1. Chapter 15
Acid-Base Equilibria

2. Solutions of Acids or Bases Containing a Common Ion—Section 15.1
A buffered solution is made by combining a weak acid and it’s conjugate base
Buffers take advantage of the common ion present in solution to “buffer” against any changes pH
Ex: Acetic acid and sodium acetate solution

3. Calculating the pH of a Buffer
What is the pH of an acetic acid/acetate buffer made with 0.4 M sodium acetate and 0.20 M acetic acid? Ka for acetic acid is 1.8  10-5

4. Henderson-Hasselbalch Equation for Buffers
Calculate the pH of the previous acetic acid/acetate buffer using the equation above
Can only be used “when x is really small”

5. Preparing Buffer Solutions
Calculate the mass of ammonium chloride that must be added to mL of 0.32 M NH3 to prepare a pH 8.50 buffer, Kb for NH3 is 1.8  10-5.

6. How Buffers Act: The Addition of Strong Acid or Strong Base
The buffer capacity for a particular buffer refers to the amount of acid or base that a buffer can neutralize before large changes in pH occur
Ex: 1 M HF/F- vs. 0.1 M HF/F-
Both solutions have the same ratio of acid/base; however 1 M solution has the higher buffer capacity because it is capable of neutralizing more acid or base

7. Acid-Base Titrations
The process of systematically adding acid or base for neutralization is known as a titration
Note: Titration is a technique and can be applied to other types of chemical reactions besides neutralization reactions

8. Titration Curves

9. Strong Acid/Strong Base Titrations
Calculate the pH in the titration of mL of M HCl after the addition of (a.) 0 (b.) 2.00, (c.) 10.00, and (d.) mL of M NaOH

10. Titration of a Strong Acid w/ Strong Base
Calculate the pH at each of the steps indicated below for the titration of 200 mL of 0.15 M HBr.
After the addition of 0 mL of 0.15 M KOH
After the addition of 100 mL of 0.15 M KOH
After the addition of 200 mL of 0.15 M KOH
After the addition of 250 mL of 0.15 M KOH

11. Titration Curves for Strong Acid/Strong Base Titrations
Divided into different regions
Region 1: No base (pH << 7)
Region 2: Addition of base up to equivalence point (pH < 7)
Region 3: Equivalence point; indicated by the inflection point of the curve (pH = pKa)
For strong acid/strong base titration, pH = 7.0
Region 4: Increased addition of base after equivalence point (pH > 7)

12. Titration Curves for Weak Acid/Strong Base Titrations
Also has 4 regions although the regions are slightly different
Region 1: Presence of weak acid only
Not 100% dissociation so ICE tables must be used to determine pH
Region 2: Buffer region
Weak acid + conjugate base (salt) = buffer; use Henderson-Hasselbalch to calculate pH
Region 3: Equivalence point
Equal amounts (# of moles) of acid and strong base; conjugate base is produced and Kb must be used to calculate pH
Region 4: Addition of excess strong base
Because excess OH- is present, it will dominate the pH; amount of conjugate base is irrelevant [OH-] is used to calculate pH

13. Titration of Weak Acid With Strong Base
Calculate the pH during the titration of 20.0 mL of M formic acid (Ka = 1.8 x 10-4) with M NaOH. Calculate the pH at the following intervals:
0 mL NaOH
10.0 mL NaOH
20.0 mL NaOH
30.0 mL NaOH

14. Titration Curves
Strong Acid w/ Strong Base
Strong Base w/ Strong Acid

15. Titration Curves (cont.)

16. Polyprotic Acid Titration Curves
Diprotic Acid
Triprotic Acid

17. Section 15.5—Acid-Base Indicators
Acid-base titrations are best observed using a pH meter which will give a digital readout of pH throughout the course of a titration
Acid-base indicators can be used however to at least determine when a titration has reached the end point

19. Solubility Equilibria
In reality, there is no such thing as a 100% insoluble compound
Only practically so
Example: AgCl in H2O
Insoluble compounds undergo an equilibrium in the same manner as gases, weak acids, or weak bases
Equilibria constants for molar solubility are represented as Ksp

20. Calculating Ksp
When lead iodate, Pb(IO3)2, is added to water, a small amount dissolves. If measurements at 25 C show that the Pb2+ concentration is 4.5 x 10-5 M, calculate the value of Ksp for Pb(IO3)2.
See Interactive Example 16.1 (Pg. 642)

21. Calculating Solubility from Given Ksp Values
Given the value of Ksp for BaCO3 (Ksp = 1.6 x 10-9), calculate the solubility of Ba(IO3)2.
See Interactive Example 16.3 (Pg. 644)

22. Solubility and the Common Ion Effect
Assuming the following equilibrium:
CaF2(s) Ca2+(aq) + 2 F-(aq)

23. Solubility and pH
Essentially just a special case of the common ion effect
Ex: Solubility of Mg(OH)2 in neutral vs. basic solution

24. Calculating Solubility in the Presence of a Common Ion
What is the solubility of calcium hydroxide (Ksp = 5.0 x 10-6) in M NaOH solution?
See Interactive Example 16.4 (Pg. 648)

25. Predicting the Precipitation of Ions
A chemist mixes 200 mL of M Pb(NO3)2 with 100 mL of M NaCl. Will lead chloride precipitate? Ksp = 1.7 x 10-5
See Interactive Example 16.5 (Pg. 650)