Kb).
 We can now calculate Kb for any base if we
know Ka for conjugate acid and vice versa.
Ka x Kb = Kw = 1.0 x 1014
Taking the – log of both sides and simplifying . . .

pKa + pKb = pKw = 14.00
Adding Salts to Water
 When
certain salts are dissolved in water
they sometimes create an acidic or basic
solution.
 Many ions can react with water to produce
H+ or OH- called a hydrolysis reaction.



In general, Anions which react with water produce basic
solutions.
Cations which react with water produce acidic solutions.
However, each salt must be examined carefully.
Salts that produce
Neutral Solutions
1.
2.
3.
Salts that consist of cations of strong
bases and the anions of strong acids
have no effect on pH when dissolved in
water.
Cations of strong bases (Na+, K+,
group 1A)
Anions of strong acids (Cl-, NO3-)
Salts that produce
Basic Solutions
1.

For any salt whose cation has
neutral properties and whose anion
is the conjugate base of a weak
acid, the aqueous solution will be
basic:
C2H3O2- + H2O  HC2H3O2 + OHbase
acid
acid
base
Salts the produce
Acidic Solutions
1.

1.
Salts in which the anion is not a base and
the cation is the conjugate acid of a
weak base produce acidic solutions
NH4+  NH3 + H+
Salts that possess a highly charged
metallic ion, such as Al+3



Aluminum ion in water is hydrated Al(H2O)6+3
High metallic charge polarizes O-H bond in
water
Hydrogens in water become acidic.
Examples

Adding NaC2H3O2 to water

NaC2H3O2 

C2H3O2- + H2O  HC2H3O2 + OHthe solution is basic

Example: NH4Cl  NH4+

NH4+
Na+ + C2H3O2-
+
Cl-
+ H2O  NH3 + H3O+
the solution is acidic
In Summary
1.
2.
3.
4.
Cl-, Br-, I-, NO3-, ClO3-, and ClO4Do not react with water = neutral solutions
Anions from a weak acid (conjugate base)
= basic solution
Cations from a weak base (conjugate acid)
= acidic solution
Metal ions (except group IA metals and
Ca+2, Sr+2, Ba+2) + water = acidic solution
In Summary
5. Ions which are amphoteric, the biggest K value
determines results
HPO4-2 can be acid or base in water,
HPO4-2 + OHPO4-3 + H3O+
HPO4-2 + H2O
H2PO4- + OHKa = 6.2 x 10-8 and Kb = 1.6 x 10-7
Kb wins, solution is basic
WHY?
Structure and Acid/Base Properties






Why are some acids weak, others strong, and
some neither?
Molecular structure
Any molecule with an H in it is a potential acid.
The stronger the X-H bond the less acidic
(compare bond dissociation energies).
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 is attached to an
oxygen atom.
 The oxygens are electronegative
 Pull electrons away from hydrogen

Strength of oxyacids
Electron Density
Cl
O
H
Hypochlorous acid
Strength of oxyacids
Electron Density
O
Cl
O
H
Chlorous acid
Strength of oxyacids
Electron Density
O
Cl
O
H
O
Chloric acid
Strength of oxyacids
Electron Density
O
O
O
Cl
O
H
Perchloric acid
Hydrated metals
• Highly charged metal
ions pull the electrons
of surrounding water
molecules toward
H
them.
+3 O
Al
• Make it easier for H+ to
H
come off.
• Metals + H2O = acidic
(except Group IA metals and
Ca+2, Sr +2, and Ba+2)
Lewis Acids and Bases
Al+3+ 6
Al
( )
( )
H
O
H
H
O
H
6
+3
Don’t confuse this with Oxides + Water
• Metal oxides dissolve in water to
produce bases.
• CaO(s) + H2O(l)
Ca(OH)2(aq)
• Non-metal oxides dissolved in water
can make acids.
• SO3 (g) + H2O(l)
H2SO4(aq)
Strategy for solving AcidBase Problems (In summary)
1.
2.
3.
List the major species in solution
Look for reaction that can be assumed to
go to completion, for a strong acid
dissociating or H+ reacting with OHFor a reaction that can be assumed to go
to completion:


Determine the concentration of the products
Write down the major species in solution after the
reaction
4.
5.
Look at each major component of the
solution and decide if it is an acid or a
base.
Pick the Equilibrium that will control pH.
Use known value of the dissociation
constants for the various species to help
decide on the dominant equilibrium
Looking into step 5







Write the equation for the reaction and the
equilibrium expression
Compute the initial concentrations (assuming
dominant equilibrium hasn’t occurred yet)
Define x
Compute equilbrium concentrations in terms of x
Substitute the concentrations into the equilibrium
expression
Check validitiy of approximation
Calculate the pH and other concentrations required
Calculating the pH of Salts
First you must decide if the salt is acidic,
basic, or neutral.
 If it is basic: Anion is conjugate base of a
weak acid (ion will undergo hydrolysis)
 Determine the pH of a 0.100 M aqueous
solution of NaCN. The Ka for HCN is
5.8x10-10.

First write the equilibrium expression for the
solution. It is basic, so Kb expression
 Use Ka and Kw to calculate Kb
 Construct an ice chart to determine expected
concentrations.
 Make assumptions and solve for x.
 Use the pOH to calculate the pH to finalize
question.

General Acid/Base Question:
The Kw for water at 25oC is 1.0x10-14, but is
1.0x10-13 at 60oC.
 Give the chemical equation for the
autoionizatoin of water.
 Determine the [OH-] for water at 60oC.
 Determine the pH of water at 60oC.
 Is the reaction endothermic or exothermic?
Support your answer.

Phosphoric acid, H3PO4 is a
triprotic acid:





Show the three equations involved in the dissociation
of this substance.
Illustrate how these three equations might be
combined to show the complete dissociation of
phosphoric acid.
If a 7.0 M H3PO4 solution is dissociated, calculate the
pH of the solution.
Ka1=7.5x10-3; Ka2=6.2x10-8; Ka3=4.8x10-13
Determine the concentration for the ions: H2PO4-1 ;
HPO4-2 ; PO4-3 ;
Determine the pOH for the same 7.0M H3PO4 solution
Chapter 17
More Equilibrium Applications
Buffers, Titrations, and Ksp
The Common Ion Effect
When the salt with the anion of a weak acid is
added to that acid, it reverses the dissociation
of the acid.
 HCN
H+ + CN
Adding NaCN to the solution shifts the
equilibrium to the left. (“Common Ion Effect”
is really LeChatelier’s Principle)
 Lowers the percent ionization of the acid.
 The same principle applies to salts with the
cation of a weak base.

Common Ion Problem
• What is the pH of a solution where 0.30 moles
of HC2H3O2 and 0.20 moles of NaC2H3O2 are
added to make a 1.00 L solution?
First, identify the species and write the equation.
HC2H3O2
H+ + C2H3O2
Recognize that adding more C2H3O2 shifts to left,
and the [H+] and [C2H3O2] are not equal.
I
C
E
HC2H3O2
0.30
-x
0.30 - x
H+ + C2H3O2
0
0.20
+x
+x
+x
0.20 + x
What is the pH of a solution where 0.30 moles
of HC2H3O2 and 0.20 moles of NaC2H3O2 are
added to make a 1.00 L solution?
HC2H3O2
H+ + C2H3O2
I
0.30
0
0.20
C
-x
+x
+x
E 0.30 - x
+x
0.20 + x
Ka =
[H+] [C2H3O2]
[HC2H3O2]
=
[x] [0.20 + x]
[0.30 -x]
= 1.8 x 10 -5
X = [H+] = 2.7 x 10-5, pH = 4.57
The resulting [H+] is very small, which justifies ignoring x.
Without the addition of 0.20 moles C2H3O2, the problem
works out to [H+] = 7.7 x 10-3, pH = 2.11
Buffers
• Buffers are solutions that resist a change in pH.
• Human blood is a complex aqueous mixture with
a pH buffered at 7.4
• Seawater is an aqueous mixture buffered at 8.1 near
the surface.
• A buffer is a weak acid and its salt, OR
a weak base and its salt.
• Buffers contain both acidic species to neutralize OHand a basic species to neutralize H+ ions.
Buffers work to maintain pH levels because of the
Common Ion Effect
Buffers
• A buffer is a weak acid and its salt, OR
a weak base and its salt.
HC2H3O2 and NaC2H3O2
NH3 and NH4Cl
• Example: buffer with equal concentrations
of weak acid and its conjugate base
HC2H3O2/C2H3O2
Buffers
• Example: buffer with equal concentrations of
weak acid and its conjugate base
HC2H3O2/C2H3O2
<---------->
add OH, neutralizes the acid to form water
HC2H3O2 + OH ---> H2O + C2H3O2
after adding OH-, more C2H3O2 than before
add H+, they react with the base component of the buffer
H+ + C2H3O2 ---> HC2H3O2
after adding H+, more acid than before
Adding OH- reduces [HC2H3O2] and increases [C2H3O2]
Adding H+ increases [HC2H3O2] and reduces [C2H3O2]
Buffer Problem (common ion problem)
• Calculate the pH of a buffer that is
0.75 M lactic acid (HC3H5O3) and
0.25 M sodium lactate (Ka = 1.4 x 10-4)
HC3H5O3 <-----> H+ + C3H5O3
I
C
E
0.75
-x
0.75 - x
0
+x
x
0.25
+x
0.25 +x
Ka = 1.4 x 10-4 = [H+] [C3H5O3] = [x] [0.25 +x]
[HC3H5O3]
[0.75 -x]
X = [H+] = 4.2 x 104 M
pH = 3.37
Buffers
• Buffers are most effective when
concentrations of weak acid and salt are
the same.
• 2 Important characteristics:
1. Buffer capacity - the amount of acid or
base the buffer can neutralize before the
pH begins to change substantially
depends on amount of acid/base it’s made from
2. Buffer pH - depends on the Ka for the acid
Why?
General Weak Acid Equation
• Ka = [H+] [A-]
[HA]
• so [H+] = Ka [HA]
[A-]
• The [H+] depends on the ratio [HA]/[A-]
• taking the negative log of both sides
• pH = -log(Ka [HA]/[A-])
• pH = -log(Ka)-log([HA]/[A-])
• pH = pKa + log([A-]/[HA])
This is called the
Henderson-Hasselbach
Equation
pH = pKa + log([A-]/[HA])
pH = pKa + log(base/acid)
• Calculate the pH of the following mixtures:
a.) 0.75 M lactic acid (HC3H5O3) and 0.25 M
sodium lactate (Ka = 1.4 x 10-4) pH =3.37
b.) 0.25 M NH3 and 0.40 M NH4Cl
pH = 9.05
(Kb = 1.8 x 10-5)
Buffer capacity
• The pH of a buffered solution is
determined by the ratio [A-]/[HA].
• As long as this doesn’t change much, the
pH won’t change much.
• The more concentrated these two are, the
more H+ and OH- the solution will be able
to absorb.
• Larger concentrations bigger buffer
capacity.
Buffer capacity
• Henderson-Hasselbach Equation:
pH = pKa + log([A-]/[HA])
• The best buffers have a ratio: [A-]/[HA] = 1
• This is most resistant to change
• True when [A-] = [HA]
• Make pH = pKa (since log1=0)
Addition of Strong Acids or
Bases to buffers:
A
buffer is made by adding 0.300 mol
HC2H3O2 and 0.300 mol of NaC2H3O2 to
enough water to make 1.00 L of solution.
The pH of the buffer is 4.74. Calculate
the pH of this solution after 0.020 mol of
NaOH is added.
First do the stoichiometry (Create a
Before and after reaction chart) with the
neutralization reaction
• B: 0.3 mol
0.02 mol
0.3 mol
•
HC2H3O2 + OH-  H2O + C2H3O2• A: 0.28 mol
0 mol
0.32 mol
• NaOH was completely consumed by the weak
acid component. Since OH- is limiting, all that
was added would be consumed.
Equilibrium: Now we focus on the
equilibrium that will determine the pH
of the buffer
(i.e. the ionization of the acetic acid)
• HC2H3O2  H+ + C2H3O2• We then use the new quantities from the
stoichiometry in the H.H. equation:
• pH = 4.74 + log (0.320 mol / 0.280 mol)
• pH = 4.80
Note: we can use mole amounts in place of
concentrations in the H.H. equation.
Prove they’re buffers:
 Determine
the pH of a solution made by
adding 0.020 mol of NaOH to 1.00 L of
pure water.
 pH = 14 – ( -log 0.020)
 pH = 12.30
 Quite a BIG difference from the
previous problem
Another practice
A
buffer is made by adding 0.300 mol
HC2H3O2 and 0.300 mol of NaC2H3O2
to enough water to make 1.00 L of
solution. The pH of the buffer is 4.74.
Calculate the pH of this solution after
0.020 mol of HCl is added.
 Do the stoichiometry
 Use equilibrium info with H.H.
 pH = 4.68
One more…

Calculate the change in pH that occurs when
0.010 mol of HCl is added to 1.0 L of each of
the following:
a.) 5.00 M HC2H3O2 and 5.00 M NaC2H3O2
b.) 0.050 M HC2H3O2 and 0.050 M NaC2H3O2
 Ka= 1.8x10-5

Remember: Make pH = pKa to find the
original pH
Last one…

What would the pH be if 0.050 mol of
solid NaOH is added to a 1.0L mixture of:
1.40 M HC2H3O2 and 1.40 M NaC2H3O2
 Ka= 1.8x10-5
 pH
= 4.74 + log (1.45/1.35)
 pH = 4.77
Titrations
• Adding a solution of known
concentration until the substance being
tested is consumed.
• This is called the equivalence point.
• The end point is a visible sign that the
equivalence point has been reached.
• Graph of pH vs. mL is a titration curve.
Helpful to Remember:
#moles
M =
# moles
# moles = M x L
L
L =
M
Strong acid with Strong Base
•
•
•
•
Do the stoichiometry. Use a BCA table.
There is no equilibrium .
They both dissociate completely.
The titration of 50.0 mL of 0.200 M HNO3
with 0.300 M NaOH requires x ml NaOH?
B
HNO3 +
0.01 mol
NaOH ----->
0.01 mol
C
-0.01
A
0
H2O
xs
+ NaNO3
0
-0.01
+ 0.01
+0.01
0
xs
0.01
The titration of 50.0 mL of 0.200 M HNO3 with 0.300 M
NaOH requires x ml NaOH?
# moles
L=
M
0.01 mol NaOH
L=
= 0.0333 L
0.300 M NaOH
= 33.3 ml
NaOH
Because they are both strong
we could also use:
MAVA = MBVB
Lets look at a titration curve of this . . .
•
•
Strong acid with strong Base
Equivalence at pH 7
pH
7
mL of Base added
Weak acid with Strong base
• There is an equilibrium.
• Do stoichiometry. Then do equilibrium.
• Titrate 50.0 mL of 0.10 M HF (Ka = 7.2 x 10-4) with 0.10 M
NaOH. How many ml of NaOH are needed?
HF
+
NaOH ---> H2O + NaF
.0050 mol
.0050 mol
.0050 mol
before
after
It requires .0050 mol NaOH to neutralize the acid;
L = .0050 mol / 0.10 M NaOH = .050 L = 50 ml NaOH
The titration is at the equivalence point.
Notice the product F- is a conjugate base.
Some F- will react with water to produce OHHow much? Do the equilibrium problem
Titrate 50.0 mL of 0.10 M HF (Ka = 7.2 x 10-4) with
0.10 M NaOH. How many ml of NaOH are needed?
F- + H2O
I 0.050
C -x
E 0.050 -x
HF + OH0
0
+x
+x
x
x
[HF] [OH-]
x2
1.39 x 10-11 =
Kb =
[ F- ]
So, at the equivalence
point, the mixture
is basic
.050 -x
X = 8.33 x 107 M OHpOH = 6.08 pH = 7.92
Weak acid with strong Base
 Equivalence at pH >7

>7
pH
7
mL of Base added

Strong base with strong acid
 Equivalence at pH 7
pH
7
mL of Base added
Weak base with strong acid
 Equivalence at pH <7

7
pH
<7
mL of Base added
• 75 mL of 1.5 M HF is titrated with
2.25 M KOH. Calculate the pH at the
equivalence point.
• pH = 8.55
Summary
For a titration, any acid-base will neutralize
and follow rules for stoichiometry. (BCA
table)
 If asked to find the pH of resulting weak acid
- strong base titration:
– Stoichiometry first
– Then Henderson-Hasselbach
 Think about the chemistry involved.
 Don’t try to memorize steps.

Indicators
• Weak acids that change color when they
become bases.
• weak acid written HIn
• Weak base
• HIn
H+ + Inclear
red
• Equilibrium is controlled by pH
• End point - when the indicator changes
color.
Indicators
• Since it is an equilibrium the color change
is gradual.
• It is noticeable when the ratio of
[In-]/[HI] or [HI]/[In-] is 1/10
• Since the Indicator is a weak acid, it has a
Ka.
• pH the indicator changes at is.
• pH=pKa +log([In-]/[HI]) = pKa +log(1/10)
• pH=pKa - 1 on the way up
Indicators
• pH=pKa + log([HI]/[In-]) = pKa + log(10)
• pH=pKa+1 on the way down
• Choose the indicator with a pKa 1 less
than the pH at equivalence point if you are
titrating with base.
• Choose the indicator with a pKa 1 greater
than the pH at equivalence point if you are
titrating with acid.
Solubility Equilibria
Will it all dissolve, and if not, how
much?
• All dissolving is an equilibrium.
• If there is not much solid it will all
dissolve.
• As more solid is added the solution will
become saturated.
• Solid
dissolved particles
• Equilibrium: the rate that the solid
dissolves is equal to the rate that the
dissolved particles rejoin the solid.
Ksp
• Metal ion bonded to nonmetal, then it dissociates.
• AgCl (s)
Ag+(aq) + Cl (aq)
[Ag+]1[Cl-]1
Keq =
[AgCl]
• But the concentration of a solid doesn’t change.
So we leave it out.
• Ksp = [Ag+] [Cl-]
• Called the solubility product expression
• Ksp called the solubility product constant
Watch out
• Solubility is not the same as
solubility product.
• Solubility product is an equilibrium
constant, (Ksp).
• It doesn’t change except by temperature.
• Solubility is how much it can dissolve
Units: g/L (set by the equilibrium position)
• A common ion can change this.
Calculating Ksp
The solubility of Li2CO3 is 4.35 g/L
What is the Ksp?
First, Always Write the Equation.
Then write the Ksp expression.
Li2CO3
2Li+ + CO3-2
Ksp = [Li+] 2 [CO3-2]
change g/L into moles/L (M)
4.35 g/L ( 1 mol / 73.8 g ) = 0.05894 M
Ksp = [ 2 x 0.05894 M] 2 [0.05894 M]
Ksp = 8.2 x 104
Calculating Solubility
• The solubility is determined by equilibrium.
• Its an equilibrium problem.
• Calculate the solubility of AgCl.
Ksp = 1.8 x 10-10
Write the equation and the Ksp expression.
AgCl (s)
Ag+(aq) + Cl (aq)
Ksp = [Ag+] [ Cl ]
5 mol/L
1.34
x
10
1.8 x 1010 = [ x ] [ x ]
x Molar Mass =
1.8 x 1010 = x2
0.00192 g/L
x = [Ag+] [ Cl ] =1.34 x 105 M
solubility
Relative solubilities
The bigger the Ksp the more soluble.
1 x 103
The smaller the Ksp, the less soluble.
1 x 1015
Ksp describes precipitates -- substances
which dissolve less than 0.1 M
pKsp =  log Ksp
Common Ion Effect
AgCl (s)
Ag+(aq) + Cl (aq)
• If we add a common ion to a saturated
solution, equilibrium shifts left, more solid
results.
• If we try to dissolve the solid in a solution
with either the cation or anion already
present less will dissolve. (more solid
results)
pH and solubility
• OH- can be a common ion.
Zn(OH)2
Zn+2 + 2OH
• Substance is more soluble in acid.
• For other anions if they come from a
weak acid they are more soluble in a
acidic solution than in water.
• CaC2O4
Ca+2 + C2O4-2
• H+ + C2O4-2
HC2O4• Reduces C2O4-2 in acidic solution.
• More solid CaC2O4 dissolves.
Predicting Precipitation
•
•
•
•
Ion Product, Q = [Ag+] [Cl-]
If Q > Ksp a precipitate forms.
If Q < Ksp No precipitate.
If Q = Ksp equilibrium.
Will a precipitate form when 100 ml of 0.050 M AgNO3 is
added to 100 ml of 0.040 M NaCl?
Equation: AgNO3 + NaCl ---> AgCl (ppt) + NaNO3
Will a precipitate form when 100 ml of 0.050 M
AgNO3 is added to 100 ml of 0.040 M NaCl?
• Equation: AgNO3 + NaCl ---> AgCl (ppt) + NaNO3
Find the M [ ] of each ion. New volume = 200 ml
.1L x .05 M = .005 mol Ag+ and .1 x .04 M = .004 mol Cl
0.200 L
0.200 L
= 0.025 M Ag+
= 0.02 M Cl
Ksp = [Ag+] [Cl]
Ion product, Q = [0.025 M] [0.02 M]
Q = .0005
1.8 x 1010 = Ksp
Since Q > Ksp, a precipitate forms
Selective Precipitations
• Used to separate mixtures of metal ions in
solutions.
• Add anions that will only precipitate certain
metals at a time.
• Used to purify mixtures.
• Often use H2S because in acidic solution
Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will
precipitate.
Selective Precipitation
• In Basic adding OH-solution S-2 will
increase so more soluble sulfides will
precipitate.
• Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3,
Al(OH)3
Selective precipitation
• Follow the steps first with insoluble
chlorides (Ag, Pb, Ba)
• Then sulfides in Acid.
• Then sulfides in base.
• Then insoluble carbonate (Ca, Ba, Mg)
• Alkali metals and NH4+ remain in solution.
Complex ion Equilibria
• A charged ion surrounded by ligands.
• Ligands are Lewis bases using their lone
pair to stabilize the charged metal ions.
• Common ligands are NH3, H2O, Cl-,CN• Coordination number is the number of
attached ligands.
• Cu(NH3)4+2 has a coordination # of 4
The addition of each ligand
has its own equilibrium
• Usually the ligand is in large excess.
• And the individual K’s will be large so we
can treat them as if they go to equilibrium.
• The complex ion will be the biggest ion in
solution.
• Calculate the concentrations of Ag+,
Ag(S2O3)-, and Ag(S2O3)-3 in a solution
made by mixing 150.0 mL of AgNO3 with
200.0 mL of 5.00 M Na2S2O3
• Ag+ + S2O3-2
Ag(S2O3)K1=7.4 x 108
• Ag(S2O3)- + S2O3-2
Ag(S2O3)-3
K2=3.9 x 104