A.P. Chemistry
Chapter 14
Acid- Base Chemistry
14.1
Arrhenius
Acid- an acid is any substance that dissolves in water to produce H+ (H3O+)
ions
Base- a base is a substance that dissolves in water to produce OH- ions
Bronsted- Lowry
Acid- an acid is a proton (H+) donor
Base- a base is a proton acceptor
Hydronium Ion- the ion that foms when water accepts a proton (H+)
General reaction that occurs when an acid is dissolved in water
HA (aq) + H2O(l)  H3O+(aq) + A-(aq)
H3O+
Conjugate base- everything that remains of an acid molecule after a proton is
lost
Conjugate acid- the molecule which is formed when the proton is transferred
to the base
Conjugate acid- base pair consists of two substances related to each other by
the donating and accepting of a single proton.
Equilibrium expression for a general reaction
when an acid is dissolved in water would be
Ka = [H3O+][A-] =
[H+][A-]
[HA]
[HA]
(water not included; pure liquid)
Ka is called the Acid Dissociation Constant
Example: Write the formula of the conjugate base
of H2SO4. (a conjugate base differs from its acid
by the lack of a proton: HSO4-)
Example: For the reaction
HSO4-(aq) + HCO3-(aq)  SO42-(aq) + H2CO3(aq)
• identify the acids and bases for the forward and
reverse reactions
(forward: HSO4- is proton donor, acid; HCO3- is
proton acceptor, a base.)
• identify the conjugate acid-base pairs
14.2 Acid Strength
Strong Acid- one in which the equilibrium lies far to
the right; it has a large value of Ka. Strong acids
also yield weak conjugate bases.
Weak Acids- one in which the equilibrium lies far to
the left; it has a small value of Ka. Weak acids
yield relatively strong conjugate bases.
See Table 14.1, p. 662
Example: The strengths of the following acids
increase in the order HCN < HF < HNO3. Arrange
the conjugate bases of these acids in order of
increasing base strength.
(CN- > F- > NO3-)
Diprotic Acid- an acid having two acidic protons;
Examples: H2SO4
Oxyacids- the acidic proton is attached to an oxygen
atom, Examples:
Organic Acids- those acids with a carbon atom
backbone; commonly contain the
carboxylic group (-COOH, or
Examples:
)
Monoprotic acid- acids having one acidic proton.
Examples:
Amphoteric- can behave either as an acid or a base; ex.,
water.
Autoionization- involves the transfer of a proton from one
water molecule to another to produce a hydroxide ion and
a hydronium ion.
2H2O(l)  H3O+(aq) + OH-(aq)
Ion-product Constant Kw (or dissociation constant for water)always refers to the autoionization of water. Using the
above reaction
Kw = [H3O+][OH-] = [H+][OH-]
At 25oC, [H+] = [OH-] = 1.0 x 10-7 M
therefore, at 25oC, Kw = (1.0 x 10-7M)2 = 1.0 x 10-14 mol2L-2
It is important to recognize the meaning of Kw:
The product of [H+] and [OH-] must always equal
1.0 x 10-14.
This lends itself to three possibilities:
• A neutral solution, where [H+] = [OH-]
• An acidic solution, where [H+] > [OH-]
• A basic solution, where [H+] < [OH-]
Ka x Kb = Kw = 1.0 x 10-14
Example:The H+ concentration in a solution is 5.0 x
10-5M. What is the OH- ion concentration?
(2.0 x 10-10M; don’t forget unit when asked for a
concentration!)
14.3 pH
The pH scale is used to determine the strength of an acid or a base.
Traditionally it ranges from 0-14. Acids have pH< 7; bases have pH >
7 and 7 on the scale is considered to be neutral. pH is defined as
pH = -log[H3O+(aq)] or pH = -log[H+(aq)]
In the case of strong acids and bases dissociation is complete and
therefore the concentration of the H+ ion or OH-ion can be
determined directly from the stoichiometric ratio in the balanced
equation and the concentration of the acid or the base. Therefore
we can simply plug in the numbers to the equation above, or, use
the fact that, @ 25oC
logKw = log[H+] + log[OH-] = 7 + 7 =
And then use
pOH = -log[OH-(aq)]
14 = pH + pOH
***Significant Figures for logarithms:
the number of decimal places in the log is equal
to
the number of significant figures in the original
number.***
Example: The OH- ion concentration in a certain ammonia
solution is 7.2 x 10-4M. What is the pOH and pH?
(Answer: 3.14; 10.86)
Problem: What is the pH and pOH of a 0.001 M solution
of HNO3?
Problem: What is the pH of a 0.045 M solution of HCl?
Problem: What is the pH and pOH of a 0.02 M solution of
H2SO4?
Example: The pH in many cola-type soft drinks is about
3.0. How many times greater is the H+ concentration in
these drinks than in neutral water? (104 or 10,000X)
(Working backwards: If you know pH and want to
find concentration, use the antilog.)
[H+] = 10-pH
[OH-] = 10-pOH
Another useful relationship in calculations is
pKa = -logKa
and
pKb = -logKb
and pKw = pH + pOH
Problem: What is the concentration of a HCl
solution for which the pH is 4.32?
14.4 Calculating the pH of Strong Acid Solutions
See Table for List of strong and weak acids and
bases. Know the List!
Example: What is the pH of a 0.1 M HCl
solution? Of a 0.0001 M HCl solution?
Example: What is the pOH of a 0.1 M NaOH
solution? The pH?
Example: What is the pOH of a 0.00001 M NaOH
solution? The pH?
14.5 Calculating the pH of a Weak Acid Solution VERY IMPORTANT!!!
Step 1- List the major species in the solution
Step 2- Choose the species that can produce H+, and write a balanced
equation for the reaction producing H+
Step 3- Using the values of the equilibrium constant for the reactions you
have written, decide which equilibrium will dominate in producing H+.
Step 4- write the equilibrium expression for the dominant equilibrium.
Step 5- List the initial concentrations of the species participating in the
dominant equilibrium
Step 6- Define the change needed to achieve equilibrium; that is, define x.
Step 7- Write the equilibrium concentrations in terms of x.
Step 8- Substitute the equilibrium concentrations into the equilibrium
expression
Step 9- Solve for x the “easy” way; that is, assume [HA]o - x = [HA]o
(assume x is negligible)
Step 10- Using the 5% rule, verify is that approximation is valid
Step 11- calculate [H+] and pH.
5% Rule: x/[HA] x 100% < 5%
then the approximation is acceptable
Percent Dissociation:
% dissociation = amount dissociated x 100%
Initial concentration
For a given weak acid, the percent dissociation
increases as the acid becomes more dilute. For
solutions of any weak acid HA, [H+] decreases as
[HA]o decreases, but the percent dissociation
increases as [HA]o decreases.
Example: Determine the pH of a of a 0.10 M solution of CH3COOH.
Ka for acetic acid is 1.8 x 10-5
Determine the percent ionization of acetic acid in the example.
Problem: Calculate the concentrations of H+, F-,
and HF in a 0.31 M HF solution. (Ka for HF is
7.1 x 10-4) Calculate the % ionization.
(Answer: [H+] = [F-] = 1.5 x 10-2 M
[HF] = 0.31 M - 0.015 M = 0.30 M; 4.8%)
14.6 Bases
Strong Base – one which dissociates completely in water and
has a large Kb value.
General reaction for a base B and water is
B(aq) + H2O(l)  BH+(aq) + OH-(aq)
Base
acid
conjugate
conjugate
acid
base
The equilibrium constant, Kb, for this equation is
Kb = [BH+][OH-]
[B]
Weak bases have small values of Kb. (see Table 14.3, p. 685)
Example: What is the H+ concentration in
0.0025 M NaOH? (4.0 x 10-12 M)
Example: What is the pH of a 0.010 M C5H5N
(pyridine) solution?(8.62)
14.7 Polyprotic Acids
Polyprotic acids- acids which can furnish more than 1 proton.
(See Table 14.4, p. 689)
Characteristics of Weak Polyprotic Acids (p. 694)
Typically, successive Ka values are so much smaller than the first value,
that only the first dissociation step makes a significant contribution
to the equilibrium concentration of H+. This means that the
calculation of the pH for a solution of a typical weak polyprotic acid
is identical to that for a solution of a weak monoprotic acid.
Sulfuric acid is unique in being a strong acid in its first dissociation step
and a weak acid in its second dissociation step. For relatively
concentrated solutions of sulfuric acid (1.0 M or higher) the large
concentration of H+ ions from the first dissociation step represses
the second dissociation step (which can then be neglected as a
contributor of H+ ions). For dilute solutions of sulfuric acid, the
second step does make a significant contribution, and the quadratic
equation must be used to obtain the total H+ concentration.
Example: Calculate the concentrations of H2A,
HA-, A2-, and H+ in a 1.0 M H2A solution.
14.8 Acid-Base Properties of Salts
Salt- another name for an ionic compound
Salts that contain highly charged metal ions
produce an acidic solution. The metal ion
becomes hydrated which then causes the
solution to become acidic. See Table 14.6, p.
700.
Example: Write net ionic equations to show which
of the following ions hydrolyze in aqueous
solution? NO3-, NO2-, NH4+
Example: calculate Kb for NO3(2.2 x 10-11)
Example: calculate Ka for NH4+
(5.6 x 10-10)
Example: Determine the pH of a 0.10 M solution of NH4Cl. (Kb
for NH3 is 1.8 x 10-5)
Example: Predict whether the following aqueous solutions will
be acidic, basic, or neutral.
KI (neutral)
NH4I (acidic)
CH3COOK (basic)
14.9 The Effect of Structure on Acid-Base Properties
There are two main factors that determine whether a
molecule containing an X-H bond will behave as a BronstedLowry acid: the strength of the bond and the polarity of the
bond. Increased polarity and high electron density typically
lends to large Ka values (strong acids). (homework
problem; has been on AP exam!!)
HF?
Example: Which is the stronger acid? HCl or HBr? (HBr) HCl or
H2S (HCl)
Example: Which is the stronger acid? HClO3 or HBrO3? (HClO3)
H3PO3 or H3PO4 (H3PO4)
14.10 Acid-Base Properties of Oxides (Important
in Equation Writing!!)
A compound containing the H-O-X group will
produce an acidic solution in water if the O-X
bond is strong and covalent. If the O-X bond is
ionic, the compound will produce a basic solution
in water. (homework problem; has been on AP
exam!!)
Acidic oxides- a covalent oxide dissolves in water,
and an acidic solution forms.
Basic oxides- an ionic oxide dissolves in water, and
oxide has a great affinity for H+, causing basic
solutions.
14.11 Lewis Acid-Base Model (encompasses the B-L model,
but the reverse is not true!)
(Be able to draw the Lewis structure of this reaction)
Acid- an acid acts as an electron pair acceptor
Base- a base acts as an electron pair donor
Example: identify the Lewis acids and bases in each of the
following reactions:
Ag+(aq) + Cl-(aq)  AgCl(s)
(acid: Ag+)
BF3(g) + NF3(g)  F3N-BF3(s) (acid: BF3)
SO2(g) + H2O(l)  H2SO3(aq)
(acid: SO2)