Acids & Bases
Acids:
 acids are sour tasting
 Arrhenius acid : Any substance that, when dissolved in water, increases the concentration of hydronium ion (H
3
O + )
 Bronsted-Lowry acid : A proton donor
 Lewis acid : An electron acceptor
Bases:
 bases are bitter tasting and slippery
 Arrhenius base : Any substance that, when dissolved in water, increases the concentration of hydroxide ion (OH )
 Bronsted-Lowery base : A proton acceptor
 Lewis acid : An electron donor

Lone Hydrogen ions do not exist by themselves in solution. H+ is always bound to a water molecule to form a hydronium ion

Brønsted-Lowry Theory of Acids & Bases
Conjugate Acid-Base Pairs

Brønsted-Lowry Theory of Acids & Bases

Brønsted-Lowry Theory of Acids & Bases

Brønsted-Lowry Theory of Acids & Bases
Notice that water is both an acid & a base = amphoteric
Reversible reaction

Electrolytes are species which conducts electricity when dissolved in water. Acids, Bases, and Salts are all electrolytes.
Salts and strong Acids or Bases form Strong
Electrolytes. Salt and strong acids (and bases) are fully dissociated therefore all of the ions present are available to conduct electricity.
HCl
(s)
+ H
2
O

H
3
O + + Cl -
Weak Acids and Weak Bases for Weak Electrolytes.
Weaks electrolytes are partially dissociated therefore not all species in solution are ions, some of the molecular form is present. Weak electrolytes have less ions avalible to conduct electricity.
NH
3
+ H
2
O

NH
4
+ + OH -

Acids & Bases
STRONG vs WEAK
_ completely ionized
_ strong electrolyte
_ ionic/very polar bonds bonds
_ partially ionized
_ weak electrolyte
_ some covalent
Strong Acids :
HClO
4
H
2
SO
4
HI
HBr
HCl
HNO
3
Strong Bases:
LiOH
NaOH
KOH
Ca(OH)
2
Sr(OH)
2
Ba(OH)
2

Acids & Bases
• One ionizable proton:
HCl → H + + Cl -
• Two ionizable protons:
H
2
SO
HSO
4
-
4
→ H + + HSO
4
-
→ H + + SO
4
2-
• Three ionizable protons:
H
2
SO
4
Combined:
→ 2H + + SO
4
2-
H
3
PO
4
H
2
PO
4
-
HPO
4
2-
→ H + + H
2
PO
4
–
→ H + + HPO
4
2-
→ H + + PO
4
-3
H
3
PO
Combined:
4
→ 3H + + PO
4
3-

Acids & Bases
For the following identify the acid and the base as strong or weak .
a. Al(OH)
3
+ HCl

Weak base Strong acid b. Ba(OH)
2
+ HC
2
H
3
O
2

Strong base c. KOH + H
2
Weak acid
SO
4

Strong base d. NH
3
Strong acid
+ H
2
O

Weak base
Weak acid

Acids & Bases
For the following predict the product. To check your answer left click on the mouse. Draw a mechanism detailing the proton movement.
a. Al(OH)
3
+ HCl

AlCl
3
+ 3 H
2
O b. Ba(OH)
2
2
2
H
3
O
2

Ba(C
2
H
3
O
2
)
2
+ 2 H
2
O
2
2
SO
4
 K
2
SO
4
+ 2 H
2
O d. NH
3
+ H
2
O

NH
4
+ + OH -

Conjugate Acid-Base Pairs

Conjugate Acid-Base Pairs

Acids & Bases
For the following Identify the conjugate acid and the conjugate base.
The conjugate refers to the acid or base produced in an acid/base reaction. The acid reactant produces its conjugate base (CB) . a. Al(OH)
3
+ 3 HCl

AlCl
3
+ 3 H
2
CB CA
O b. Ba(OH)
2
+ 2 HC
2
H
3
O
2

Ba(C
2
H
3
O
2
)
2
+ 2 H
2
O
CB CA c. 2 KOH + H
2
SO
4

K
2
SO
4
+ 2 H
2
CB CA
O d. NH
3
+ H
2
O

NH
4
+ + OH -
CA CB

Titration of a strong acid with a strong base
ENDPOINT = POINT OF NEUTRALIZATION =
EQUIVALENCE POINT
At the end point for the titration of a strong acid with a strong base, the moles of acid (H + ) equals the moles of base (OH ) to produce the neutral species water (H
2
O). If the mole ratio in the balanced chemical equation is 1:1 then the following equation can be used.
MOLES OF ACID = MOLES OF BASE n acid
= n base
Since M=n/V
M
A
V
A
= M
B
V
B

TITRATION
M
A
V
A
= M
B
V
B
1. Suppose 75.00 mL of hydrochloric acid was required to neutralize 22.50 mLof 0.52 M NaOH. What is the molarity of the acid?
HCl + NaOH

H
2
O + NaCl
M a
V a
= M b
V b rearranges to M a
= M b
V b
/ V a so M a
= (0.52 M) (22.50 mL) / (75.00 mL)
= 0.16 M
Now you try:
2. If 37.12 mL of 0.843 M HNO
3 neutralized 40.50 mL of KOH, what is the molarity of the base?
M b
= 0.773 mol/L

TITRATION
Titration of a strong acid with a strong base
ENDPOINT = POINT OF NEUTRALIZATION =
EQUIVALENCE POINT
At the end point for the titration of a strong acid with a strong base, the moles of acid (H + ) equals the moles of base (OH ) to produce the neutral species water (H
2
O). If the mole ratio in the balanced chemical equation is NOT 1:1 then you must rely on the mole relationship and handle the problem like any other stoichiometry problem.
MOLES OF ACID = MOLES OF BASE n acid
= n base

TITRATION
1. If 37.12 mL of 0.543 M LiOH neutralized 40.50 mL of H
2
SO
4
, what is the molarity of the acid?
2 LiOH + H
2
SO
4

Li
2
SO
4
+ 2 H
2
O
First calculate the moles of base:
0.03712 L LiOH (0.543 mol/1 L) = 0.0202 mol LiOH
Next calculate the moles of acid:
0.0202 mol LiOH (1 mol H
2
SO
4
/ 2 mol LiOH)= 0.0101 mol
H
2
SO
4
Last calculate the Molarity:
M a
= n/V = 0.010 mol H
2
SO
4
/ 0.4050 L = 0.248 M
2. If 20.42 mL of Ba(OH)
2 solution was used to titrate29.26 mL of 0.430 M HCl, what is the molarity of the barium hydroxide solution?
M b
= 0.308 mol/L

• A student finds that 23.54 mL of a 0.122 M
NaOH solution is required to titrate a 30.00-mL sample of hydr acid solution. What is the molarity of the acid?
• A student finds that 37.80 mL of a 0.4052 M
NaHCO
3 solution is required to titrate a 20.00mL sample of sulfuric acid solution. What is the molarity of the acid?
• The reaction equation is:
H
2
SO
4
+ 2 NaHCO
3
→ Na
2
SO
4
+ 2 H
2
O + 2 CO
2

K w
= [H + ] [OH ] = 1.0 x 10 -14
Equilibrium constant for water
 Water or water solutions in which [H + ] = [OH ] = 10 -7 M are neutral solutions.
 A solution in which [H+] > [OH-] is acidic
 A solution in which [H+] < [OH-] is basic

•
The scale for measuring the hydronium ion concentration
[H
3
O + ] in any solution must be able to cover a large range. A logarithmic scale covers factors of 10. The “p” in pH stands for log.
• A solution with a pH of 1 has [H
3
O + ] of 0.1 mol/L or 10 -1
• A solution with a pH of 3 has [H
3
O + ] of 0.001 mol/L or 10 -3
• A solution with a pH of 7 has [H
3
O + ] of 0.0000001 mol/L or 10 -7 pH = - log [H
3
O + ]

The pH scale ranges from 1 to 10 -14 mol/L or from 1 to 14.
pH = - log [H
3
O + ]
1 2 3 4 5 6
8 9 10 11 12 13 14 acid
base

Algebraic manipulation of: pH = - log [H
3
O + ] allows for:
[H
3
O + ] = 10 -pH
If pH is a measure of the hydronium ion concentration then the same equations could be used to describe the hydroxide (base) concentration.
[OH ] = 10 -pOH pOH = - log [OH ] thus:

1. How many milliliters of 1.25 M LiOH must be added to neutralize
34.7 mL of 0.389 M HNO
3
?
10.8 mL
2. What mass of Sr(OH)
2 will be required to neutralize 19.54 mL of
0.00850 M HBr solution?
0.0101 g
3. How many mL of 0.998 M H
2
SO
4 mL of 1.233 M KOH?
must be added to neutralize 47.9
29.6 mL
4. What is the molar concentration of hydronium ion in a solution of pH 8.25?
5.623 x 10 -9 M
5. What is the pH of a solution that has a molar concentration of hydronium ion of 9.15 x 10 -5 ?
pH = 4.0
6. What is the pOH of a solution that has a molar concentration of hydronium ion of 8.55 x 10 -10 ?
pOH = 4.9

______1. How many milliliters of 0.75 M KOH must be added to neutralize
50.0 mL of 2.50 M HCl?
______2. What mass of Ca(OH)
2
0.170 M HCl solution?
will be required to neutralize 100 mL of
______3. How many mL of 0.554 M H
2
SO
4
25.0 mL of 0.9855 M NaOH?
must be added to neutralize
______ 4. What is the molar concentration of hydronium ion in a solution of pH 2.45?
______ 5. What is the pH of a solution that has a molar concentration of hydronium ion of 3.75 x 10 -9 ?
______ 6. What is the pOH of a solution that has a molar concentration of hydronium ion of 4.99 x 10 -4 ?