Acid-Base Titrations
Section 17.3
Introduction
• Definition:
– In an acid-base titration, a solution containing a
known concentration of a base is slowly added to
an acid.
– An indicator is used to signal the equivalence
point of the titration.
• This is the point at which stoichiometrically equivalent
amounts of acid and base have been mixed.
– A pH meter can also be used to find the
equivalence point.
Introduction
• The typical titration apparatus includes:
– a buret to hold the titrant
– a beaker to hold the analyte
– a pH meter to measure the pH
Introduction
• In this section, we will be looking at a series of
titrations in detail to understand why acids
and behave the way they do.
– Strong acid-strong base titration
– Weak acid-strong base titration
– Polyprotic acid-strong base titration
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong
base titration has the following shape.
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong
base titration has the following shape.
We divide the curve into
four regions:
1. Initial pH
2. Initial pH to eq. point
3. Equivalence point
4. After eq. point
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong
base titration has the following shape.
We divide the curve into
four regions:
1. Initial pH
The pH of the solution is
determined by the
concentration of the strong
acid.
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong
base titration has the following shape.
We divide the curve into
four regions:
2. Initial pH to eq. point
As base is added, pH
increases slowly and then
rapidly. The pH is
determined by the
concentration of the acid
that is not neutralized.
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong
base titration has the following shape.
We divide the curve into
four regions:
3. Equivalence point
At the equivalence point,
[OH−] = [H+]. The pH = 7.00.
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong
base titration has the following shape.
We divide the curve into
four regions:
4. After eq. point
As more base is added, pH
increases rapidly and then
slowly. pH is determined by
the concentration of the
excess base.
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong
base titration has the following shape.
Let’s see how this works
in practice.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
b) 51.0 mL
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
This is between the initial point and the
equivalence point.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
This is between the initial point and the
equivalence point. pH is determined by the
amount of acid that has not been neutralized.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
Therefore, we need to determine the number of
mols of acid remaining, nacid, and the total
volume, Vtotal, of the solution. (Remember,
adding the NaOH increases the total volume.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,i = Macid,i × Vacid,i
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,i = Macid,i × Vacid,i = (0.100 M)(0.0500 L)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = Mbase,added × Vbase,added
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = (0.100 M)(0.0490 L)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = 4.90 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = 4.90 × 10−3 mol
nacid,remaining = nacid,i − nbase,added
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = 4.90 × 10−3 mol
nacid,remaining = (5.00 − 4.90) × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = 4.90 × 10−3 mol
nacid,remaining = 0.10 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = 4.90 × 10−3 mol
nacid,remaining = 1.0 × 10−4 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,remaining = 1.0 × 10−4 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,remaining = 1.0 × 10−4 mol
Vtotal = Vacid + Vbase
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,remaining = 1.0 × 10−4 mol
Vtotal = Vacid + Vbase = 0.0500 L + 0.0490 L
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,remaining = 1.0 × 10−4 mol
Vtotal = Vacid + Vbase = 0.0990 L
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
[H+] = nacid,remaining/Vtotal
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
[H+] = (1.0 × 10−4 mol)/(0.0990 L)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
[H+] = 1.0 × 10−3 M
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
[H+] = 1.0 × 10−3 M
pH = −log[H+]
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
[H+] = 1.0 × 10−3 M
pH = −log(1.0 × 10−3)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
nacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
[H+] = 1.0 × 10−3 M
pH = 3.00
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
pH = 3.00
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
a) 49.0 mL
b) 51.0 mL
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
This is beyond the equivalence point.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
This is beyond the equivalence point. All of the
strong acid has been used up.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
This is beyond the equivalence point. All of the
strong acid has been used up. The pH is
determined by the excess base that has been
added.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
We already determined nacid,i = 5.00 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nacid,i = 5.00 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nacid,i = 5.00 × 10−3 mol
nbase,added = Mbase,added × Vbase,added
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nacid,i = 5.00 × 10−3 mol
nbase,added = (0.100 M)(0.0510 L)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nacid,i = 5.00 × 10−3 mol
nbase,added = 5.10 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nacid,i = 5.00 × 10−3 mol
nbase,added = 5.10 × 10−3 mol
nbase,remaining = nbase,added − nacid,i
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nacid,i = 5.00 × 10−3 mol
nbase,added = 5.10 × 10−3 mol
nbase,remaining = (5.10 − 5.00) × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nacid,i = 5.00 × 10−3 mol
nbase,added = 5.10 × 10−3 mol
nbase,remaining = 0.10 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Vtotal = Vacid + Vbase
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Vtotal = Vacid + Vbase = 0.0500 L + 0.0510 L
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Vtotal = Vacid + Vbase = 0.1010 L
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = nbase,remaining/Vtotal
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = (1.0 × 10−4 mol)/(0.1010 L)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 M
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 M
pOH = −log[OH−]
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 M
pOH = −log(9.9 × 10−4)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 M
pOH = 3.00
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 M
pOH = 3.00 ⇒ pH = 14.00 − 3.00
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 M
pOH = 3.00 ⇒ pH = 14.00 − 3.00 = 11.00
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
nbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 M
pH = 11.00
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities
of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
pH = 11.00
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration
is similar to that for a strong acid-weak base.
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration
is similar to that for a strong acid-weak base.
1. The initial pH is
determined by the Ka
of the acid.
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration
is similar to that for a strong acid-weak base.
2. To determine the pH
from the initial point
to the eq. point, we
first neutralize the
weak acid and then
use the HendersonHasselbach equation.
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration
is similar to that for a strong acid-weak base.
3. At the eq. point, we
have no HX, only X−.
We need to use the Kb
value to find pH.
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration
is similar to that for a strong acid-weak base.
4. Beyond the eq. point,
we use the excess
base to calculate pH.
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration
is similar to that for a strong acid-weak base.
Let’s see how this works
in practice.
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
First, calculate the concentrations of materials
before neutralization reaction.
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
First, calculate the concentrations of materials
before neutralization reaction.
CH3COOH + OH− ➙ CH3COO− + H2O
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
First, calculate the concentrations of materials
before neutralization reaction.
CH3COOH + OH− ➙ CH3COO− + H2O
ninitial,acid = Macid × Vacid
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
First, calculate the concentrations of materials
before neutralization reaction.
CH3COOH + OH− ➙ CH3COO− + H2O
ninitial,acid = Macid × Vacid = (0.100 M)(0.0500 L)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
First, calculate the concentrations of materials
before neutralization reaction.
CH3COOH + OH− ➙ CH3COO− + H2O
ninitial,acid = Macid × Vacid = 5.00 × 10−3 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
First, calculate the concentrations of materials
before neutralization reaction.
CH3COOH + OH− ➙ CH3COO− + H2O
ninitial,acid = 5.00 × 10−3 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
First, calculate the concentrations of materials
before neutralization reaction.
CH3COOH + OH− ➙ CH3COO− + H2O
ninitial,acid = 5.00 × 10−3 mol
nbase = Mbase × Vbase
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
First, calculate the concentrations of materials
before neutralization reaction.
CH3COOH + OH− ➙ CH3COO− + H2O
ninitial,acid = 5.00 × 10−3 mol
nbase = Mbase × Vbase = (0.100 M)(0.0450 L)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
First, calculate the concentrations of materials
before neutralization reaction.
CH3COOH + OH− ➙ CH3COO− + H2O
ninitial,acid = 5.00 × 10−3 mol
nbase = Mbase × Vbase = 4.50 × 10−3 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
First, calculate the concentrations of materials
before neutralization reaction.
CH3COOH + OH− ➙ CH3COO− + H2O
ninitial,acid = 5.00 × 10−3 mol
nbase = 4.50 × 10−3 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
ninitial,acid = 5.00 × 10−3 mol
nbase = 4.50 × 10−3 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
ninitial
5.0 × 10−3 mol
4.5 × 10−3 mol
0.0 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
ninitial
5.0 × 10−3 mol
4.5 × 10−3 mol
nchange
−4.5 × 10−3 mol −4.5 × 10−3 mol
0.0 mol
+4.5 × 10−3 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
ninitial
5.0 × 10−3 mol
4.5 × 10−3 mol
nchange
−4.5 × 10−3 mol −4.5 × 10−3 mol
+4.5 × 10−3 mol
nfinal
5.0 × 10−4 mol
4.5 × 10−3 mol
0.0 mol
0.0 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
nacid,final = 5.0 × 10−4 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
nacid,final = 5.0 × 10−4 mol
[acid] = nacid,final/Vtotal
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
nacid,final = 5.0 × 10−4 mol
[acid] = (5.0 × 10−4 mol)/(0.0950 L)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
nacid,final = 5.0 × 10−4 mol
[acid] = 5.3 × 10−3 M
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
[acid] = 5.3 × 10−3 M
[base] = nbase,final/Vtotal
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
[acid] = 5.3 × 10−3 M
[base] = (4.50 × 10−3 mol)/(0.0950 L)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
[acid] = 5.3 × 10−3 M
[base] = 4.74 × 10−2 M
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 M
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 M
pKa = −logKa
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 M
pKa = −log(1.8 × 10−5)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 M
pKa = 4.74
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
pH = pKa + log([base]/[acid)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
pH = 4.74 + log([4.74 × 10−2 M]/[5.3 × 10−3 M])
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
pH = 4.74 + log(9.00)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
pH = 4.74 + 0.954
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when
45.0 mL of 0.100 M NaOH is added to 50.0 mL
of 0.100 M CH3COOH.
Next, use the values to determine the
concentrations after the neutralization.
CH3COOH + OH− ➙ CH3COO− + H2O
pH = 5.69
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
At the equivalence point, all of the weak acid is
converted to its conjugate weak base.
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
At the equivalence point, all of the weak acid is
converted to its conjugate weak base. This
means that we will be using the base hydrolysis
expression to find [OH−], pOH, and pH.
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
The number of mols of acetate in solution at the
equivalence point is equal to the number of
mols of acetic acid at the start of the titration.
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
nbase,eq. pt. = ninitial,acid
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
nbase,eq. pt. = ninitial,acid = (0.100 M)(0.0500 L)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
nbase,eq. pt. = ninitial,acid = 5.00 × 10−3 mol
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 mol
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 mol
The concentration of the acetate is given by:
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 mol
The concentration of the acetate is given by:
[base] = (nbase,eq. pt.)/(Vtotal)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 mol
The concentration of the acetate is given by:
[base] = (5.00 × 10−3 mol)/(0.100 L)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 mol
The concentration of the acetate is given by:
[base] = 5.00 × 10−2 M
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M
Next, we do an i-c-e table on the base
hydrolysis.
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M
CH3COO− + H2O ➙ CH3COOH + OH−
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M
CH3COO− + H2O ➙ CH3COOH + OH−
i
5.00 × 10−2
0.0
0.0
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M
CH3COO− + H2O ➙ CH3COOH + OH−
i
5.00 × 10−2
0.0
0.0
c
−x
+x
+x
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M
CH3COO− + H2O ➙ CH3COOH + OH−
i
5.00 × 10−2
0.0
0.0
c
−x
+x
+x
e
5.00 × 10−2 − x
x
x
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = x
CH3COO− + H2O ➙ CH3COOH + OH−
i
5.00 × 10−2
0.0
0.0
c
−x
+x
+x
e
5.00 × 10−2 − x
x
x
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = x
Kb = Kw/Ka = (1.0 × 10−14)/(1.8 × 10−5)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = x
Kb = Kw/Ka = 5.6 × 10 −10
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = x
Kb = 5.6 × 10 −10
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = x
Kb = 5.6 × 10 −10 = ([acid][OH−])/[base]
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = x
Kb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = x
Kb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)
x2 = (5.6 × 10 −10)(5.00 × 10−2) = 2.8 × 10−11
x = (2.8 × 10−11)½
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = x
Kb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)
x2 = (5.6 × 10 −10)(5.00 × 10−2) = 2.8 × 10−11
x = (2.8 × 10−11)½ = 5.3 × 10−6
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = x
Kb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)
x2 = (5.6 × 10 −10)(5.00 × 10−2) = 2.8 × 10−11
x = (2.8 × 10−11)½ = 5.3 × 10−6 = [OH−]
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 M
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 M
pOH = −log[OH−]
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 M
pOH = −log[OH−] = −log(5.3 × 10−6)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 M
pOH = −log[OH−] = −log(5.3 × 10−6) = 5.28
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 M
pOH = 5.28
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 M
pOH = 5.28
pH = 14.00 – pOH
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 M
pOH = 5.28
pH = 14.00 – pOH = 14.00 − 5.28
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 M
pOH = 5.28
pH = 14.00 – pOH = 14.00 − 5.28 = 8.72
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in
the titration of 50.0 mL of 0.100 M CH3COOH
with 0.100 M NaOH.
pH = 8.72
• The pH of the titration at the equivalence
point depends on the type of acids and bases
being titrated.
– Strong acid-Strong base: eq. pt. = 7.0
– Weak acid-Strong base: eq. pt. > 7.0
– Strong acid-Weak base: eq. pt. < 7.0
Titrations of Polyprotic Acids
• When weak acid contain more than one
ionizable H atom, as in phosporous acid,
H3PO3, reaction with OH− occurs in a series of
steps.
– H3PO3(aq) + H2O(l) ➙ H2PO3−(aq) + H3O+(aq)
– H2PO3−(aq) + H2O(l) ➙ HPO32−(aq) + H3O+(aq)
– HPO32−(aq) + H2O(l) ➙ PO33−(aq) + H3O+(aq)
Titrations of Polyprotic Acids
• When the neutralization steps are sufficiently
separated, the substance exhibits multiple
equivalence points.