I. A/B General
A. Definitions
1. Arrhenius
2. B-L
3. Lewis
*H+=H3O+
B. Properties
1. Acid
2. Base
II. ACIDS
A. Types
1. Binary
2. Ternary
3. Acid Anhydride
B. Protic
1. mono
2. di
3. tri
C. Strengths
1. Determining Factors
a. Degree of Dissociation
b.Molecular Shape and composition
2. Strong Acids
3. Weak Acids
4. Ka, Kb, Kw
D. Calculation of Hydronium from Ka
1. Strong Acids
2. Weak acids
a. < 5%
b. > 5%
1. Method of successive approximations
2. Quadratic Formula
III. Bases
A. Types of bases
1. hydroxy bases (-OH)
2. carbonates (-CO3)
3. molecular bases (NH3)
4. Strong Bases
IV. Amphoterism
V. Acid Base Reactions
A. Arrhenius (Neutralization) STRONG + STRONG
B. B-L (Conjugates) STRONG + Weak
C. B-L Incomplete Neutralization
D. Carbonates + Acid
E. Basic and Acid Anhydrides
VI. pH and pOH
A. Definition, Kw
B. Formula
C. pH + pOH = 14
D. calculating [H3O+] and [OH-] from pH or pOH
VII. Titrations
A. 1:1
B. Unbalanced
VIII. EXTRA MATERIAL
A. pH of salts
B. buffers
Resources:
http://www.files.chem.vt.edu/RVGS/ACT/notes/Notes_on_acids_and_bases.html
http://www.doctortang.com/Honour%20Chemistry%20(Old)/Unit%208%20Acids%20and%20Bases%20No
tes%20(answers).pdf
http://replay.web.archive.org/20061202235618/http:/millard.esu3.org/mnhs/apchem/assignments/Notes/PDF
%20Notes/notes14.PDF
http://phet.colorado.edu/en/simulation/ph-scale
http://www.shodor.org/unchem/basic/ab/ -Henderson Hesselbach Equation
Acids, Bases and Salts
I. Definitions
A. Definitions:
1. Arrhenius
Acid
A substance which
A substance which
increases [H+]
increases [OH-]
2. Bronsted-Lowry A substance that
3. Lewis
Base
A substance that
donates [H+]
accepts [H+]
An e- pair acceptor
An e- pair donor
***** H+(aq) = H3O+
hydrogen ion = a proton = hydronium ion
B. Properties
1. Acids
A. dilute acids taste sour
B. acids react with most metals to liberate H2 gas (S.R.rxn)
C. are neutralized by strong bases to form:
Acid + Base(-OH)  a salt + HOH
Acid + Base(-CO3)  a salt + HOH + CO2
D. cause color changes in indicators Table M
E. are electrolytes in proportion to their [ ]
(keep in mind [ ], solubility (Ksp) and degree of dissociation (Ka)
2. Bases
A. bases taste bitter
B. bases feel slippery
C. are neutralized by strong acids.
D. cause color changes in indicators Table M
E. are electrolytes in proportion to their [ ]
(keep in mind [ ], solubility (Ksp) and degree of dissociation (Kb)
II. ACIDS
A. Types
1. Binary “two element” acid,
(H + nm)
A. Naming Rule: “hydro_____ic acid”
B. Examples:
1. HCl – hydrochloric acid
2. HBr – hydrobromic acid
3. H2S – hydrosulfuric acid
2. Ternary “three or more element” acid, (H + PAI)
TABLE E!!!!!
A. Naming Rule: “ate - ic acid”
“ite – ous acid”
B. Examples:
1. H2SO3 – sulfurous acid
2. H2SO4 – sulfuric acid
3. HNO3 – nitric acid
4. H3PO4 – phosphoric acid
C. Acid Anhydride
Definition: any nonmetallic substance that becomes acidic when
mixed with water.
Examples: CO2 + H2O  H2CO3
SO3 + H2O  H2SO4
B. Protic- indicates the # of protons which can potentially be donated
1. monoprotic – 1 H+ examples: HCl, HNO3
2. diprotic - 2 H+ examples: H2S, H2SO4
3. triprotic - 3 H+ examples: H3PO4
C. Strengths
1. Determining Factors
a. Degree of Dissociation–
1. The more and acid dissociates the stronger it is. This is
determined through Ka values.
2. Acids that dissociate less are weaker.
b. Molecular Shape and composition
1. The more oxygenated an ion is, the larger the oxidation number
of the central atom and thus the less weakly held the H+ ion is
(stronger).
HNO2 v HNO3
HClO vs. HClO4
2. Inorganic acids are typically stronger than organic ones (carbonbased)
2. STRONG acids.
1. The following acids are considered strong because they have a
large degree of dissociation:
3. WEAK Acids
1. Acids that do not dissociate readily are considered weak.
2. Examples: HF, HC2H3O2 aka CH3COOH (vinegar), HC6H7O7 (citric)
4. Ka , Kb, Kw values
1. Ka – dissociation constant for acids
*For strong acids, Ka = extremely large therefore we assume
roughly 100% dissociation.
**For weak acids Ka is a more useful value that will allow us to
determine pH later on.
a. General format: HA(aq)   H+(aq) + A-(aq)
Ka = [H+] [A-]
[HA]
b. Specific Examples:
1. HC2H3O2(aq)  H+(aq) + C2H3O2-(aq)
Ka = [H+] [C2H3O2-]
[HC2H3O2]
2. HC5H3N4O3 (aq)  H+(aq) + C5H3N4O2-(aq)
Ka = [H+] [C5H3N4O2-]
[HC5H3N4O2-]
2. Kb - dissociation constant for bases
*For strong bases, Kb = extremely large therefore we assume
roughly 100% dissociation.
**For weak bases Kb is a more useful value that will allow us to
determine pH later on.
a. General format: A- (aq) + H2O (l)  HA(aq) + OH-(aq)
Kb = [HA] [OH-]
[A-]
b. Specific Examples:
1. NH3(aq) + H2O  NH4+(aq) + OH-(aq)
Kb = [NH4+] [OH-]
[NH3]
3. Kw = dissociation constant for water = 1 x 10-14
H2O (l)   H+ (aq) + OH- (aq)
D. Calculation of [H3O+] from Ka
1. 0.1M HCl = 0.1M [H3O+]
2. 0.033M HBr = 0.033M [H3O+]
3. 0.2 M HC2H3O2 = X [H3O+]
[H3O+] = square root (Ka x [HA])
[H3O+] = 0.0019M
Ka = 1.8x10-5
5% rule check = 0.0019M/0.2M x100 =
ASSUMPTION VALID
4. 0.05M HClO = X [H3O+]
Ka = 3.0x10-8
[H3O+] = sq. rt (Ka x [HA])
[H3O+] = 3.875 x 10-5M
5% Rule = 0.08%
ASSUMPTION VALID
5. 0.05 M HF = X [H3O+]
Ka = 6.5x10-4
[H3O+] = .0057M
5% Rule = 11.4%
THEREFORE ASSUMPTION INVALID
We need to use 1. Method of Successive Approximations or
2. Quadratic Formula
6. 0.10M HClO2 = X [H3O+]
Ka = 1.1x10-2
[H3O+] = 0.033M
5% Rule = 33%
THEREFORE ASSUMPTION INVALID
New [H3O+] = .027M
III. Bases
A. Types
1. hydroxy bases (m-OH)
2. carbonates (m-CO3)
3. molecular (NH3) – covalently bonded. Typically have unused electron pairs.
4. Strong Bases - these are the hydroxide family and will fully neutralize strong ac
IV. Amphoteric Substances !!!!!!!!!!!!!
1. Definition: Substances that can act as either an acid or a base!!!
2. Examples: H2O, HSO4-, H2PO4-2 (any partially neutralized acid)
V. Acid/Base Reactions
A. Complete Neutralization
(a salt + H2O)
1. HCl + NaOH -->
2. H2SO4 + Mg(OH)2 -->
3. H2SO4 + KOH -->
4. HClO4 + Ca(OH)2 -->
B. Incomplete Neutralization
1. H2SO4 + (1 mole)NaOH -->
2. H3PO4 + (1 mole) KOH -->
C. B-L Conjugate Acid/Base Pairs
1. HNO3 + H2O -->
2. HC2H3O2 + H2O -->
3. HC2H3O2 + NH3 -->
4. HClO2 + H2O 
Important Information:
1. only 1 H+ is transferred in these reactions.
2. A STRONG ACID/BASE will become a weak conjugate base/acid.
A weak acid/base will become a STRONG CONJUGATE BASE/ACID
D. Carbonates
(a salt, H2O, CO2)
1. HC2H3O2 + NaHCO3 -->
2. HCl + CaCO3 -->
E. Acid/Basic Anhydrides + Base/Acid (a salt + H2O)
1. MgO + H2SO4 -->
2. CO2 + Ca(OH)2 -->
VI. Kw, H+, OH-, pH, pOH - CLICK HERE for a cool animation
A. Kw = [H+] [OH-]
1. Kw = 1x10-14 = [H+] [OH-]
acidic
[H+] > [OH-]
basic
[H+] < [OH-]
neutral [H+] = [OH-]
Therefore in a neutral solution 1x10-14 = (1x10-7)(1x10-7)
2. Calculating [H+] or [OH-]
[H+] = Kw / [OH-]
[OH-] = Kw / [H+]
example: 1. Determine the hydronium ion concentration if the
hydroxide concentration = .0001M
[H+] = Kw / [OH-]
[H+] = 1x10-14 / [1x10-4]
[H+] = 1x10-10
2. Determine [OH-] if [H3O+] = 4.8x10-11?
[OH-] = Kw / [H+]
[OH-] = 1x10-14 / [4.8x10-11]
[OH-] = 2.08x10-4
3. pH = -log [H3O+]
Example:
a. What is the pH of a .00000001M HCl sln?
pH = -log (1x10-8)
pH = 8
b. What is the pH of a .000482M H2SO4 sln?
pH = -log (4.82 x 10-4)
pH = 3.32
c. What is the pH of a .00078M NaOH sln?
We need to first determine the [H3O+] from [OH-]
[H3O+] = Kw / [OH-]
[H3O+] = 1x10-14 / [7.8x10-4]
[H3O+] = 1.28x10-11
pH = -log [H3O+]
pH = -log 1.28x10-11
pH = 10.89
4. pOH calcs
pOH = - log [OH-] or
14 = pH + pOH
COMPLETING THE CALCULATION CIRCUIT!!!!!!!!!!
5. [H3O+] = inv log -pH
[OH-] = inv log -pOH
VII. Titrations
A. Definition: The laboratory procedure where a known concentration
of an acid or base is used to neutralize the other to and end point
in order to determine the unknown's [ ]. Typically NaOH is added to the
acid and an indicator like phth is used.
B. In a titration;
moles acid = moles base
Assuming that the # H+ = #OH-
available then you can use the following formula:
Ma x Va = Mb x Vb
Ma = molarity of acid
Va = volume of acid
Mb = molarity of base
Vb = volume of base
1. Examples
a. 40 mls. of a 0.50M HCl are titrated with 20 mls of NaOH.
What is the [NaOH]?
MaVa=MbVb
HCl and NaOH are 1:1 so use formula
directly.
Mb = MaVa / Vb
Mb = (0.50M x 40mls) / 20 mls
Mb = 1.0M NaOH
b. 60 mls of HNO3 are neutralized by 20 mls of 0.1M NaOH.
What is the [HNO3]?
MaVa = MbVb
H+:OH- = 1:1 ratio
Ma = MbVb / Va
Ma = (0.1M x 20 mls) / 60 mls
Ma = 0.033M HNO3
c. 10 mls H2SO4 are neutralized by 25 mls of 0.04M Ca(OH)2.
What is the [H2SO4]?
MaVa = MbVb
H+:OH- = 2:2 or 1:1 ratio
Ma = MbVb / Va
Ma = (0.04M x 25mls) / 10mls
Ma = 0.1M H2SO4
2. If the H+:OH- is not 1:1 you need to introduce the equivalents factor
into the equation and then solve!!!!
a. 10 mls of 0.1M H2SO4 are neutralized by 20 mls of NaOH. What
is the [NaOH]?
MaVa = MbVb
H+:OH- = 2:1 So we need to
introduce those #'s to the equation
2 MaVa = MbVb
Mb = 2 MaVa / Vb
Mb = (2 x 0.1M x 10 mls) / 20 mls
Mb = 0.1M NaOH
b. 10 mls of 0.1M H3PO4 are neutralized by 40 mls of Ca(OH)2.
What is the [Ca(OH)2]?
MaVa = MbVb
H+:OH- = 3:2 So we need to
introduce those #'s to the equation
3MaVa = 2MbVb
Mb = 3MaVa / 2Vb
Mb = (3 x 0.1M x 10 mls) / (2 x 40 mls)
Mb = 0.0375M Ca(OH)2
VIII. Additional Information
A. pH of Salts
The pH of a salt depends on the types of acids and bases that were used
to make it.
The quick way of determining the pH of a salt is:
1. If the cation comes from a strong base and the anion from a weak
acid the salt will be basic.
Examples:
NaC2H3O2
strong base / weak acid = basic
KClO2
strong base / weak acid = basic
2. If the cation comes from a weak base and the anion from a strong
acid the salt will be acidic.
Examples:
NH4SO4
weak base / strong acid = acidic
NH4Cl
weak base / strong acid = acidic
3. If the cation comes from a strong base and the anion from a
strong acid the salt will be neutral.
Examples:
NaBr
strong base / strong acid = neutral
K2SO4
strong base / strong acid = neutral
B. Buffers
A buffer solution is one in which the pH of the solution is "resistant" to
small additions of either a strong acid or strong base. Buffers usually
consist of a weak acid and its conjugate base, in relatively equal and
"large" quantities.
Example: acetic acid + sodium acetate
HC2H3O2 + H2O <--> C2H3O2-(aq) + H3O+
According to LeChatelier's Principle if you add acid, the excess
acetate ion will absorb
it and shift to the left. If you add a base the OH- will be
neutralized by the H3O+ and shift the reaction to the right.