Acid-Base Chemistry
Arrhenius acid: Substance that dissolves in water and
provides H+ ions
Arrhenius base: Substance that dissolves in water and
provides OH- ions
Examples:
HCl  H+ and Cl-
Acid
NaOH  Na+ + OH- Base
Bronsted Acid: Substance that donates proton to another substance
Bronsted base: Substance that accepts proton from another substance
Example: HCl + H2O  H3O+ + ClHCl acts as acid; H2O acts as base
In the Reverse Reaction,
H3O+ acts as an acid; Cl- acts as a base
Note:
(H3O+ = hydronium ion = H+ = proton)
•A monoprotic acid contains one acidic proton.
HCl
•A diprotic acid contains two acidic protons.
H2SO4
•A triprotic acid contains three acidic protons.
H3PO4
•A Brønsted–Lowry acid may be neutral or it may
carry a net positive or negative charge.
HCl, H3O+, HSO4−
Conjugate acid: Species formed after base accepts a proton
Conjugate base: Species remaining after an acid donates its proton
Conjugate acid-base pair: an acid and base on opposite sides of the
equation that correspond to each other
Examples: HNO3 + H2O
acid
base
H3O+ + NO3acid
Conjugate pairs: HNO3 and NO3H2O and H3O+
base
Example: HS- + H2O  H3O+ + S2Conjugate pairs:
HS- and S2H2O and H3O+
Practice: HClO4 + H2O  H3O+ + ClO4What are the conjugate pairs?
HClO4 and ClO4H2O and H3O+
gain of H+
H Br
acid
+
H2O
base
Br−
base
+
H3O+
acid
loss of H+
•HBr and Br− are a conjugate acid–base pair.
•H2O and H3O+ are a conjugate acid–base pair.
Note: The net charge must be the same on both sides
of the equation.
Water can act as both an acid and a base (amphiprotic)!
HClO4 + H2O  H3O+ + ClO4 (base)
NH3 + H2O OH- + NH4+ (acid)
Strengths of Acids and Bases
Strong acids/bases: dissociate completely when dissolved in
solution
Weak acids/bases: dissociate only partially when dissolved in
solution
Examples:
Strong Acid: HCl  H+ + Cl-
(100% dissociation)
Strong Base: NaOH  Na+ + OH (100% dissociation)
Weak Acid: CH3COOH  H+ + CH3COO- (1.3% dissociation)
Weak Base: NH3 + H+  NH4+
Table 9.1
Naming Acids
Binary Acids: hydo + root of anion + ic + “acid”
ex. HCl hydrochloric acid, HBr hydrobromic acid
HI
Hydroiodic acid
Polyatomic-based Acids: root of polyatomic ion + ic + “acid”
ex. H2SO4 sulfuric acid, H3PO4 phosphoric acid
H2CO3
carbonic acid
HNO3
nitric acid
The Self-Ionization of Water
H2O + H2O  H3O+ + OHPure water: [H3O+]=[OH- = 10-7 M (at 250C)
Neutral Solution: Any solution in which the concentrations of H3O+
and OH- ions are equal (10-7 M)
Acidic Solution: Solutions having a greater concentration of H3O+
than OH- ions ([H3O+] greater than 10-7 M)
Example: A solution with [H3O+] = 10-5 M
Basic Solution: solution having a greater concentration of OH- than
H3O+ions ([H3O+] less than 10-7 M)
Example: A solution with [H3O+] = 10-12 M
The pH Scale
pH is a measure of acidity
Scale ranges from 0-14
pH = 7 Neutral
pH < 7 Acidic
pH > 7 Basic
pH represents the concentration of H+ ions in solution
Pure water: 1 x 10-7 moles H+ per liter and1 x 10-7 moles OH- per liter
Solutions with equal concentrations of and ions are called Neutral
Solutions with more than 1 x 10-7 moles H+ per liter are Acidic
Solutions with less than 1 x 10-7 moles H+ per liter are Basic
Note: [H+] x [OH-] = 10-14 (always!)
pH Scale Summary
pH scale refers to amount of H+ ions in solution
pH 7 is neutral, less than 7 is acidic, greater than 7 is basic
Lower pH = more acidic = more H+ ions
Higher pH = more basic = less H+ ions
Each pH unit represents a 10-fold change in H+ ion concentration!
pH 4 has 10 times more H+ ions than pH 5
pH 9 has 10 times fewer H+ ions than pH 8
Mathematical equation for pH:
pH is the negative log of the H3O+ concentration
pH = -log [H3O+]
pH = -log [H3O+]
Any number can be expressed as 10 raised to some exponent: y = 10x
Examples:
100 = 10 2
1000 = 103
0.10 = 10 -1
The log is that exponent!
100 = 10 2; Log of 100 =2
1000 = 103; Log of 1000 = 3
0.10 = 10 –1; Log of 0.10 = -1
We can also take the log of non-whole numbers, but we use our
calculators for this.
Example: Find the log of 2.4 x 10-3
Enter 2.4 x 10-3into calculator
Press the “log” key
0.0024 “log” = -2.62
Therefore, 10-2.62 = 2.4 x 10-3
Calculating pH from [H3O+]
pH = -log [H3O+]
Enter [H3O+] into calculator
Press the “log” key
Change the sign
Example:
[H3O+] = 1.0 x 10-7 M
pH = -log [H3O+]
pH = -log [1 x 10-7] = 7
Example:
[H3O+] = 1 x 10-11M
pH = -log [H3O+]
pH = -log [1 x 10-11] = 11
Example:
[H3O+] = 1 x 10-3 M
pH = -log [H3O+]
pH = -log [1 x 10-3] = 3
Example:
[H3O+] = 4.2 x 10-5
pH = -log [H3O+]
Enter [H3O+] into calculator (4.2 x 10-5)
Press the “log” key (-4.3767507)
Change the sign (4.3767507)
pH = 4.3767507 = 4.4
Example:
[H3O+] = 8.1 x 10-9
pH = -log [H3O+]
Enter [H3O+] into calculator (8.1 x 10-9)
Press the “log” key (-8.091515)
Change the sign (8.091515)
pH = 8.091515 = 8.1
Reactions Between Acids and Bases
Neutralization: reaction between an acid and a base; always
produces salt and water
Example: Write a balanced equation for the reaction of
hydrochloric acid with sodium hydroxide.
HCl(aq) + NaOH(aq)
H—OH(l) + NaCl(aq)
Example: Write a balanced equation for the reaction of
hydrochloric acid with magnesium hydroxide.
2 HCl + Mg(OH)2
2 H2O + MgCl 2
Titration
Titration: a technique used to determine the concentration of an acid
or base in a solution
•If we want to know the concentration of an acid
solution, a base of known concentration is
added slowly until the acid is neutralized.
•When the acid is neutralized:
# of moles of acid = # of moles of base
•This is called the end point of the titration.
Acid-Base Titration
• Titration is a laboratory procedure
Base
used to determine the molarity of an (NaOH)
acid.
• In a titration, a base such as NaOH is
added to a specific volume of an acid.
Acid
solution
• A few drops of an indicator is
added to the acid in the flask.
• The indicator changes color
when the base (NaOH) has
neutralized the acid.
• At the end point, the indicator
has a permanent color.
• The volume of the base used to
reach the end point is measured.
• The molarity of the acid is
calculated using the
neutralization equation for the
reaction.
Determining an unknown molarity from titration data
requires three operations:
mole–mole
conversion
factor
Moles of
base
M (mol/L)
conversion
factor
[1]
Volume of
base solution
[2]
Moles of
acid
[3]
Molarity of
acid solution
M (mol/L)
conversion
factor
HOW TO Determine the Molarity of an Acid Solution
from Titration
Example: What is the molarity of an HCl solution if 22.5 mL of a
0.100 M NaOH solution are needed to titrate a 25.0 mL sample of
the acid?
volume of base (NaOH)
22.5 mL
conc. of base (NaOH)
0.100 M
volume of acid (HCl)
25.0 mL
conc. of acid (HCl)
?
Step [1]
Determine the number of moles of base
used to neutralize the acid.
M (mol/L)
conversion factor
Volume of
base solution
22.5 mL NaOH
x
1L
x
1000 mL
mL–L
conversion factor
0.100 mol NaOH
1L
=
0.00225 mol NaOH
Step [2]
Determine the number of moles of acid that react
from the balanced chemical equation.
HCl(aq) + NaOH(aq)
0.00225 mol NaOH
x
H2O(l) + NaCl(aq)
1 mol HCl
1 mol NaOH
mole–mole
conversion
factor
= 0.00225 mol HCl
Step [3]
M
=
Determine the molarity of the acid from the number
of moles and the known volume.
mol
L
=
0.00225 mol HCl
25.0 mL solution
x
1000 mL
1L
mL–L
conversion factor
=
0.0900 M HCl
Answer
Buffers
Buffer: a solution whose pH changes very little when acid or
base is added.
Most buffers are solutions composed of roughly
equal amounts of
•a weak acid
•the salt of its conjugate base
The buffer resists change in pH because
•added base, −OH, reacts with the weak acid
•added acid, H3O+, reacts with the conjugate base
Buffers contain 2 compounds:
Compound with the ability to react with H+ ions
Compound with the ability to react with OH- ions
Example: HCO3- + H+  H2CO3
If acids (H+) are added, react with HCO3H2CO3 + OH-  HCO3- + H2O
If OH- ions are added, react with H2CO3
H2CO3 is unstable: H2CO3  H2O + CO2
More Examples of a Buffer
If an acid is added to the following buffer equilibrium,
Adding more
product…
CH3COOH + H2O
H3O+ + CH3COO−
conjugate
base
weak acid
…drives the reaction to the left.
then the excess acid reacts with the conjugate base,
so the overall pH does not change much.
If a base is added to the following buffer equilibrium,
Adding more
reactant…
CH3COOH + −OH
H2O + CH3COO−
conjugate
base
weak acid
…drives the reaction to the right.
then the excess base reacts with the weak acid, so
the overall pH does not change much.