1. Section 18.3 Hydrogen Ions and pH
Explain pH and pOH.
Relate pH and pOH to the ion product constant for water.
Calculate the pH and pOH of aqueous solutions.
pH and pOH are logarithmic scales that express the concentrations of hydrogen ions and hydroxide ions in aqueous solutions.
Section 18-3

2. Acids Have a pH less than 7
The pH Scale
Acids Have a pH less than 7
Bases have a pH greater than 7

3. pH and pOH (cont.)
Litmus paper and a pH meter with electrodes can determine the pH of a solution.
Section 18-3

4. Self-Ionization of Water
H2O + H2O  H3O+ + OH-
At 25, [H3O+] = [OH-] = 1 x 10-7
Kw is a constant at 25 C:
Kw = [H3O+][OH-]
Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

5. pH and pOH
Concentrations of H+ ions are often small numbers expressed in exponential notation.
pH is the negative logarithm of the hydrogen ion concentration of a solution. pH = –log [H+]
Section 18-3

6. The sum of pH and pOH equals 14.
pH and pOH (cont.)
pOH of a solution is the negative logarithm of the hydroxide ion concentration.
pOH = –log [OH–]
The sum of pH and pOH equals 14.
Section 18-3

7. Relationship between pH and pOH
Calculating pH, pOH
pH = -log10[H3O+]
pOH = -log10[OH-]
Relationship between pH and pOH
pH + pOH = 14
Finding [H3O+], [OH-] from pH, pOH
[H3O+] = 10-pH
[OH-] = 10-pOH

8. Calculating pH from solution concentration…

9. pH and pOH (cont.)
For all strong monoprotic acids, the concentration of the acid is the concentration of H+ ions.
For all strong bases, the concentration of the OH– ions available is the concentration of OH–.
Weak acids and weak bases only partially ionize and Ka and Kb values must be used.
Section 18-3

10. Dissociation Constants: Strong Acids
Formula
Conjugate Base Ka
 Perchloric 
 HClO4 
 ClO4- 
 Very large 
 Hydriodic 
 HI 
 I- 
 Hydrobromic 
 HBr 
 Br- 
 Hydrochloric 
 HCl 
 Cl- 
 Nitric 
 HNO3 
 NO3- 
 Sulfuric 
 H2SO4 
 HSO4- 
 Hydronium ion 
 H3O+ 
 H2O 
 1.0 

11. Dissociation Constants: Weak Acids
Formula
Conjugate Base Ka
 Iodic 
 HIO3 
 IO3- 
 1.7 x 10-1 
 Oxalic 
 H2C2O4 
 HC2O4- 
 5.9 x 10-2 
 Sulfurous 
 H2SO3 
 HSO3- 
 1.5 x 10-2 
 Phosphoric 
 H3PO4 
 H2PO4- 
 7.5 x 10-3 
 Citric 
 H3C6H5O7 
 H2C6H5O7- 
 7.1 x 10-4 
 Nitrous 
 HNO2 
 NO2- 
 4.6 x 10-4 
 Hydrofluoric 
HF 
 F- 
 3.5 x 10-4 
 Formic 
 HCOOH 
 HCOO- 
 1.8 x 10-4 
 Benzoic 
 C6H5COOH 
 C6H5COO- 
 6.5 x 10-5 
 Acetic 
 CH3COOH 
 CH3COO- 
 1.8 x 10-5 
 Carbonic 
 H2CO3 
 HCO3- 
 4.3 x 10-7 
 Hypochlorous 
 HClO 
 ClO- 
 3.0 x 10-8 
 Hydrocyanic 
 HCN 
 CN- 
 4.9 x 10-10 

12. A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #1: Write the dissociation equation
HC2H3O2  C2H3O2- + H+

13. A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #2: ICE it!
HC2H3O2  C2H3O2- + H+ I C E
0.50
- x +x +x x x x

14. A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #3: Set up the law of mass action
HC2H3O2  C2H3O2- + H+ E x x x

15. A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #4: Solve for x, which is also [H+]
HC2H3O2  C2H3O2- + H+ E x x x
[H+] = 3.0 x 10-3 M

16. A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #5: Convert [H+] to pH
HC2H3O2  C2H3O2- + H+ E x x x

17. Dissociation of Strong Bases
MOH(s)  M+(aq) + OH-(aq)
Strong bases are metallic hydroxides
Group I hydroxides (NaOH, KOH) are very soluble
Group II hydroxides (Ca, Ba, Mg, Sr) are less soluble
pH of strong bases is calculated directly from the concentration of the base in solution

18. Reaction of Weak Bases with Water
The base reacts with water, producing its conjugate acid and hydroxide ion:
CH3NH2 + H2O  CH3NH3+ + OH- Kb = 4.38 x 10-4

19. Kb for Some Common Weak Bases
Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?
Base
Formula
Conjugate Acid Kb
Ammonia 
 NH3
 NH4+
 1.8 x 10-5 
 Methylamine
 CH3NH2
 CH3NH3+
 4.38 x 10-4 
 Ethylamine
 C2H5NH2
 C2H5NH3+
 5.6 x 10-4 
 Diethylamine
 (C2H5)2NH
 (C2H5)2NH2+
 1.3 x 10-3 
 Triethylamine
  (C2H5)3N
  (C2H5)3NH+
 4.0 x 10-4 
 Hydroxylamine
 HONH2 
 HONH3+  
 1.1 x 10-8 
 Hydrazine
H2NNH2 
H2NNH3+  
 3.0 x 10-6 
 Aniline
 C6H5NH2 
 C6H5NH3+  
 3.8 x 10-10 
 Pyridine
 C5H5N 
 C5H5NH+  
 1.7 x 10-9 

20. Reaction of Weak Bases with Water
The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion:
B + H2O  BH+ + OH-
(All weak bases do this.)

21. A Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #1: Write the equation for the reaction
NH3 + H2O  NH4+ + OH-

22. A Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #2: ICE it!
NH3 + H2O  NH4+ + OH- I C E
0.50
- x +x +x x x x

23. A Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #3: Set up the law of mass action
NH3 + H2O  NH4+ + OH- E x x x

24. A Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #4: Solve for x, which is also [OH-]
NH3 + H2O  NH4+ + OH- E x x x
[OH-] = 3.0 x 10-3 M

25. A Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #5: Convert [OH-] to pH
NH3 + H2O  NH4+ + OH- E x x x

26. A B C D Section 18.3 Assessment
In dilute aqueous solution, as [H+] increases:
A. pH decreases
B. pOH increases
C. [OH–] decreases
D. all of the above A B C D
Section 18-3

27. A B C D Section 18.3 Assessment
What is the pH of a neutral solution such as pure water?
A. 0
B. 7
C. 14
D. 1.0 × 10–14 A B C D
Section 18-3