1. pH of Strong Acid Solutions
**In attacking acid-base problems, the importance of writing the major species in the solution as the first step cannot be overemphasized. This single step is the key to solving these problems successfully.
When a strong acid dissociates in water, the [H+] of the strong acid will be the only negligible contributor to pH (the water will produce almost no H+ ions).

2. Example Calculate the pH and [OH-] of a 5 X 10-3 M HClO4 solution.
Answer: pH = 2.3, [OH-] = 2 X M

3. pH of Weak Acid Solutions
BRING BACK THE ICE TABLES!
Determine the pH of a 1.00 M solution of HF (Ka = 7.2 X 10-4)…this is a weak acid…
**FIRST: write the major species in the solution!
HF and H2O
Determine what can donate H+ ions
HF(aq) <-> H+(aq) + F-(aq) Ka = 7.2 X 10-4
H2O(l) <-> H+(aq) + OH-(aq) Kw = 1.0 X 10-14
Greater K will be a stronger acid and will donate more H+ ions. Therefore, HF donates more.

4. HF(aq) <-> H+(aq) + F-(aq)
Set up the ICE table and write the equilibrium expression where Ka = 7.2 X Make an assumption and solve for X. Check your assumption (<5%).
Determine [H+] at equilibrium and pH
Answer: X = 2.7 X 10-2 so [H+] = 2.7 X 10-2 M and pH = 1.57

5. Example
Ka = 7.45 X 10-4 for citric acid, C6H10O8 (we’ll call it “HCA”). Calculate the pH of a M HCA solution.
Answer: 1.93

6. pH of a Mixture of Weak Acids
Follow the same general process as before!
Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2 X 10-10) and 5.00 M HNO2 (Ka = 4.0 X 10-4). Also calculate the concentration of cyanide ion (CN-) in this solution at equilibrium.
Answer: pH = 1.35, [CN-] = 1.4 X 10-8 M

7. Example
Calculate the pH of a mixture of 2.00 M formic acid (HCOOH, Ka = 1.77 X 10-4) and 1.50 M hypobromous acid (HOBr, Ka = 2.06 X 10-9). What are the concentrations of both the hypobromite ion (OBr-) and hydroxide (OH-) ion at equilibrium?
Answer: pH = 1.73, [OBr-] = 1.6 X 10-7 M, [OH-] = 5.4 X M

8. % dissoc. = amt dissociated (mol/L) X 100
Percent Dissociation
AKA percent ionization
% dissoc. = amt dissociated (mol/L) X 100
Initial [ ] (mol/L)
*For a weak acid, the percent dissociation increases as the acid becomes more dilute
Earlier, 1.00 M solution of HF, [H+] = 2.7 X 10-2 M…% dissociation = 2.7%

9. Examples
Calculate the percent dissociation of acetic acid (Ka = 1.8 X 10-5) in each of the following:
1.00 M HC2H3O2
Answer: 0.42%
0.100 M HC2H3O2
Answer: 1.3%

10. Results
For solutions of any weak acid HA, [H+] decreases as [HA]0 decreases, but the percent dissociation increases as [HA]0 decreases.
As an acid is diluted, the percent dissociation increases.

11. Backward Calculation We can calculate Ka from Percent Dissociation!
Example: In a M solution, uric acid (HC5H3N4O4) is 1.6% dissociated. Calculate the value of Ka for uric acid.
Answer: 1.3 X 10-4
Example #2: Lactic acid (HC3H5O3) is a waste product that accumulates in muscle tissue during exertion, leading to pain and a feeling of fatigue. In a M aqueous solution, lactic acid is 3.7% dissociated. Calculate the value of Ka for this acid.
Answer: 1.4 X 10-4

12. Helpful Hints!
See the “Problem-Solving Strategies” tables in your book for helpful steps to follow when solving problems.

13. Bases
Review definitions: Arrhenius (OH- donator) vs. Bronsted-Lowry (H+ acceptor) vs. Lewis (electron donator)
Strong Bases:
Hydroxides of group IA (LiOH, NaOH, KOH, RbOH, CsOH) but we really only use NaOH and KOH in lab
Hydroxides of some group IIA (Ca(OH)2, Ba(OH)2, and Sr(OH)2) are strong, but not very soluble
Calcium hydroxide (“slacked lime”) is frequently used as a cheap way to soften hard water, remove sulfur dioxide from power plant exhaust

14. pH of Strong Bases List species present Determine [OH-] ions
Determine [H+] ions
Use pH = -log[H+]
Example: Calculate the pH of a solution made by putting 4.63 g of LiOH into water and diluting to a total volume of 400. mL.
Answer: 13.68

15. Example #2 Calculate the pH of a 5.0 X 10-2 M NaOH solution
Answer: 12.70

16. NH3(aq) + H2O(l) <-> NH4+(aq) + OH-(aq)
Weak Bases
A base does NOT have to contain an OH- ion.
NH3(aq) + H2O(l) <-> NH4+(aq) + OH-(aq)
Bases like these usually have an unshared pair of electrons (many of them contain a nitrogen atom)

17. B(aq) + H2O(l) <-> BH+(aq) + OH-(aq)
Bases Continued
General Reaction for a base with water:
B(aq) H2O(l) <-> BH+(aq) + OH-(aq)
Base Acid CA CB
We can also write an equilibrium constant:
Kb = [BH+][OH-]
[B]
Kb always refers to the reaction of a base with water to form the conjugate acid and the hydroxide ion (Some Kb values are in Table 14.3).

18. B(aq) + H2O(l) <-> BH+(aq) + OH-(aq) Base Acid CA CB
Weak base (B) competes with OH- for the H+ ion
pH calculations surrounding these are similar to those done for weak acids
Example: Calculate the pH for a 15.0 M solution of NH3 (Kb = 1.8 X 10-5).
Answer: 12.20

19. Example #2
Calculate the pH of a M solution of methylamine, CH3NH2 (Ka = 4.38 X 10-4).
Answer: pOH = 1.91, pH = 12.09