Acids and Bases
Arrhenius (limited definition)
 Acid- substance that dissolves in H2O to produce H+ ions
 H+ ion or the H3O+(hydronium ion) represents an acid

Base- a substance that dissolves in H2O to produce OH- ions.
 OH- represents a base
Bronsted Lowry(more general definition) ***** Main Theory*****
 Acid- proton (hydrogen) donor in water
 Base- proton (hydrogen) acceptor in water
HA
+
H 2O
Proton donor
Proton acceptor
ACID
BASE
Lewis



H3O+
hydronium ion
conjugate acid
+
Aconjugate base
Acid- electron pair acceptor
Base- electron pair donor
Further Look into Bronsted-Lowry Acids and Bases
Conjugate Acid-Base Pair: an acid/conjugate base or base/conjugate acid partnering
of substances related to one another by donating/accepting
a single hydrogen ion
Conjugates are found on the right side of an equation
 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
+
H2O
ACID
↔
NH4 +
CONJUGATE
ACID
3. H2PO4-1- + H2O ↔
ACID
BASE
BASE
+
H2O
ACID
↔
+
OHCONJUGATE
BASE
H3O+
+ HPO42CONJUGATE CONJUGATE
ACID
BASE
CONJUGATE
HCl and Cl- and H2O and H3O+ are the conjugate acid-base pair
BASE
4. CO32-
Cl-
HCO3 1CONJUGATE
ACID
BASE
+ OHCONJUGATE
BASE
Amphoteric- a substance that can act as an acid or a base. See water above!
Example 2: Identify the acid and the conjugate base pair & the base and conjugate acid
pair
HNO3 + H2O
↔
H3O+
+ NO3Example 3: What is the conjugate acid of PO43-? What is the conjugate base of H2S?
Example 4: Write the dissociation reaction and then identify the acid, base, conjugate
acid and conjugate base
a. Formic acid, HCOOH
b. Perchloric acid, HClO4
There is a competition between the two bases:
 the stronger the acid- the weaker the conjugate base
 the stronger the base- the weaker the conjugate acid
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 acids- equilibrium lies extremely far to the right- all original acid has
dissociated(ionized) at equilibrium Ka is large
[H+] = [HA0]
Examples to memorize – only 6!!
HCl
+
HI, HBr, HCl, HNO3, H2SO4, HClO4

H2O
H3O+ +
Cl-
Weak Acids – 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);
Ka <<<<< 1
[H+] ≠ [HA0]
Examples: organic acids and ALL others not in the above list
HC2H3O2
+
H2 O

H3O+ +
C2H3O2-
Strong Bases- 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
 Na+
+ H2O
+ OH-
Weak Bases- 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(methylamine) and all other
NH3
+ H2O

NH4+ + OH-
To write an equilibrium expression for a WEAK acid or WEAK base:
*****Remember the pure liquid water is left out of expression- concentration will remain
the same in a dilute solution
Acid Dissociation constant
HC2H3O2 (aq) ↔ H+(aq) + C2H3O2-(aq)
Ka = [H3O+][ C2H3O2-]
HC2H3O2
or
Ka = [H+][ C2H3O2-]
HC2H3O2
 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]
Example 5: Write the equilibrium expression for each:
a. Formic acid, HCOOH
b. Nitrous acid
HCOOH(aq) ↔ H+(aq) + COOH-(aq)
HNO2(aq) ↔ H+(aq) + NO2 -(aq)
Auto Ionization of Water
 One water molecule can donate a hydrogen ion(proton) to another water molecule
 Only 1 out of every 109 molecules will self 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 [H] = [OH-]
Acidic
[H+] > [OH-]
Basic
[H+] < [OH-]
= 1.0 x 10-14 M2
H+ = 1.0 x 10-7 M
H+ > 1.0 x 10-7 M
H+ < 1.0 x 10-7 M
Example 6: Determine acidity by looking at the [H+]!
a. Calculate [H+] if [OH-]= 9.3 x 10-4 M at 25 ºC. Is the solution acidic, basic, or
neutral?
b. Calculate [OH-] if [H+]= 6.7 x 10-11 M at 25 ºC. Is the solution acidic, basic, or
neutral?
c. Calculate both the [H+] or [OH-] concentrations in a neutral solution at 60 ºC.
Kw at 60 ºC = 1.0 x 10-13 M2
pH SCALE
Indicate the strengths of acid or base (range from 0-14)
Acid (0-6.999)
Neutral (7.000)
Base (7.001-14)
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 sig figs in the original concentration is
equal to the number of places to the right of the decimal point in the pH
Other Relationships connected to pH
pOH = -log [OH-]
and
Example 7: Fill in the blanks.
pH
pOH
6.88
[H+]
14= pH + pOH
[OH-]
Acid, Base, Neutral
13.08
1.3 x 10-7 M
1.0 x 10-7 M
To calculate pH of Strong Acids and Bases
 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
auto-ionization 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 two hydroxide ions
when it dissociates
ie. 0.30 M NaOH is 0.30 M Na+ and 0.30 M OH- where as
0.30 Ba(OH)2 is 0.30 M Ba+ and 0.60 M OHTo calculate the pH of bases, use pOH = -log[OH-] and then 14 = pH + pOH
Example 8: Calculate the pH of a 5.0 x 10-2 M NaOH solution.
Example 9: Calculate the pH of a solution made by putting 4.63 g LiOH into water and
diluting to a total volume of 400.0 ml.
Example 10: Calculate the pH and the [OH-] of a 5.0 x 10-3 M 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 Acids– 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 11: Calculate the pH of 1.00 M solution of hypochlorous acid.(HOCl, Ka = 3.5 x 10-8)
Example 12: Calculate the pH of a 0.500 M solution of formic acid.(HCOOH Ka = 1.77 x 10-4)?
Example 13: The value for Ka = 7.45 x 10-4 for citric acid, C6H10O8. Calculate the pH of
0.200 M of citric acid.
Calculating the pH of a Mixture of Weak Acids
 Follow same steps as before; ONLY the predominant species will produce the
[H+] ions so look for the acid with the largest Ka value; Ignore the others.
Example 14: Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2 x 10-10)
and 5.00 M HNO2 (Ka = 4.0 x 10-4). Also calculate the concentration of the cyanide ions
(CN-) in this solution at equilibrium.
Example 15: Calculate the pH of a mixture of 2.00 M formic acid, HCOOH (Ka = 1.77 x
10-4) and 1.50 M hypobromous 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?
Percent Dissociation Relationship to Ka
% = amount dissociated (M) / initial concentration (M) x 100
Example 16: Calculate the percent 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!
Example17: In a 1.00 M HC3H5O3, the % dissociation is 3.7%, calculate the Ka for the
acid.
Polyprotic Acids
 H2SO4, H3PO4……
 Always dissociate stepwise. Each dissociation has its own Ka
 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
HCO3-1  H+ + CO3-2
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+] ; will need to use the quadratic equation
Example 18:
Calculate the pH of a 5.0 M H3PO4 solution and the equilibrium concentrations of
H3PO4, H2PO4-, HPO4-2, PO4-3 . Ka H3PO4 = 7.5 x 10-3 ; Ka H2PO41- = 6.2 x 10-8; Ka
HPO42-= 4.8 x 10-13
Example 19:
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 20: 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 21: Calculate the pH of 1.00 x 10-2 M H2SO4
To Calculate pH of Weak Bases– 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 22: Calculate the pH of a 15.0 M solution of NH3 (Kb = 1.8 x 10-5 )
Exmple 23: 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 24: 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. C6H5NH2 (Ka C6H5NH3+= 3.8 x 10-10)
Acid-Base Properties of Salt Solutions (Hydrolysis)
 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
DEPENDS ON Ka and Kb
+ ion form a weak base
Is a strong conjugate acid
If Ka > Kb its acidic
Example: NH4F
If Ka = Kb its neutral
If Ka < Kb its basic
Example 25: Predict whether each of the following salts is ACIDIC, BASIC, or Neutral
a. Na3PO4
b. KI
c. C5H5NCl
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 26: Calculate pH of a 0.100 M salt solution of NH4Cl ( Ka= 5.6 x 10-10)
Example 27: Calculate the pH of a .500 M NaNO2 solution (Ka= 4.0 x 10-4)
Example 28: Calculate the pH of a .10 M NH4Cl solution (Kb= 1.8 x 10-5)
The easier it is to break an H-X bond, the more acidic
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. the strength of the H-X bond
In Binary Acids (Hydrogen and one other element)
 DOWN a group, bond strength is the most important factor: As you go down a
group, the size of the atom gets larger and the bond strength decreases and
acidity increases easier to separate the H from the nm
smaller nm
H-F > H-Cl> H-Br > H-I
larger nm

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
Molecule
H-F
H-Cl
H-Br
H-I

Bond strength
(kJ/mol)
565
427
363
295
Acid Strength in
water
Weak
Strong
Strong
strong
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
H-X bond becomes more polar & acidity increases.  greater tendency for
electrons to move closer to the nm weakening the H-X bond.
Less acidic PH3 < H2S .< HCl more acidic
In OxyAcids (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)
DIFFERENT central atom, but NM varies down a group
 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+
SAME central atom, different numbers of O
 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, and the easier it is 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