Transcript Slide 1

Chemistry
Chapter 15
Acid-Base
Titration and
pH
• In the self-ionization of water, two water molecules
produce a hydronium ion and a hydroxide ion by
transfer of a proton.
Self-Ionization of Water
H2O + H2O

H3O+ + OH-
Ion Concentration in Water
Test Yourself
A neutral aqueous solution
A. has a 7.0 M H3O+ concentration.
B. contains neither hydronium ions nor hydroxide
ions.
C. has an equal number of hydronium ions and
hydroxide ions.
D. None of the above
Kw – Ionization Constant for Water
In pure water at 25 oC:
[H3O+] = 1 x 10-7 mol/L
[OH-] = 1 x 10-7 mol/L
Kw is a constant at 25 oC:
Kw = [H3O+][OH-]
Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14
The Value of Kw Changes With
Temperature
Strong Acids Ionize Almost Completely in Solution!

–
H N O 3 ( l ) + H 2O ( l )  H 3O (a q ) + N O 3 (a q )
1 mol
1 mol
1 mol
1 mol
Weak Acids Reach Equilibrium Points Early.
HC2H3O2 + H2O
H3O+ + CH3COO-
The strength of a base depends on the
extent it dissociates.

–
N aO H( s )  2 
 N a ( aq ) + O H ( aq )
H O
1 mol
1 mol
1mol
Hydronium Ions and Hydroxide Ions
Calculating [H3O+] and [OH–]
• Strong acids and bases are considered completely
ionized or dissociated in weak aqueous solutions.

–
2
N a O H( s )  

 N a (a q ) + O H (a q )
H O
1 mol
1 mol
1 mol
• 1.0 × 10−2 M NaOH solution has an [OH−] of 1.0 × 10−2 M
• The [H3O+] of this solution is calculated using Kw.
Kw = [H3O+][OH−] = 1.0 × 10−14

[H 3 O ]

1 .0  1 0
–
-1 4
[O H ]

1 .0  1 0
-1 4
1 .0  1 0
-2
 1 .0  1 0
-1 2
M
Sample Problem
A 1.0 X 10–4 M solution of HNO3 has been prepared for a
laboratory experiment.
a. Calculate the [H3O+] of this solution.

(IONIZES COMPLETELY)
–
H N O 3 ( l ) + H 2O ( l )  H 3O (a q ) + N O 3 (a q )
1 mol
1 mol
1 mol
1 mol
m ol H N O 3
L solution

1 m ol H 3 O


1 m ol H N O 3
m ol H 3 O

L solution
 m olarity of H 3 O
b. Calculate the [OH–].
[H3O+][OH−] = 1.0 × 10−14
–
[O H ] 
1.0  10

[H 3 O ]
–14

1.0  10
–14
1.0  10
-4
 1.0  10
-10
M

Hydronium Ions and Hydroxide Ions
Calculating [H3O+] and [OH–]
• If the [H3O+] of a solution is known, the [OH−] can be
calculated using Kw.
[HCl] = 2.0 × 10−4 M
[H3O+] = 2.0 × 10−4 M
Kw = [H3O+][OH−] = 1.0 × 10−14
–
[O H ] 
1.0  10

-14
[H 3 O ]

1.0  10
-14
2.0  10
-4
 5.0  10
-10
M
Determine the hydronium and hydroxide ion concentration in
a solution that has 1.0 X 10 -4 M Ca(OH)2
1 mol
1 mol
.
2 mol
1.0 × 10−2 M NaOH solution has an [OH−] of (2) X 1.0 × 10−4 M
– The [OH-] concentration is 2.0 X 10-4 M
• The [H3O+] of this solution is calculated using Kw.
• Kw = [H3O+][OH−] = 1.0 × 10−14
[H3O+][2.0 X 10-4 M] = 1.0 × 10−14
[H3O+] = 5.0 X 10-11 M
Pause for a Cause
Student Practice Pg. 522 Problem #8
Calculate the [H3O+] and [OH-] for each of the following:
a. 0.030 M HCl
b. 1.0 X 10-4 M NaOH
c. 5.0 X 10-3 M HNO3
d. 0.010M Ca(OH)2
Answers: a. [H3O+] = 3.0 X 10-2 M, [OH−] = 3.3 × 10−13 M; b. [H3O+] = 1.0 X 10-10 M, [OH−] =
1.0 × 10−4 M; c. [H3O+] = 5.0 X 10-3 M, [OH−] = 2.0 × 10−12 M; d. [H3O+] = 5.0 X 10-13 M,
[OH−] = 2.0 × 10−2 M
pH
• pH comes from French words meaning “hydrogen
power”. In chemistry, pH is a scale that indicates the
concentration of hydronium ions [H3O+] in solutions.
pH
Scale
Calculating pH, pOH
pH = -log10(H3O+)
pOH = -log10(OH-)
Relationship between pH and pOH
pH + pOH = 14
Finding [H3O+], [OH-] from pH, pOH
[H3O+] = antilog -pH
[OH-] = antilog -pOH
pH
+
pOH
=
14
When is a solution an acid, a base,
or a neutral solution?
• Solutions in which [H3O+] = [OH−] are
neutral.
• Solutions in which the [H3O+] > [OH−] are
acidic.
– [H3O+] > 1.0 × 10−7 M
• Solutions in which the [OH−] > [H3O+] are
basic.
– [OH−] > 1.0 × 10−7 M
Calculating pH
• The pH of a solution is defined as the negative of the
common logarithm of the hydronium ion concentration,
[H3O+].
• pH = −log [H3O+]
– example: a neutral solution has a [H3O+] = 1×10−7
– The logarithm of 1×10−7 is −7.0.
• pH = −log [H3O+] = −log(1 × 10−7) = −(−7.0) = 7.0
• What is the pH of 1.0 X 10-3 M HCl?
– pH = −log [H3O+] = −log(1.0 × 10−3) = −(−3.0) = 3.0
• What is the pH of a 1.0 X 10-2 M NaOH solution?

[H 3 O ] 
1 .0  1 0
–
-1 4
[O H ]

1 .0  1 0
-1 4
1 .0  1 0
-2
 1 .0  1 0
-1 2
M
– pH = −log [H3O+] = −log(1.0 × 10−12) = −(−12) = 12
• What is the pH of 3.4 X 10-5 M HNO3?
– pH = −log [H3O+] = −log(3.4 × 10−5) = −(−4.47) = 4.47
Pause for a Cause
Student Practice Pg. 506 Practice Problems 1-4
1.
2.
3.
4.
What is the pH of a solution if the [H3O+] is 6.7 X 10-4?
What is the pH of a solution with a hydronium ion
concentration of 2.5 X 10-2 M?
Determine the pH of a 2.5 X 10-6 M HNO3 solution.
Determine the pH of a 2.0 X 10-2 M Sr(OH)2 solution.
Answers: 1) pH = 3.17 2) pH =1.60 3) pH = 5.60 4) pH = 12.60
Determining Hydronium Ions from pH
Determine the hydronium ion
concentration of an aqueous solution
that has a pH of 4.0.
[H3O+] = antilog –pH
[H3O+] = antilog – 4.0
[H3O+] = 1.0 X 10-4
The pH of a solution is measured
and determined to be 7.52
• a. What is the hydronium ion concentration?
• b. What is the hydroxide ion concentration?
• c. Is the solution acidic or basic?
a. [H3O+] = antilog –pH
[H3O+] = antilog – 7.52
[H3O+] = 3.0 X 10-8
b. Kw = [H3O+][OH−] = 1.0 × 10−14
[3.0 X 10-8][OH-] = 1.0 × 10−14
[OH-] = 3.3 X 10-7 M
c. slightly basic
Pause for a Cause
Practice Problems Pg. 508 # 1-4
1.
2.
3.
4.
The pH of a solution is determined to be 5.0. What is the
hydronium ion concentration of this solution?
The pH of a solution is determined to be 12.0. What is the
hydronium ion concentration of this solution?
The pH of an aqueous solution is measured as 1.50. Calculate
the [H3O+] and [OH-] concentration.
The pH of an aqueous solution is 3.67. Determine the [H3O+].
1.
[H3O+] = 1.0 X 10-5
2.
[H3O+] = 1.0 X 10-12
3.
[H3O+] = 3.2 X 10-2
[ OH-] = 3.2 X 10-12
4.
[H3O+] = 2.1 X 10-4
The pH of a Substance Can be
Found Using Indicators, pH Paper,
or pH Meters.
• pH meter
– Determines the pH of a solution by measuring the voltage
between the two electrodes that a placed in the solution.
• Acid-base indicators
– Compounds whose colors are sensitive to pH.
• Transition interval
– The pH range over which an indicator changes color.
Measuring pH with Wide-Range Paper/
Narrow Range Paper
pH Indicators and
Their Ranges
Problem: An unknown
solution is colorless
when tested with
phenolphthalein, but
causes the indicator
phenol red to turn red.
Use this information to
determine the possible
pH of this solution.
Steps in Determining the
Molarity of a Solution.
1. Start with the balanced equation for the neutralization
reaction, and determine the chemically equivalent
amounts of the acid and base.
2. Determine the moles of acid (or base) from the known
solution used during the titration.
3. Determine the moles of solute of the unknown solution
used during the titration.
4. Determine the molarity of the unknown solution.
Titration Calculations
Let’s Work this Sample Problem on the Board!
In a titration, 27.4 mL of 0.0154 M Ba(OH)2 is added to a 20.0
mL sample of HCl solution of unknown concentration until
the equivalence point is reached. What is the molarity of the
acid solution?
Answer: 4.22 X 10-2 M HCl
Let’s Work Another Sample Problem on the Board!
A 15.5 mL sample of 0.215 M KOH solution required21.2 mL of
aqueous acetic acid solution in a titration experiment. Calculate the
molarity of the acetic acid solution.
Answer: 0.157 M HC2H3O2
One Last Sample Problem on the Board!
By titration, 17.6 mL of aqueous H2SO4 neutralized 27.4 mL
of 0.165 M LiOH solution. What is the molarity of the
aqueous acid solution?
Answer: 0.0128 M H2SO4
Pause for a Cause
Student Practice Pg. 524 # 26, 26, 36
25. Suppose that 15.0 mL of 2.50 X 10-2 M aqueous H2SO4
is required to neutralize 10.0 mL of an aqueous solution
of KOH. What is the molarity of the KOH solution?
26. In a titration experiment, a 12.5 mL sample of 1.75 M
Ba(OH)2 just neutralized 14.5 mL of HNO3 solution.
Calculate the molarity of the HNO3 solution.
36. Find the molarity of a Ca(OH)2 solution given that 428
mL of the solution is neutralized in a titration by 115 mL
of 6.7 X 10-3 M HNO3
Answers: 25. 7.50 X 10-2 M KOH; 26. 3.02 X 10-2 M HNO3;
36. 9.0 X 10-4 M Ca(OH)2
Challenge Problem
Pg. 902 # 359
• A chemist wants to produce 12.00 grams of
barium sulfate by reacting a 0.600 M BaCl2
solution with excess H2SO4, as shown in the
reaction below. What volume of BaCl2
solution should be used?
– BaCl2 + H2SO4  BaSO4 + 2 HCl
Instructions for Titrations
Filling the Acid Buret
Adding the Indicator
Filling the Base Buret
The Experiment Begins
Titration Completed!
Titration Curves
Test Yourself
1. Distilled water contains
A. H2O.
B. H3O+.
C. OH−.
D. All of the above
Test Yourself
2. What is the pH of a 0.0010 M HNO3?
A. 1.0
B. 3.0
C. 4.0
D. 5.0
Test Yourself
3. Which of the following solutions would have a
pH value greater than 7?
A. [OH−] = 2.4 × 10−2 M
B. [H3O+] = 1.53 × 10−2 M
C. 0.0001 M HCl
D. [OH−] = 4.4 × 10−9 M
Test Yourself
4. If the pH of a solution of the strong base NaOH
is known, which property of the solution can be
calculated?
A. molar concentration
B. [OH−]
C. [H3O+]
D. All of the above
Test Yourself
5. A neutral aqueous solution
A. has a 7.0 M H3O+ concentration.
B. contains neither hydronium ions nor
hydroxide ions.
C. has an equal number of hydronium ions
and hydroxide ions.
D. None of the above
Test Yourself
6. Identify the salt that forms when a
solution of H2SO4 is titrated with a
solution of Ca(OH)2.
A. calcium sulfate
B. calcium hydroxide
C. calcium oxide
D. calcium phosphate
Test Yourself
7. The pH of a solution is 6.32. What is
the pOH?
A. 6.32
B. 4.8 × 10−7
C. 7.68
D. 2.1 × 10−8
Test Yourself
8. The Kw value for water can be affected
by:
A. dissolving a salt in the solution.
B. changes in temperature.
C. changes in the hydroxide ion
concentration.
D. the presence of a strong acid
Test Yourself
9. Which of the pH levels listed below
is the most acidic?
A. pH = 1
B. pH = 5
C. pH = 9
D. pH = 13
The
End!
The
End!
The
End!
The
End!
The End!
End!
The