ACIDS and BASES
DEFINITIONS of Acids and Bases:
Arrhenius Theory
Acid: A molecular substance that ionizes in
aqueous solution to form hydrogen ions (H+).
Base: A substance that produces hydroxide ions
(OH-) in aqueous solution.
Brønsted-Lowry Concept
An alternative definition:
Brønsted-Lowry Acid = Proton donor
Brønsted-Lowry Base = Proton acceptor
Works for non-aqueous solutions and explains why NH3 is
basic:
NH3(g) + H2O(l)
Base:
H+ acceptor
Acid:
H+ donor
NH4+(aq) + OH-(aq)
Brønsted-Lowry Concept
Strong acids and bases almost completely ionize.
H3O+(aq) + NO3-(aq)
HNO3(aq) + H2O(l)
Weak acids and bases do not fully ionize.
HF(aq)
+ H2O(l)

H2O(l)
+ NH3(aq)

Acid:
H+ donor
Base:
H+ acceptor
H3O+(aq)
+
NH4+(aq) +
Conj.Acid:
H+ donor
Conjugate acid-base pairs
F-(aq)
OH-(aq)
Conj.Base:
H+ acceptor
Acid Ionization Constants, Ka
When an acid ionizes in water:
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
The acid ionization constant is used to report the
degree of ionization:
Ka =
[A-][H3O+]
[HA]
(Water omitted, as usual)
Strong acids have large Ka values
Weak acid have small Ka values
Dissociation Constants: Strong Acids
Acid
Perchloric
Hydriodic
Hydrobromic
Hydrochloric
Nitric
Sulfuric
Hydronium ion
Formula
HClO4
HI
HBr
HCl
HNO3
H2SO4
H3O+
Conjugate
Base
ClO4-
Ka
NO3HSO4-
Very large
Very large
Very large
Very large
Very large
Very large
H2O
1.0
IBrCl-
Dissociation Constants: Weak Acids
Acid
Formula
Conjugate
Base
Ka
Iodic
HIO3
IO3-
1.7 x 10-1
Oxalic
H2C2O4
HC2O4-
5.9 x 10-2
Sulfurous
H2SO3
HSO3-
1.5 x 10-2
Phosphoric
H3PO4
H2PO4-
7.5 x 10-3
Citric
H 3 C 6 H 5 O7
H2C6H5O7-
7.1 x 10-4
Nitrous
HNO2
NO2-
4.6 x 10-4
Hydrofluoric
HF
F-
3.5 x 10-4
Formic
HCOOH
HCOO-
1.8 x 10-4
Benzoic
C6H5COOH
C6H5COO-
6.5 x 10-5
Acetic
CH3COOH
CH3COO-
1.8 x 10-5
Carbonic
H2CO3
HCO3-
4.3 x 10-7
Hypochlorous
HClO
ClO-
3.0 x 10-8
Hydrocyanic
HCN
CN-
4.9 x 10-10
Base Ionization Constants, Kb
For a base in water:
B:(aq) + H2O(l)
BH+(aq) + OH-(aq)
The base ionization constant, Kb, is:
[BH+][OH-]
Kb =
[B:]
If the base is an anion:
A-(aq) + H2O(l)
[HA][OH-]
Kb =
[A-]
HA(aq) + OH-(aq)
Dissociation of Strong Bases
MOH(s)  M+(aq) + OH-(aq)
 Strong bases are metallic hydroxides (MOH)
Group I hydroxides (NaOH, KOH) are very
soluble
Group II hydroxides (Mg, Ca, Ba, Sr) are less
soluble
Kb for Some Common Weak Bases
Many students struggle with identifying weak
bases and their conjugate acids.What patterns
do you see that may help you?
Formula
Conjugate
Acid
Kb
NH3
NH4+
1.8 x 10-5
Methylamine
CH3NH2
CH3NH3+
4.38 x 10-4
Ethylamine
C2H5NH2
C2H5NH3+
5.6 x 10-4
Diethylamine
(C2H5)2NH
(C2H5)2NH2+
1.3 x 10-3
Triethylamine
(C2H5)3N
(C2H5)3NH+
4.0 x 10-4
Hydroxylamine
HONH2
HONH3+
1.1 x 10-8
Hydrazine
H2NNH2
H2NNH3+
3.0 x 10-6
Aniline
C6H5NH2
C6H5NH3+
3.8 x 10-10
C 5H 5N
C5H5NH+
1.7 x 10-9
Base
Ammonia
Pyridine
Relative Strength of Acids & Bases
Strong acids are better H+ donors than weak acids
Strong bases are better H+ acceptors than weak bases
• Stronger acids have weaker conjugate bases.
• Weaker acids have stronger conjugate bases.
HCl(aq) + H2O(ℓ) → H3O+(aq) + Cl-(aq)
strong acid
(100 % ionized)
HF(aq) + H2O(ℓ)
weak acid
(many HF are un-ionized)
very weak base
(no tendency to form
HCl)
H3O+(aq) + F-(aq)
F- readily forms
HF
12
Water can be Acid or Base
Water acts as a base when an acid dissolves in water:
HBr(aq) + H2O(l)
acid
base
H3O+(aq) + Br-(aq)
acid
base
But water also acts as an acid for some bases:
H2O(l) + NH3(aq)
acid
base
NH4+(aq) + OH-(aq)
acid
base
Water is amphiprotic/amphoteric - it can donate or
accept a proton (act as acid or base).
Autoionization of Water
Two water molecules can react to form ions.
Autoionization occurs:
Acid
H2O(l) + H2O(l)
Base
Base
H3O+(aq) + OH-(aq)
Acid
Heavily reactant favored.
Only a very small fraction is ionized:
Kw = [ H3O+ ] [ OH- ]
ionization
constant for
water
= (1.0 x 10-7)(1.0 x 10-7)
= 1.0 x 10-14
(at 25°C)
[H2O]2 is
omitted…
Ionization Constant for Water
Kw, like all equilibrium constants, is T-dependent.
T = 25°C (77°F) is usually used as the standard T.
T (°C)
10
15
20
25
30
50
Kw
0.29 x 10-14
0.45 x 10-14
0.68 x 10-14
1.01 x 10-14
1.47 x 10-14
5.48 x 10-14
The pH Scale
pH is a means of expressing the acidity or basicity
of a solution.
The pH scale
Acid concentration can vary over a very large
range. A logarithmic scale is more convenient:
pH = −log10[H3O+]
At 25°C a neutral aqueous solution has:
pH = −log10[1.0 x 10-7] = −(−7.00) = 7.00
Acidic solutions: pH < 7.00
Basic solutions: pH > 7.00
The pOH scale
Base concentrations:
pOH = −log10[OH-]
A neutral solution (25°C) has:
pOH = −log10[1.0 x 10-7] = −(−7.00) = 7.00
Since Kw = [ H3O+ ][ OH- ] = 1.0 x 10-14
−log(KW)= −log[H3O+] + (−log[OH-]) = −log(1.0 x 10-14)
pKw = pH + pOH = 14.00
(Valid in all aqueous solns. at 25°C: acidic, neutral or basic)
pH of
common
solns:
A. Given two aqueous solutions (25°C).
Solution A: [OH-] = 4.3 x 10-4 M,
Solution B: [H3O+] = 7.5 x 10-9 M.
Which has the higher pH? Which is more acidic?
B. Calculate the hydronium and hydroxide ion
concentration of an aqueous solution with pH 4.21
Measuring pH
H3O+ concentrations can be measured with an:
• Electronic pH meter:
 fast and accurate
 preferred method.
• Acid-base Indicator:
 substance that changes color within a narrow pH range
 may have multiple color change (e.g. bromthymol blue)
 one “color” may be colorless (e.g. phenolphthalein)
 cheap and convenient.
Relationship between Ka and Kb
values
For an acid-base conjugate pair: HA and AKa x Kb =
[H3O+][A-]
[HA][OH-]
[HA]
[A-]
= [H3O+][OH]
= Kw
Phenol, C6H5OH, is a weak acid, Ka = 1.3 x 10-10 at
25°C. Calculate Kb for the phenolate ion C6H5O-
Reactions of Acids, Bases
Strong acid + strong base → salt solution, pH = 7
Strong acid + weak base
→ salt solution, pH < 7
Weak acid + strong base
→ salt solution, pH > 7
Weak acid + weak base
Need K’s
→ salt solution, pH = ?
“strongest wins”
The problem-solving approach
Make a habit of applying the following steps:
 Write the balanced equation and Ka expression.
 Define x as the unknown change in concentration that occurs
during the reaction. Frequently, x = [HA]dissoc, the concentration of
HA that dissociates, which, through the use of certain assumptions,
also equals [H3O+] and [A-] at equilibrium.
 Construct a reaction table (ICE) that incorporates the unknown.
 Make assumptions that simplify the calculations: usually that x is
very small relative to the initial concentration.
 Substitute the values into the Ka expression, and solve for x.
 Check that the assumptions are justified. (Apply the 5% rule) If
they are not justified, use the quadratic formula to find x.
A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic
acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #1: Write the dissociation equation
HC2H3O2  C2H3O2- + H+
A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic
acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #2: ICE it!
HC2H3O2  C2H3O2- + H+
0.50
I
C
-x
E
0.50 - x
0
0
+x
+x
x
x
A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic
acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #3: Set up the law of mass action
HC2H3O2  C2H3O2- + H+
E
0.50 - x
x
x
2
( x)( x)
x
1.8 x 10 

(0.50  x) (0.50)
5
A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic
acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #4: Solve for x, which is also [H+]
HC2H3O2  C2H3O2- + H+
E
0.50 - x
x
2
x
1.8 x 10 
(0.50)
5
x
[H+] = 3.0 x 10-3 M
A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic
acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #5: Convert [H+] to pH
HC2H3O2  C2H3O2- + H+
E
0.50 - x
x
x
5
pH   log( 3.0 x10 )  4.52
A Weak Base Equilibrium
Problem
What is the pH of a 0.50 M solution of
ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #1: Write the equation for the reaction
NH3 + H2O  NH4+ + OH-
A Weak Base Equilibrium
Problem
What is the pH of a 0.50 M solution of
ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #2: ICE it!
NH3 + H2O  NH4+ + OHI
C
0.50
-x
E 0.50 - x
0
0
+x
+x
x
x
A Weak Base Equilibrium
Problem
What is the pH of a 0.50 M solution of
ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #3: Set up the law of mass action
NH3 + H2O  NH4+ + OHE
0.50 - x
x
x
2
( x)( x)
x
1.8 x 10 

(0.50  x) (0.50)
5
A Weak Base Equilibrium
Problem
What is the pH of a 0.50 M solution of
ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #4: Solve for x, which is also [OH-]
NH3 + H2O  NH4+ + OHE
0.50 - x
x
2
x
1.8 x 10 
(0.50)
5
x
[OH-] = 3.0 x 10-3 M
A Weak Base Equilibrium
Problem
What is the pH of a 0.50 M solution of
ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #5: Convert [OH-] to pH
NH3 + H2O  NH4+ + OHE 0.50 - x
x
x
5
pOH   log( 3.0 x10 )  4.52
pH 14.00  pOH  9.48
Acid-base Titrations
In a titration, a solution of accurately known concentration
is added gradually added to another solution of unknown
concentration until the chemical reaction between the two
solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
the indicator
changes color
35
Chapter 4 Three Major Classes of Chemical Reactions
A
At end
point
Near end
point
Before
titration
B
H+ (aq ) + X –(aq ) + M+ (aq ) + OH – (aq )
C
H2O(l ) + M+(aq ) + X–(aq )
Figure 4.9 An acid-base titration. A, In this procedure, a measured
volume of the unknown
acid solution is placed in a flask beneath a buret containing the known (standardized) base solution. A few drops of indicator are added to the flask; the indicator used here is phenolphthalein,
which is colorless in acid and pink in base. After an initial buret reading, base (OH ions) is added
36
Acid-base Titrations
 Primary standard – A chemical compound which can
be used to accurately determine the concentration of
another solution. Examples include KHP and sodium
carbonate.
 Standard solution – A solution whose concentration
has been determined using a primary standard.
 Standardization – The process in which the
concentration of a solution is determined by accurately
measuring the volume of the solution required to react
with a known amount of a primary standard.
37
Titrations can be used in the analysis of
Acid-base reactions
H2SO4 + 2NaOH
2H2O + Na2SO4
EXAMPLE:
A sodium hydroxide solution was standardized and used to
titrate 25.00 mL of a sulfuric acid solution. The titration requires
43.79 mL of the 0.1172 M NaOH solution to completely
neutralize the acid. What is the concentration of the H2SO4
solution?
38
Think!
• Potassium hydrogen phthalate is a
very good primary standard.
– It is often given the acronym, KHP.
– KHP has a molar mass of 204.2
g/mol.
• In a titration experiment, a student finds that
23.48 mL of a NaOH solution are needed to
neutralize 0.5468 g of KHP. What is the
concentration (in molarity) of the NaOH solution?