1. Chapter 17 Additional Aspects of Aqueous Equilibria

2. Buffered Solutions and the Common Ion Effect Sections 17.1-2  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? K a for acetic acid is 1.8  10 -5

4. Henderson-Hasselbalch Equation for Buffers  Henderson-Hasselbalch equation:  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 500.0 mL of 0.32 M NH 3 to prepare a pH 8.50 buffer, K b for NH 3 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. Calculating pH Change in Buffered Solutions  Consider the buffer solution mentioned previously (slide 3). The pH of this buffer is 5.05. Calculate the change in pH when 1.5 mL of 0.067 M HCl is added to 100.0 mL of this buffer solution.  For sake of comparison, calculate the change in pH when 1.5 mL of 0.067 M HCl is added to 100.0 mL of HCl solution with a pH of 5.05.

8. 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

9. Titration Curves

10. Strong Acid/Strong Base Titrations  Calculate the pH in the titration of 20.00 mL of 0.125 M HCl after the addition of (a.) 0 (b.) 2.00, (c.) 10.00, and (d.) 20.00 mL of 0.250 M NaOH

11. 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

12. 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 = pK a )  For strong acid/strong base titration, pH = 7.0 Region 4: Increased addition of base after equivalence point (pH > 7)

13. 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 K b 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

14. Titration of Weak Acid With Strong Base  Calculate the pH during the titration of 20.0 mL of 0.500 M formic acid (K a = 1.8 x 10 -4 ) with 0.500 M NaOH. Calculate the pH at the following intervals: 0 mL NaOH 10.0 mL NaOH 20.0 mL NaOH 30.0 mL NaOH

15. Solubility Equilibria  In reality, there is no such thing as a 100% insoluble compound Only practically so  Example: AgCl in H 2 O  Insoluble compounds undergo an equilibrium in the same manner as gases, weak acids, or weak bases Equilibria constants for molar solubility are represented as K sp

16. Calculating K sp When lead iodate, Pb(IO 3 ) 2, is added to water, a small amount dissolves. If measurements at 25  C show that the Pb 2+ concentration is 4.5 x 10 -5 M, calculate the value of K sp for Pb(IO 3 ) 2.  See Sample Exercise 17.10 (Pg. 739)

17. Calculating Solubility from Given K sp Values Given the value of K sp for Ba(IO 3 ) 2 (K sp = 1.5 x 10 -9 ), calculate the solubility of Ba(IO 3 ) 2.  See Sample Exercise 17.11 (Pg. 740)

18. Solubility and the Common Ion Effect  Assuming the following equilibrium: CaF 2 (s) Ca 2+ (aq) + 2 F - (aq)

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

20. Calculating Solubility in the Presence of a Common Ion What is the solubility of calcium hydroxide (K sp = 5.0 x 10 -6 ) in 0.050 M NaOH solution?  See Sample Exercise 17.12 (Pg. 742)

21. Predicting the Precipitation of Ions A chemist mixes 200 mL of 0.010 M Pb(NO 3 ) 2 with 100 mL of 0.0050 M NaCl. Will lead chloride precipitate? K sp = 1.7 x 10 -5  See Sample Exercise 17.15 (Pg. 751)