1. Principles of Reactivity: Other Aspects of Aqueous Equilibria
Chapter 17

2. Learning Objectives Students understand Students will be able to
The common ion effect
The control of pH in aqueous solutions with buffers
Students will be able to
Calculate the pH of buffer solutions
Evaluate the pH in the course of acid-base titrations
Apply equilibrium concepts to the solubility of ionic compounds

3. 17.1 Common Ion Effect
The common ion effect is the limiting of the ionization of an acid (or base) by the presence of a significant concentration of its conjugate base (or acid).
The extent to which the acid can ionize is affected, therefore affecting the pH of the solution.

4. Practice Problem
Assume you have a 0.30M solution of formic acid (HCO2H) and have added enough sodium formate (NaHCO2) to make the solution 0.10M in the salt. Calculate the pH of the formic acid solution before and after adding sodium formate.

5. Practice Problem
What is the pH of the solution that results from adding 30.0mL of 0.100M NaOH to 45.0mL of 0.100M acetic acid?

6. 17.2 Controlling pH: Buffers
A buffer causes solutions to be resistant to a change in pH on addition of a strong acid or base.
Two substances are needed: an acid capable of reacting with added OH- ions and a base that can consume added H3O+ ions.
The acid and base must not react with each other

7. Buffers A buffer is usually prepared from a conjugate acid-base pair.
The action of a buffer is a special case of the common ion effect.

8. Buffers Weak acid and its conjugate base:
[H3O+] = [acid]/[conjugate base] ∙ Ka
Another form of the same equation:
pH = pKa + log [conjugate base]/[acid]
known as the Henderson-Hasselbalch equation

9. Henderson-Hasselbalch equation
To use this equation, you assume that the equilibrium concentrations of the acid and its conjugate base are approximately equal to their initial concentrations
The pH of the buffer falls within 3 to 11
The initial concentrations of the acid and the conjugate base are large

10. Practice Problem
What is the pH of a buffer solution composed of 0.50M formic acid (HCO2H) and 0.70M sodium formate (NaHCO2)?

11. Practice Problem
Suppose you dissolve 15.0 g of NaHCO3 and 18.0g of Na2CO3 in enough water to make 1.00L of solution. Use the H-H equation to calculate the pH of the solution. (Consider this buffer as a solution of the weak acid HCO3- with CO32- as its conjugate base.)

12. Preparing Buffer Solutions
pH control: The solution should control the pH at the desired value. Choose an acid with Ka near to the intended value of [H3O+]. The exact value of [H3O+] can be achieved by adjusting the acid/conjugate base ratio.
Buffer capacity: The buffer should be able to control the pH after the addition of reasonable amounts of acid and base.

13. Practice Problem – Not Tested!
Using an acetic acid/sodium acetate buffer solution, what ratio of acid to conjugate base will you need to maintain the pH at 5.00?

14. Preparing Buffer Solutions
It is the relative number of moles of acid and conjugate base that is important in determining the pH of a buffer solution. (the solution volume is the same for both components)
Diluting a buffer solution will not change its pH!

15. Practice Problem – Not Tested!
Calculate the pH of 0.500L of a buffer solution composed of 0.50M formic acid (HCO2H) and 0.70M sodium formate (NaHCO2) before and after adding 10.0mL of 1.0M HCl.

16. Homework
After reading sections , you should be able to do the following…
P. 677 (7-25 odd)

17. 17.3 Acid-Base Titrations
A titration is one of the most important ways of determining accurately the quantity of an acid, a base, or some other substance in a mixture.
The pH at the equivalence point of a strong acid/strong base titration is 7.
If the substance is a weak acid or base, then the pH at equivalence is not 7; depends on conjugate base or acid.

18. Titration: Strong Acid/Strong Base
The equivalence point in any acid-base titration is identified as the midpoint in the vertical portion of the pH versus volume of titrant curve. (titrant refers to the substance being added, analyte is the substance being tested)
The pH of the solution at the equivalence point in a strong acid/strong base reaction is always 7. (at 25oC)

20. Practice Problem
What is the pH after 25.0 mL of 0.100M NaOH has been added to 50.0mL of 0.100M HCl? What is the pH after mL of NaOH has been added?

21. Titration: Weak Acid/Strong Base
Before titration begins, the pH is found from the weak acid Ka value and the acid concentration.
At the equivalence point, the pH is controlled by the conjugate base.
The pH at the halfway point of the titration is equal to the pKa of the weak acid.

22. At the halfway point in the titration of a weak acid with a strong base [H3O+] = Ka and pH = pKa

23. Practice Problem
The titration of 0.100M acetic acid with 0.100M NaOH is described in the text. What is the pH of the solution when 35.0mL of the base has been added to 100.0mL of 0.100M acetic acid?

24. Titration of Weak Polyprotic Acids
Multiple equivalence points in the graph as each successive H ion is titrated.

25. Titration: Weak Base/Strong Acid
Similar to Weak Acid/Strong Base, but the titration curve goes from high pH to low (opposite).

26. Practice Problem
Calculate the pH after 75.0mL of 0.100M HCl has been added to 100.0mL of 0.100M NH3. See Figure 17.7 on page 651.

27. pH Indicators
An acid-base indicator is a weak acid or a weak base whose color is sensitive to pH. The acid form (HInd) has one color and the conjugate base (Ind-) has another.
Although the indicator reacts with substances in solution, so little indicator is present that the analysis is not significantly affected.

28. Homework
After reading section 17.3, you should be able to do the following…
P. 677a-b (27-35 odd)

29. 17.4 Solubility of Salts
The equilibrium constant that reflects the solubility of a compound is referred to as its solubility product constant, Ksp.
The solubility of a salt is the amount present in some volume of saturated solution. The Ksp is an equilibrium constant.

30. Practice Problem
Write Ksp expressions for the following insoluble salts and look up numerical values for the constant in Appendix J.
AgI
BaF2
Ag2CO3

31. Practice Problem
The barium ion concentration [Ba2+] in a saturated solution of barium fluoride is 3.6 x 10-3M. Calculate the value of the Ksp for BaF2.
BaF2(s)  Ba2+(aq) + 2F-(aq)

32. Practice Problem
Using the value of Ksp = 5.5 x 10-5, calculate the solubility of Ca(OH)2 in moles per liter and grams per liter.

33. Solubility and Ksp
You can compare Ksp for two salts, but only if they have the same ion ratio!
Larger Ksp means that the salt is more soluble.

34. Practice Problem
Using Ksp values, predict which salt in each pair is more soluble in water.
AgCl or AgCN
Mg(OH)2 or Ca(OH)2
Ca(OH)2 or CaSO4

35. Solubility and the Common Ion Effect
The ionization of weak acids and bases is affected by the presence of an ion common to the equilibrium process and the effect of adding an ion to a saturated solution (with the same ion) will shift the equilibrium back to the formation of the compound (decrease the solubility).

36. Practice Problem
Calculate the solubility of BaSO4 (a) in pure water and (b) in the presence of 0.010M Ba(NO3)2, Ksp for BaSO4 is 1.1x10-10.

37. Practice Problem
Calculate the solubility of Zn(CN)2 at 25oC (a) in pure water and (b) in the presence of 0.10M Zn(NO3)2. Ksp for Zn(CN)2 is 8.0x10-12.

38. Effect of Basic Anions on Salt Solubility
Any salt containing an anion that is the conjugate base of a weak acid will dissolve in water to a greater extent than given by Ksp.
due to reaction of the anion with water, which decreases its concentration and shifts the equilibrium
Insoluble salts in which the anion is the conjugate base of a weak acid will dissolve in strong acids.

39. Homework
After reading section 17.4, you should be able to do the following…
P. 677b-c (41-61 odd)

40. 17.5 Precipitation Reactions
Recall that the difference between Q (the reaction quotient) and K (the equilibrium constant) is that the concentrations used in the reaction quotient may or may not be those at equilibrium.
If Ksp=Q the solution is saturated.
If Ksp>Q the solution is not saturated.
If Ksp<Q the solution is supersatured.

41. Practice Problem
Solid PbI2 (Ksp = 9.8x10-9) is placed in a beaker of water. After a period of time, the lead (II) concentration is measured and found to be 1.1x10-3M. Has the system yet reached equilibrium? That is, is the solution saturated? If not, will more PbI2 dissolve?

42. Practice Problem
If the concentration of strontium ion, Sr2+, is 2.5 x 10-4M, will precipitation of SrSO4 occur when enough of the soluble salt Na2SO4 is added to make the solution 2.5 x 10-4M in SO42-? Ksp for SrSO4 is 3.4x 10-7.

43. Practice Problem
What is the minimum concentration of I- that can cause precipitation of PbI2 from a 0.050M solution of Pb(NO3)2? Ksp for PbI2 is 9.8 x What concentration of Pb2+ ions remains in solution when the concentration of I- is M?

44. Practice Problem
You have 100.0mL of M silver nitrate. Will AgCl precipitate if you add 5.0mL of 0.025M HCl?

45. 17.6 Solubility and Complex Ions
Metal ions exist in aqueous solution as complex ions.
The equilibrium constant for the formation of a complex ion such as Ag(NH3)2+ is called a formation constant, Kform.
Knet = KspKform

46. Practice Problem
Silver nitrate ( mol) is added to 1.00 L of 1.00 M NH3. What is the concentration of Ag+ ions at equilibrium?
Ag+(aq) + 2NH3(aq)  [Ag(NH3)2]+(aq)
Kf = 1.1 x 107

47. 17.7 Solubility, Ion Separations, and Qualitative Analysis
It is often necessary to use Ksp information in order to identify ions in solution.
You will need to find a reagent that will form a precipitate with one or more of the cations while leaving the others in solution.
Also consider which salts are soluble in water and which aren’t.
You can then add chemicals in order to identify the salts that have known colors.

48. Homework
After reading sections , you should be able to do the following…
P. 677c (65-73 odd)