ACIDS AND BASES
CHEMISTRY
CHAPTER 12
12.1 Acids and Bases: Introduction
A. All aqueous solutions contain hydrogen ions (H+) and
hydroxide (OH-).
1. An acidic solution contains more (H+) ions than
(OH-) ions.
2. A basic solution contains more (OH-) ions than (H+)
ions.
3. A neutral solution will contain equal concentrations
of (H+) and (OH-) ions.
4. A hydronium ion (H3O+) is a hydrated hydrogen ion.
5. The symbols H+ and OH- can be used
interchangebly in chemical reactions.
B. Arrhenius model – of acids and bases
1. An acid is a substance that contains hydrogen and
ionizes in aqueous solutions to produce hydrogen
ions.
2. A base is a substance that contains a hydroxide
group and dissociates in aqueous solution to
produce hydrogen ions.
3. Example: HCl
H2+ + Cl4. Sulfuric acid, a polyprotic compound, ionizes in two
steps.
H2SO4
H2+ + HSO4HSO4
H 2+
+ SO4-2
5. a base ionizes or dissociates in solution to produce
hydroxide ions, OHNaOH
Na+ + OH-
C. Bronstead-Lowery model
1. An acid is a hydrogen donor and a base is a hydrogen-ion
acceptor.
a) When a acid donates a hydrogen-ion, a conjugate
base is formed.
b) When a base accepts a hydrogen-ion, a
conjugate acid formed.
c) Two substances related to each other by
donating and accepting of a hydrogen ion are a
conjugate acid-base pair.
2. A base includes any substance that would accept a proton,
and an acid is a proton donor.
a) A product that results from an acid-base reaction are
called the conjugate acid and conjugate base.
3. Example:
HF + HCO3
H2CO3 + F
4. The conjugate base is a particle that remains after a
proton is donated by an acid. The conjugate acid is
formed when a base accepts a proton from an acid.
The hydrogen carbonate ion, HCO3, behaves as a base
but does not contain ionizable hydroxide.
Practice Problems Assigned WS-19A
D. Monoprotic and Polyprotic Acids
1. An acid that can donate only one hydrogen ion is called a
monoprotic acid.
a) Example: hydrochloric acid (HCl) and formic acid
(HCOOH) are monprotic acids, because they
each contain only one ionizable hydrogen.
(Note that only those hydrogen atoms that are
bonded to electronegative elements are
ionizable.)
2. An acid that can donate more than one hydrogen ion.
a) Example: sulfuric acid (H2SO4) contains two
ionizable hydrogen atoms called a diprotic acid.
b) Example: boric acid (H3BO3) contains three
ionizable hydrogen atoms, so it is a triprotic acid.
3. Examples:
H3BrO3 + H2O
H2BrO3 + H2O
HBrO3 + H2O
H3O + H2BrO3
H3O + HBrO3
H3O + BrO3
12.2 Strengths of Acids and Bases
A. Strength of Acids
1. An acid that ionizes completely in dilute aqueous solution
is a strong acid.
a) Examples: hydrochloric acid (HCl), nitric acid
(HNO3), sulfuric acid (H2SO4), and perchloric
acid (HClO4)
2. An acid that only partially ionizes in aqueous solution is a
weak acid.
a) Examples: carbonic acid (H2CO3), boric acid
(H3BO3), phosphoric acid (H3PO4), and acetic
acid (HC2H3O2).
3. The ionization of a weak acid reaches a state of
equilibrium in which the forward and reverse reactions
occur at equal rates.
a) Example: Formic acid (HCOOH); a weak organic
acid
HCOOH + H2O
H3O + HCOO
b) The equilibrium constant expression for the
ionization of formic acid.
c) Ka is called the acid ionization constant , which
is the value of the equilibrium constant
expression for the ionization of a weak acid.
d) The value of Ka indicates the extent of
ionization of the acid. The weakest acids have
the smallest Ka value.
e) In the case of a polyprotic acid, there is a Ka
value for each ionization, and the Ka values
decrease for each successive ionization.
Example: the first and second ionization of phosphoric acid
is represented by this equation.
H3PO4 +
H2 O
H3O + H2PO4
Ka = (H3O) (H2PO4)
(H3PO4)
H2PO4 + H2O
Ka = (H3O) (HPO4)
(H2PO4)
Practice Problems Assigned WS-19C
H3O + HPO4
B. Strength of Bases
1. Metallic hydroxides, such as potassium hydroxide, are
strong bases because they dissociate entirely into metal
ions and hydroxide ions in aqueous solution.
KOH
K+ + OH2. A weak base ionizes only partially in dilute aqueous
solution to form an equilibrium mixture. (Aniline equals
C6H5NH2)
C6H5NH2 + H2O
C6H5NH3 + OH
a) The equilibrium constant ionization expression for
the ionization of aniline in water is as follows.
Kb = (C6H5NH3) (OH)
(C6H5NH2)
3. Kb is called the base ionization constant, which is the value of
the equilibrium constant expression for the ionization of a
weak base.
Practice Problems Assigned WS-19D
Calculating Ionization Constants
When ionic compounds are dissolved in water the ions separate from each
other in a process called dissociation. Many molecular compounds when
dissolved in water, react with the water to produce ions in a process called
ionization.
Calculating the (OH-) or (H3O+) when the Kb or Ka is known.
Example: Calculate the (OH-) of a .500 M solution of aqueous ammonia. The Kb
of NH3 = 1.77 x 10-5
Write the equation:
NH3 + H2O
NH4 + OH
Write the ionization constant expression of ammonium hydroxide:
Kb = (NH4) (OH) .
(NH3)
So, (NH4+) = (OH-) = x
1.77 x 10-5 =
x2
.500
So, x2 = (1.77 x 10-5) (0.500)
x2 = 8.85 x10-6
x = 8. 85 𝑥 10 − 6
x = 2.97 x 10-3
x = (NH4) = (OH) = 2.97 x 10-3 M OH
Calculating the ionization constant of Ka and Kb of a solution.
Example: Determine the experimental ionization constant of an acetic acid
solution, if .100 mole of acetic acid is dissolved in enough water to make 1.00
dm3 of solution in which (H3O+) equals a 0.00135 mol / dm3.
Write the Ionization equation: CH3COOH + H2O
Write the ionization constant:
So,
Ka = (H3O) (CH3COO)
(CH3COOH)
Ka = (0.00135) (0.00135) = 1.85 x 10-5
(0.100 – 0.00135)
Ka = 1.85 x 10-5 Acetic Acid Solution
ASSIGN WORKSHEET 12B
H3O+ + CH3COO-
12.3 What is pH
A. Pure water self-ionizes slightly to form H3O and OH ions as shown
in the equation.
H2O + H2O
H3O + OH
B. The equation for the equilibrium can be simplified by removing one
water molecule from each side.
H2O
H + OH
C. A special equilibrium expression for the self-ionization of water is
defined as:
Kw = (H+) (OH-)
D. Kw is called the ion product constant for water, which is the value
of the equilibrium content expression for the self-ionization of
water. In pure water at 298K, the concentrations of H+ ions and
OH- ions both equal 1.0 x10-7, so the value of Kw can be
calculated.
Kw = (H+) = (1.0 x 10-7) & (OH-) = 1.0 x 10-7
Kw = 1.0 x 10-14
E. Using Kw to calcualte (H+) and (OH-)
At 298K, the OH- ion concentrations of an aqueous solution is 1.0 x 10-11 M.
Find the H+ ion concentration in the solution and determine whether the
solution is acidic, basic or neutral.
First: Write the ion product constant expression.
Kw = (H+) (OH-) = 1.0 x 10-14
Second: Divide both sides of the equation by (OH-)
(H+) = Kw / (OH-)
Third: Substitute the value for Kw and (OH-) and solve.
(H+) = 1.0 x 10-14 / 1.0 x 10-11 = 1.0 x 10-3 M
(H+) is greater than (OH-),so the solution is acidic
Practice Problems Assigned WS-19E
F. pH and pOH
1. The pH scale has a value of 0 to 14.
a) Acidic solutions have pH values between 0-7,
with the value of 0 being the most acidic.
b) The pH if a basic solution is between 7 and14,
with 14 representing the most basic solution.
c) A neutral solution has a pH of 7.
2. A pOH scale expresses the basicity of a solution. The
pOH of a solution is the negative log rhythm of the
hydroxide ion concentration.
pOH = -log (OH-)
a) If either pH or pOH is known, the other may be
determined by using the following relationship.
pH + pOH = 14.00
b) The pH and pOH values for a solution may be
determined if either (H+) or (OH-) is known.
3. Sample Problem: Calculating pH and pOH from (H+)
If a certain carbonated soft drink has a hydrogen ion concentration of
7.3 x 10-4 M, what are the pH and pOH of the soft drink?
Because (H+) is given, it is easier to calculate pH frist.
pH = -log (H+)
pH = -log (7.3 x 10-4)
pH = -(log 7.3 + log 10-4)
pH = -(0.86 + (-4)) = 3.14
The pH of the soft drink is 3.14. Note that the number of the decimal places
retained in the pH value equals the number of significant figures in the H+ ion
concentration.
To find the pOH, recall that pH + pOH = 14.00. Isolate pOH by subtracting pH
from both sides of the equation.
pOH = 14.00 – 3.14 = 10.86
(Therefore the soft drink is acidic)
Practice Problems Assigned WS-19F
G. Calculating ion concentration from pH
1. When the pH of a solution is known, you can determine
the concentration of H+ and OH-.
pH = -log (H+)
Multiply both sides of the equation by -1
-pH = log (H+)
Now take the antilog of both sides of the eqaution
antilog (-pH) = (H+)
Rearrange equation
(H+) = antilog (-pH)
Similar for (OH+)
(OH-) = antilog (-pOH)
3. Example Problem- Calculating (H+) and (OH-) from pH
What are (H+) and (OH-) in an antacid solution with a pH of 9.70?
Use pH to find (H+)
(H+) = antilog (-pH)
(H+) = antilog (-9.70)
(H+) = 2.0 x 10-10 M
To determine (OH-), first use the pH value to calculate pOH
pOH = 14.00 – pH
pOH = 14.00 – 9.70 = 4.30
(OH-) = antilog (-pOH)
(OH-) = antilog (-4.30)
(OH-) = 5.0 x 10-5 M
As expected (OH-) is greater than (H+) , therefore a basic solution
Practice Problems Assigned WS-19G
H. Using pH to calculate Ka
1. If you know the pH and the concentration of a solution of a
weak acid, you can calculate Ka for the acid. The following
example problem illustrates this types of calculation.
2. Example Problem – Calculating Ka from pH
The pH of a 0.200M solution of acetic acid (CH3COOH) is
2.72. What is Ka for acetic acid?
CH3COOH
Ka = (H+) (CH3COO-)
CH3COOH
pH = -log (H+)
(H+) = antilog (-pH)
(H+) = antilog (-2.72)
(H+) = 1.9 x10-7 M
H+ + CH3COO-
(CH3COO-) = (H+) = 1.9 x 10-3 M
(CH3COOH) = 0.200 M – (H+)
(CH3COOH) = 0,200 M – 1.9 x 10-3 M = 0.198M
Ka = (1.9 x 10-3) (1.9 x 10-3) = 1.8 x 10-5
(.198)
Weak Acid
Practice Problems Assigned WS-19H
12.4 Neutralization
A. The reaction of an acid and a base in aqueous solution is called a
neutralization reaction.
1. Products of a neutralization reaction are a salt and water.
a) Salt – is an ionic compound of a positive ion from
a base and a negative ion from an acid.
b) An example of an acid-base neutralization is the
reaction of nitric acid and calcium hydroxide to
form calcium nitrate and water.
2HNO3 + Ca(OH)2
Ca(NO3)2 + 2H2O
2. Acid-base neutralizations are used in the procedure called
titration, which is a method of determining the
concentration of a solution by reacting it with another
solutions known concentration.
a) to find the concentration of an acid solution, you
would slowly add a basic solution of known
concentration.
b) the neutralization reaction would proceed until it
reaches the equivalence point, where the moles
of H+ ion from the acid equal the moles of OHion form the base.
c) At the equivalence point, a large change in pH
occurs that can be detected by a pH meter or an
acid-base indicator, which is a chemical dye
whose color is affected by pH changes.
B. Example Problem – Calculating concentration from Titration Data
In a titration, 53.7-mL 0.100M HCl solution is needed to
neutralize 80.0-mL of KOH solution. What is the molarity of
the KOH solution?
1. Write the balanced equation for the reaction.
HCl + KOH
KCl + H2O
2. Convert millileters if HCl solution to liters.
53.7-mL HCl x 1-L HCl = 0.0537-L HCl
1000-mL
3. Determine the moles of HCl used to multiplying the volume
of the solution by its molarity, or mol/L.
0.0537-L HCl x 0.100 mol HCl = 5.37 x 10-3
1-L HCl
mol HCl
4. Use the mole ratio in the balanced equation to calculate
the moles of KOH in the unknown solution.
5.37 x 10-3 x 1 mol KOH = 5.37 x 10-3 mol KOH
1 mol HCl
5. Convert millimeters of KOH solution to liters
80-mL KOH x 1-L KOH = 0.0800-L KOH
1000-mL
6. Determine molarity of the KOH solution
M = 5.37 x 10-3 mol KOH = 6.71 x 10-2 M
0.0800-L KOH
Practice Problems Assigned WS-12I
12.5 Percent of Ionization
A. When weak acids or bases dissolve in water, they ionize slightly.
Most of the acid or base remains in the molecular form. The
amount of original acid or base that ionizes is expressed as the
percent of ionization.
% of ionization = amount ionized
original acid or base
x 100
B. The percent of ionization is used as an indicator of acid or base
strength. The stronger acids usually have a higher percent of
ionization. If the percent ionization of the weak electrolyte is
known, the Ka or Kb can be calculated if the concentration of the
solution is given.
C. Example Problem A:
If an acetic acid solution has an initial concentration of 0.0800M
and is 1.50% ionized, determine the experimental Ka of the
CH3COOH.
Solving Process:
If the initial CH3COOH concentration is 0.0800, the H3O+ and the
CH3COO can be obtained by the multiplying the concentration by
the percent ionization.
(H3O+) = (CH3COO-) = (0.0800) (0.0150) = 0.00120M
At equilibrium (CH3COOH) = 0.0800 – 0.00120 = 0.0788M
From the ionization equation CH3COOH + H2O
Ka = (H3O+) (CH3COO-)
(CH3COOH)
H3O+ + CH3COO-
= (1.20 x 10-3)2 = 1.83 x 10-5
7.88 x 10-2
D. Example Problem B:
If the Ka of hydrochloric acid is 3.53 x 10-4 and the HF solution
has an initial concentration of 0.150M, calculate the percentage of
ionization.
Solving Process:
From the ionization equation determine the amount ionized by
solving the Ka expression for x.
HF + H20
Ka =
H 3 O + + F(H3O+) (F-)
(HF)
.
The (HF) is assumed to be 0.150 instead if 0.150 –x, as x is very
small number. Substitute the Ka value and solve for x.
The (HF) is assumed to be 0.150 instead if 0.150 –x, as x is very
small number. Substitute the Ka value and solve for x. (Repeat)
3.53 x 10-4 =
(x) (x) .
0.150
x2 = 5.30 x 10-5
x = 7.28 x 10-3M
Check your answer using iteration.
x2 = (3.53 x 10-4) (0.150 – 7.28 x10-3)
x2 = 5.04 x 10-5
x = 7.10 x 10-3
Since there is a difference between your two values for x, continue to
interate until your value for x does not change.
x = (3.53 x 10-4) (0.150 – 7.10 x 10-3)
x = 7.10 x 10-3
Now that you have corrected value for x, continue to solve for percent
of ionization.
% ionization = amount ionized
original acid
= 7.10 x 10-3 M
0.150 M
x 100
x 100 = 4.73%
Practice Problems Assigned WS-12J