1. Chapter 15 Applications of Aqueous Equilibria
Chemistry: McMurry and Fay, 6th Edition
Chapter 15: Applications of Aqueous Equilibria
4/26/ :15:57 PM
Chapter 15 Applications of Aqueous Equilibria
Copyright © 2011 Pearson Prentice Hall, Inc.

2. Neutralization Reaction
General Formula
Acid Base - Water Salt

3. Neutralization Reactions
Chapter 15: Applications of Aqueous Equilibria
4/26/2017
Neutralization Reactions
Strong Acid-Strong Base
HCl(aq) +
NaOH(aq)
H2O(l) +
NaCl(aq)
Assuming complete dissociation:
2H2O(l)
H3O1+(aq) + OH1-(aq)
H2O(l)
H1+(aq) + OH1-(aq) or
Recall that we’re considering H1+ and H3O1+ equivalent.
HCl is a strong acid. NaOH and NaCl are soluble ionic compounds.
Water and NaCl are the products after neutralization. Since neither sodium nor chloride ions has any substantial acidic/basic properties, the solution is neutral (pH=7) upon neutralization. The titration generally goes to 100%.
(net ionic equation)
After neutralization: pH = 7
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4. Strong acid-Strong base neutralization
2H2O(l)
H3O1+(aq) + OH1-(aq)
When the number moles of acid and base are mixed together
[H3O+] = [-OH] = 1.0 x 10-7M
Reaction proceeds far to the right

5. Neutralization Reactions
Chapter 15: Applications of Aqueous Equilibria
Neutralization Reactions
4/26/2017
Weak Acid-Strong Base
CH3CO2H(aq)
NaOH(aq)
H2O(l)
NaCH3CO2(aq)
Assuming complete dissociation:
H2O(l) + CH3CO21-(aq)
CH3CO2H(aq) + OH1-(aq)
CH3CO2H is a weak acid. NaOH and NaCH3CO2 are soluble ionic compounds.
Water and NaCH3CO2 are the products after neutralization. Sodium ion does not have substantial acidic/basic properties; but, the solution is basic (pH>7) upon neutralization because of the presence of the conjugate base, CH3CO21-, upon neutralization. The titration generally goes to 100%.
(net ionic equation)
After neutralization: pH > 7
Copyright © 2008 Pearson Prentice Hall, Inc.

6. Weak acid-strong base neutralization
Neutralization of any weak acid by a strong base goes 100% to completion
-OH has a great infinity for protons
H2O(l) + CH3CO21-(aq)
CH3CO2H(aq) + OH1-(aq)

7. Neutralization Reactions
Chapter 15: Applications of Aqueous Equilibria
4/26/2017
Neutralization Reactions
Strong Acid-Weak Base
HCl(aq)
+ NH3(aq)
NH4Cl(aq)
Assuming complete dissociation:
NH41+(aq)
H1+(aq) + NH3(aq) or
(net ionic equation)
H3O1+(aq) + NH3 (aq)
H2O(l) +
NH41+(aq)
HCl is a strong acid. NH4Cl is a soluble ionic compound.
Water and NH4Cl are the products after neutralization. Chloride ion does not have substantial acidic/basic properties; but, the solution is acidic (pH<7) upon neutralization because of the presence of the conjugate acid, NH41+, upon neutralization. The titration generally goes to 100%.
After neutralization: pH < 7
Copyright © 2008 Pearson Prentice Hall, Inc.

8. Strong acid-weak base neutralization
Neutralization of any weak base by a strong acid goes 100% to completion
H3O+ has a great infinity for protons
H3O1+(aq) + NH3 (aq)
H2O(l) +
NH41+(aq)

9. Neutralization Reactions
Chapter 15: Applications of Aqueous Equilibria
Neutralization Reactions
4/26/2017
Weak Acid-Weak Base
CH3CO2H(aq) + NH3(aq)
NH4CH3CO2(aq)
NH41+(aq) + CH3CO21-(aq)
CH3CO2H(aq) + NH3(aq)
(net ionic equation)
After neutralization: pH = ?
Since both the acid and the base are weak, it complicates the titration. Typical general chemistry titrations are not done with both of them weak.
The titration generally does not go to 100%. The pH of the solution can be acidic or basic and depends on the weak acid and weak base.
Copyright © 2008 Pearson Prentice Hall, Inc.

10. Weak acid-weak base neutralization
Less tendency to proceed to completion than neutralization involving strong acids and strong bases

11. Examples
Write balanced net ionic equations for the neutralization of equal molar amounts of the following acids and bases. Indicate whether the pH after the neutralization is greater than, equal to or less than 7
HNO2 and KOH Ka HNO2 = 4.5 x 10-4

12. Chemistry: McMurry and Fay, 6th Edition
Chapter 15: Applications of Aqueous Equilibria
4/26/ :15:57 PM
The Common-Ion Effect
Common-Ion Effect: The shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium
CH3CO2H(aq) + H2O(l)
H3O+(aq) + CH3CO2–(aq)
Copyright © 2011 Pearson Prentice Hall, Inc.

13. Chemistry: McMurry and Fay, 6th Edition
Chapter 15: Applications of Aqueous Equilibria
4/26/ :15:57 PM
The Common-Ion Effect
The pH of 0.10 M acetic acid is Calculate the pH of a solution that is prepared by dissolved 0.10 mol of acetic acid and 0.10 mol sodium acetate in enough water to make 1.00 L of solution.
25 °C p
This assumes the complete dissociation of sodium acetate as shown previously.
Copyright © 2011 Pearson Prentice Hall, Inc.

14. The Common-Ion Effect Chemistry: McMurry and Fay, 6th Edition
Chapter 15: Applications of Aqueous Equilibria
4/26/ :15:57 PM
The Common-Ion Effect
Copyright © 2011 Pearson Prentice Hall, Inc.

15. The Common-Ion Effect CH3CO2H(aq) + H2O(l) H3O+(aq) + CH3CO2–(aq)
Chemistry: McMurry and Fay, 6th Edition
Chapter 15: Applications of Aqueous Equilibria
4/26/ :15:57 PM
The Common-Ion Effect
Le Châtelier’s Principle
CH3CO2H(aq) + H2O(l)
H3O+(aq) + CH3CO2–(aq)
The addition of acetate ion to a solution of acetic acid suppresses the dissociation of the acid. The equilibrium shifts to the left.
Copyright © 2011 Pearson Prentice Hall, Inc.

16. The Common-Ion Effect Chemistry: McMurry and Fay, 6th Edition
Chapter 15: Applications of Aqueous Equilibria
4/26/ :15:57 PM
The Common-Ion Effect
Copyright © 2011 Pearson Prentice Hall, Inc.

17. Example
In 0.15 M NH3, the pH is and the percent dissociation is 1.1%. Calculate the concentrations of NH3, pH and percent dissociation of ammonia in a solution that is 0.15M and 0.45 MNH4Cl

18. Chapter 15: Applications of Aqueous Equilibria
Buffer Solutions
4/26/2017
Buffer Solution: A solution which contains a weak acid and its conjugate base and resists drastic changes in pH.
Weak acid +
Conjugate base
CH3CO2H + CH3CO21-
HF + F1-
NH41+ + NH3
H2PO42- + HPO42-
For
Example:
Acidic/basic salts with the weak base/acid.
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19. Buffer Solutions Add a small amount of base (-OH) to a buffer solution
Acid component of solution neutralizes the added base
Add a small amount of acid (H3O+) to a buffer solution
Base component of solution neutralizes the added acid
The addition of –OH or H3O+ to a buffer solution will change the pH of the solution, but not as drastically as the addition of –OH or H3O+ to a non-buffered solution

20. Chapter 15: Applications of Aqueous Equilibria
4/26/2017
Buffer Solutions
H3O1+(aq) + A1-(aq)
HA(aq) + H2O(l)
Weak acid
Conjugate base
(M+A-)
Addition of OH1- to a buffer:
H2O(l) + A1-(aq)
HA(aq) + OH1-(aq)
100%
Both the acid and the conjugate base are present.
For a buffer made from a weak acid and conjugate base, the conjugate base is in the form of a salt.
For a buffer made from a weak base and conjugate acid, the conjugate acid is in the form of a salt.
Addition of H3O1+ to a buffer:
H2O(l) HA(aq)
A1-(aq) + H3O1+(aq)
100%
Copyright © 2008 Pearson Prentice Hall, Inc.

21. Buffer Solutions Chemistry: McMurry and Fay, 6th Edition
Chapter 15: Applications of Aqueous Equilibria
4/26/ :15:57 PM
Buffer Solutions
Copyright © 2011 Pearson Prentice Hall, Inc.

22. Example
pH of human blood (pH = 7.4) controlled by conjugated acid-base pairs (H2CO3/HCO3-). Write the neutralization equation fro the following effects
With addition of HCl
With addition of NaOH

23. Example
Calculate the pH of 0.100L of a buffer solution that is 0.25M in HF and 0.50 M in NaF.
What is the change in pH on addition of mol HCl
What is the change in pH on addition of moles KOH
Calculate the pH after addition of moles HBr
* Assume the volume remains constant

24. Example
Calculate the pH of the buffer that results from mixing 60.0mL of M HCHO2 and 15.0 mL of 0.500M NaCHO2
Ka = 1.8 x 10-4
Calculate the pH after addition of 10.0 mL of MHBr. Assume volume is additive

25. The Henderson-Hasselbalch Equation
Chapter 15: Applications of Aqueous Equilibria
4/26/2017
The Henderson-Hasselbalch Equation
H3O1+(aq) + CH3CO21-(aq)
CH3CO2H(aq) + H2O(l)
Weak acid
Conjugate base
H3O1+(aq) + Base(aq)
Acid(aq) + H2O(l)
[H3O1+][Base]
[Acid]
[Acid]
[Base]
Ka =
[H3O1+] = Ka
Copyright © 2008 Pearson Prentice Hall, Inc.

26. The Henderson-Hasselbalch Equation
Chapter 15: Applications of Aqueous Equilibria
4/26/2017
The Henderson-Hasselbalch Equation
[Acid]
[Base]
[H3O1+] = Ka
[Acid]
[Base]
-log([H3O1+]) = -log(Ka) - log
[Base]
[Acid]
-log(x/y) = log(y/x)
pH = pKa + log
Copyright © 2008 Pearson Prentice Hall, Inc.

27. Examples
Calculate the pH of a buffer solution that is 0.50 M in benzoic acid (HC7H5O2) and M in sodium benzoate (NaC7H5O2). Ka = 6.5 x 10-5
How would you prepare a NaHCO3-Na2CO3 buffer solution that has pH = Ka2 = 4.7 x 10-11