Acids and Bases
Arrhenius
 Acid- substance that dissolves in H2O to produce H+ ions
H+ ion or theH3O+(hydronium ion) represents an acid

Base- dissolves in H2O to produce OH- ions.
OH- represents a base
Bronsted Lowry (more general definition)
 Acid- proton (hydrogen) donor;
 Base- proton (hydrogen) acceptor
Example:
HA + H2O 
H3O+ +
AProton donor
Proton acceptor
hydronium ion
Lewis
 Acid- electron pair acceptor
 Base- electron pair donor
Further Look into Bronsted-Lowry Acids and Bases
 Conjugate base – species that remains when H+ is removed from the acid
 Conjugate acid – species that is formed when the H+ is accepted by the base
1. HCl + H2O
ACID
↔
BASE
H3O+
+
CONJUGATE
ACID

2. NH3
BASE


CONJUGATE
BASE
HCl and Cl- and H2O an H3O+ are the conjugate acid-base pair
+
H2O
ACID
↔
NH4 +
CONJUGATE
ACID


Cl-
+
OHCONJUGATE
BASE
.
Amphoteric- a substance that can act as an acid or a base. See water above!
Conjugate acid-base pair is 2 substances related to one another by the donating
and accepting of a single hydrogen ion (proton)
the stronger the acid- the weaker the conjugate base
the stronger the base- the weaker the conjugate acid
Examples
Write the dissociation reaction and then identify the acid, base, conjugate acid and
conjugate base
a. Formic acid, HCOOH
b. Perchloric acid
There is a competition between the two bases:
HA


+
H2O

H3O+ +
A-
If H2O is a much stronger base than A-, equilibrium will move to the right, most
of the acid will dissolve into ions.
If A- is much stronger than H2O, equilibrium will move to the left, most of the
acid will remain as molecules.
Acid and Base Strengths
 Strength is determined by the equilibrium position of the reaction
Strong acid- equilibrium lies far to the right- all original acid has dissociated(ionized) at
equilibrium
Examples to memorize – only 7!!
HCl
+
HI, HBr, HCl, HNO3, H2SO4, HClO3, HClO4

H2O
H3O+ +
Cl-
Weak Acid – equilibrium lies far to the left- almost all the original acid is in molecule
form and it only ionizes or dissociates to a small extent ( mixture of molecules and ions)
Examples: organic acids and ALL others not in the above list
HC2H3O2
+
H2 O

H3O+ +
C2H3O2-
Strong base- equilibrium lies far to the right- all the original base has
dissociated(ionized) at equilibrium
Examples to memorize- group I and II soluble hydroxides (Li, Na, K,…..) (Ba, Sr, Ca)
NaOH
+ H2O
 Na+
+ OH-
Weak base- equilibrium lies far to the left- almost all the original base is in molecule
form and it only ionizes or dissociates to a small extent
Examples: ammonia, organic bases and all other
NH3 + H2O  NH4+ + OHTo write an equilibrium expression for a WEAK acid or WEAK base: from above
examples

Remember the pure liquid water is left out of expression- concentration will
remain the same in a dilute solution
Example: Write the equilibrium expression for each:
a. Formic acid, HCOOH
b. Perchloric acid
+
-(
HCOOH(aq) ↔ H (aq) + COOH aq)
HClO4(aq) ↔ H+(aq) + ClO4 -(aq)
Acid dissociation constant
HC2H3O2 (aq) ↔ H+(aq) + C2H3O2-(aq)
Ka = [H3O+][ C2H3O2-]
HC2H3O2
Ka = [H+][ C2H3O2-]
HC2H3O2
or
 if Ka > 1 the equilibrium position favors the products
 if Ka < 1 the equilibrium position favors the reactants
Base dissociation constant
NH3(aq) + H2O(l) ↔ NH4+(aq) + OH- (aq)
Kb = [NH4+][OH-]
[NH3]
Auto ionization of water
 1 water molecule can donate a hydrogen ion(proton) to another water molecule
 only 1 out of every 109 molecules will ionize
 SO pure water – almost entirely water molecules
H2O
+ H2O
 H3O+
+ OH-
Kw = [H3O+][OH-]
or Kw = [H+][OH-]
Ion product constant

Ion Product Constant does vary with temperature Kw at 25 ºC



Pure water
Acidic
Basic
[H] = [OH-]
[H+] > [OH-]
[H+] < [OH-]
H+ =1.0 x 10-7 M
H+ > 1 x 10-7 M
H+ < 1 x 10-7 M
= 1.0 x 10-14 M2
Examples
a. Calculate [H+] if [OH-]= 9.3 x 10-4 M Is the solution acidic, basic, or neutral?
b. Calculate both the [H+] or [OH-] concentrations if the solution is neutral
c. Calculate [OH-] if [H+]= 6.7 x 10-11 M Is the solution acidic, basic, or neutral?
pH SCALE
Indicate the strengths of acid or base (range from 1-14); easier number to work with
Acid (1-6)
Neutral (7)
Base (7-14)

pH= -log [H+]
Example: What is the pH of [H+] = 1.0 x 10-7 M HCl
PH = - log(1.0 x 10-7)

- (-7.00)
= 7.00
**** sig figs on logs are special: the number of decimal places in the log equals to the
number of sig figs in the original number
Other Relationships connected to pH
pOH = -log [OH-]
and
Example: Fill in the blanks
pH
pOH
6.88
13.08
[H+]
14= pH + pOH
[OH-]
Acid, Base, neutral
1.3 x 10-7
1.0 x 10-7
To calculate pH of strong acid and base
 complete ionization; do the above to get your answer
 focus on the major species - minor species like water can be ignored if their H+
contribution is very small compared to the major species and vice versa
in calculating the pH of .10 M HCl the contribution of water (10-7) can be
ignored
in calculating the pH of 1.0 x 10-12 HNO3, the [H+] contributed by the
autoionization of water
will be much larger and the contribution of the 1.0 x 10-12 HNO3 can be
ignored

be careful with bases because some strong bases will release 2 ions when it
dissociates
ie. .30 M NaOH is .30 M Na+ and .30 M OH- where as
.30 Ba(OH)2 is .30 M Ba+ and .60 M OH
with bases, you use pOH = -log[OH-] and then 14 = pH + pOH
Example
Calculate the pH of a 5.0 x 10-2 M NaOH solution.
Example
Calculate the pH of a solution made by putting 4.63 g LiOH into water and diluting to a
total volume of 400 ml.
Example
Calculate the pH and the [OH-] of a 5.0 x 10-3M HClO4 solution
Ka values
Ka >>>> 1
Ka <<<< 1
strong acid [H+] = [ HA]
weak acid [H+] <<<< [ HA]
Property
Ka value
Strong Acid
large
Weak Acid
small
Position of the
dissociation (ionization)
equilibrium
Far to right
Far to left
Equilibrium [H+]
compared to[HA]
[H+] ≈ [HA]0
[H+] << [HA]0
Strength of conjugate
base compared to water
A- is a much weaker
conjugate base than water
A- is a much stronger
conjugate base than water
To Calculate pH of weak acid– Must use equilibrium expression
1. Determine the major species (ions) in solution
2. Choose which one that can produces the most H+ and write down the equations
for the reaction.
3. Using the equilibrium constants, decide which one wills dominate- compare Ka to
Kw.
4. Write down the equilibrium expression of dominant species
5. Use ICE table: List (initial), define changes, determine [equilibrium] by
subtracting into expression.
6. Solve for x “easy way” by assuming [HA]0 -x = [HA]0
 5% rule - if X / [HA]0 x 100 is less than 5% then we can safely assume
[HA]0- x ≈ [HA]0
 This actually has another name in this chapter:
% ionization = [H+]/ [initial HA] x 100
7. Calculate [H+] and pH using the equation pH = -log[H+]
Example:
Calculate the pH of 1.00 M solution of hypochlorous acid (HOCl, Ka = 3.5 x 10-8)
Example:
Calculate the pH of a .500 M solution of formic acid, HCOOH (Ka = 1.77 x 10-4)?
Example:
The value for Ka = 7.45 x 10-4 for citric acid, C6H10O8. Calculate the pH of .200 M of
citric acid.
Calculating the pH of a mixture of weak acids
 Follow same steps as before; pick the predominant species that will produce the
[H+] ions
Example
Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2 x 10-2) and 5.00 M
HNO2 (Ka = 4.0 x 10-4). Also calculate the concentration of cyanide ions (CN-) in this
solution at equilibrium.
Example
Calculate the pH of a mixture of 2.00 M formic acid, HCOOH (Ka = 1.77 x 10-4) and 1.50
M bromic acid, HOBr (Ka = 2.06 x 10-9). What is the concentration of both the
hypobromite ion (OBr-) and the hydroxide ions(OH-) at the equilibrium?
% dissociation relationship to Ka
% = amount dissociated (M) / initial concentration (M) x 100
Calculate the % dissociation of acetic acid (Ka = 1.8 x 10-5) for each of the following:
a. 1.00 M HC2H3O2
b. .100 M HC2H3O2
For a weak acid solution, as [HA] decreases, [H+] decreases BUT % dissociation
increases!
One more example: In a 1.00 M HC3H5O3, the % dissociation is 3.7%, calculate the Ka
for the acid.
Polyprotic Acids
 H2SO4, H3PO4……
 Always dissociate stepwise
 First H+ ion comes off easy - Ka for first H+ is much larger than second; denoted
as Ka1 > Ka2 > Ka3 …..



Example: H2CO3  H+ + HCO3-1
Ka = 4.3 x 10-7
-1
+
-2
HCO3  H + CO3
Ka = 5.6 x 10-11
Usually only the 1st Ka value makes a significant contribution of [H+]; exception
is H2SO4 (STRONG)
only in dilute (less than 1.0 M) solutions of H2SO4 does the second dissociation
contribute significantly to [H+]
Example:
H2CO3 H+ + HCO3-1 Ka = 4.3 x 10-7
HCO3-1 H+ + CO3-2 Ka = 5.6 x 10-11
Example
Calculate the pH of a 5.0 M H3PO4 solution and the equilibrium concentrations of
H3PO4, H2PO4-, HPO4-2, PO4-3
Example
Using the following Ka values, calculate the pH of a 1.40 M H2C2O4 solution and the
equilibrium concentrations of H2C2O4, HC2O4-1, C2O4-2 and OH-
Sulfuric Acid***** unique because it’s the only polyprotic acid that is strong
Strong  1st dissociation
Weak 2nd dissociation
H2SO4  HSO4-1 + H+
HSO4-1  SO4-2 + H+
Ka = large
Ka= 1.2 x 10-2
Example: Calculate the pH of a 1.0 M H2SO4 solution.
***** For a concentrated solution of sulfuric acid (1.0 M or higher) Ka1 provides
enough information – [H+] is mostly from 1st step.
For dilute solutions, 2nd step does make a significant contribution to H+ ions
Example- Calculate the pH of 1.00 x 10-2 M H2SO4
To Calculate pH of weak base– Must use equilibrium expression
 Weak bases react with water removing protons from water
 Use Kb dissociation constants
 Complete the same set of rules set for the calculation of pH of a weak acid
 Use pOH to convert to pH
Example
Calculate the pH of a 15.0 M solution of NH3 (Kb = 1.8 x 10-5 )
Calculate the pH of 1.0 M solution of methylamine CH3NH2 (Kb = 4.38 x 10-4 )
Relationship between Ka and Kb
NH4+  NH3 + H+
NH3 + H2O  NH4+ + OH------------------------------------------H2O  H + + OHKa = [NH3][ H+]
[NH4+]


Kb = [NH4+][ OH-]
[NH3]
By multiplying Ka and Kb you will get Kw = [H+][OH-]
So Kw = Ka x Kb
Example
Calculate Ka or Kb and write the reaction with water for each of the following aqueous
ions
a. NO2-1 (Ka HNO2= 4.0 x 10-4)
b. C6H5NH3 (Ka C6H5NH3= 3.8 x 10-10)
Acid-Base Properties of Salt Solutions
 Salts can be acidic or basic and/or neutral
 Remember when salts dissolve in water, they completely dissociate; all salts are
strong electrolytes
 Some ions can react with water- hydrolysis to form a weak base or acid


Whether the anion of a salt reacts with water to produce hydroxide ions depends
on the strength of the acid to which it is the conjugate. To identify the acid, add a
H and determine its strength
Whether the cation of a salt reacts with water to produce hydrogen ions depends
on the strength of the base to which it is the conjugate. To identify the base, add a
OH and determine its strength
Neutral Solution
Is a weak conjugate base
- ion from a strong acid
SALT
+ ion form a strong base
NO EFFECT ON pH -do not hydrolyze
pH = 7
Is a weak conjugate acid
Example: KCl
Basic Solution
Is a strong conjugate base
- ion from a weak acid
SALT
+ ion form a strong base
WILL BE BASIC
pH > 7
no effect
Example: NaF
F- + H2O  HF + OH-
Acidic Solution
No effect
- ion from a strong acid
SALT
+ ion form a weak base
WILL BE ACIDIC
pH < 7
Is a strong conjugate acid
Example: NH4Cl
NH4+ + H2O  NH3 + H3O+
WHAT IF:
Is a strong conjugate base
- ion from a weak acid
SALT
+ ion form a weak base
Is a strong conjugate acid
DEPENDS ON Ka and Kb
If Ka > Kb its acidic
If Ka = Kb its neutral
If Ka < Kb its basic
Example: NH4F
Example
Predict whether each of the following salts will create an ACIDIC, BASIC, or Neutral
soln
a. Na3PO4
b. KI
c. HC5H5NCl
d. NH4F
To determine the pH of a salt solution
 Determine the type of salt- add H ion to anion and OH ion to cation
 Use equation to write equilibrium expression and then ice tables
Example: Calculate pH of a .100 M salt solution of NH4Cl ( Ka= 5.6 x 10-10)
Example
Calculate the pH of a .500 M NaNO2 solution (Ka= 4.0 x 10-4)
Example
Calculate the pH of a .10 M NH4Cl solution (Kb= 1.8 x 10-5)
Factors that Affect Acid Strength
 Any molecule with a H is a potential acid
 2 factors: the polarity of the H-X bond, and strength of the H-X bond
1. the polarity of the H-X bond
2. strength of the H-X bond
Molecule
H-F
H-Cl
H-Br
H-I
Bond strength
(kJ/mol)
565
427
363
295
Bond polarity
Most polar
2nd most polar
2nd least polar
Least polar
Acid Strength in
water
Weak
Strong
Strong
strong
In binary Acids ( hydrogen and one other element)
 DOWN a group, strength is the most important factor: As you go down a group,
the size of the atom gets larger and the strength decreases and acidity increases

The stronger the H-X bond in a binary acid, the less acidic. Explains HF which is
highly polar- more energy needed to break the bonds than in the other hydrogen
halides which are all considered strong
More polar

H-F > H-Cl> H-Br > H-I less polar
ACROSS a row, bond strengths do not change as much so Bond polarity is more
important. As you go across a row, the electronegativity increases and acidity
increases
Oxy acids ( one or more O-H bonds)
Molecule
HClO
HClO2
HClO3
HClO4


Ka value
3.5 x 10-8
1.2 x 10-2
~1
Large (~107)
The greater the electronegativity of the central atom that has the O-H bond
attached to it, the more acidic because the O-H bond is more polar, favoring the
loss of the H+
Also, the more oxygen added to the central atom of an oxyacid, the greater the
pull of electron density away from the O-H bond, easier to remove the hydrogen
HClO is less acidic than HClO4
Which is the stronger acid? H2TeO4 or H2SO4
Acid Base Properties of Oxides
 acidic oxides - form from elements with high electronegativities - nonmetallic
(covalent) oxides
- SO2 : SO2(g) + H2O(l) H2SO3(aq)
- CO2 : CO2(g) + H2O(l) H2CO3(aq)
 basic oxides - metallic oxides
- CaO(s) + H2O(l) Ca(OH)2(aq)
- K2O(s) + H2O(l) 2KOH(aq)
The Lewis Acid-Base Model
 a Lewis acid is an electron pair acceptor and a Lewis base is an electron pair donor
example: NH3