1. Chapter 17 Additional Aspects of Aqueous Equilibria
Lecture Presentation
Chapter 17 Additional Aspects of Aqueous Equilibria
John D. Bookstaver
St. Charles Community College
Cottleville, MO
© 2012 Pearson Education, Inc.

2. Buffers crash course

3. © 2012 Pearson Education, Inc.
17.1 The Common-Ion Effect
Consider a solution of acetic acid:
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO(aq)
If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.
NaC2H3O2 (s)  Na+ (aq) + C2H3O2- (aq)
© 2012 Pearson Education, Inc.

4. © 2012 Pearson Education, Inc.
The Common-Ion Effect
“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”
© 2012 Pearson Education, Inc.

5. Calculating the pH When a Common Ion is Involved
Sample Exercise 17.1
Calculating the pH When a Common Ion is Involved
What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution?
Solution
Analyze
We are asked to determine the pH of a solution of a weak electrolyte (CH3COOH) and a strong electrolyte (NaCH3COO) that share a common ion, CH3COO-
Plan In any problem in which we must determine the pH of a solution containing a mixture of solutes, it is helpful to proceed by a series of logical steps:
1. Consider which solutes are strong electrolytes and which are weak electrolytes, and identify the major species in solution.
2. Identify the important equilibrium that is the source of H+ and therefore determines pH.
3. Tabulate the concentrations of ions involved in the equilibrium.
4. Use the equilibrium-constant expression to calculate [H+] and then pH. 5

6. Notice where the “initial” concentrations come from:
What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution?
Notice where the “initial” concentrations come from:
The strong electrolyte 0.30 M NaCH3COO dissociates “immediately” (NaCH3COO(s)  Na+(aq) + CH3COO-(aq))
so its concentrations are:
*[Na+] = 0.30 M
[CH3COO-] = 0.30 M
The weak electrolyte, 0.30 M CH3COOH starts entirely as CH3COOH molecules so the initial concentrations of CH3COOH, H+ and CH3COO- for the acetic acid equilibrium,
CH3COOH  CH3COO- + H+
are,
[CH3COOH] = 0.30 M
[CH3COO-] = 0 M
[H+] = 0 M
* Na + is a spectator of course so it won’t be needed – of course!.
This is the most important concept you can understand and apply in Ch17!

7. Solve
First, because CH3COOH is a weak electrolyte and NaCH3COO is a strong electrolyte, the major species in the solution are CH3COOH (a weak acid), Na+ (which is neither acidic nor basic and is therefore a spectator in the acid–base chemistry), and CH3COO– (which is the conjugate base of CH3COOH).
Second, [H+] and, therefore, the pH are controlled by the dissociation equilibrium of CH3COOH:
[We have written the equilibrium using H+ (aq) rather than H3O+(aq), but both representations of the hydrated hydrogen ion are equally valid.]
Third, we tabulate the initial and equilibrium concentrations as we did in solving other equilibrium problems in Chapters 15 and 16. Let x = the H+ concentration we’re looking for:
Again notice
Starting conc.
of acetate
comes from
the NaC2H3O2

8. the equilibrium-constant expression:
The equilibrium concentration of CH3COO- (the common ion) is the initial concentration that is due to NaCH3COO (0.30 M) plus the change in concentration (x) that is due to the ionization of CH3COOH.
Now we can use
the equilibrium-constant expression:
The dissociation constant for CH3COOH at 25 C is from Appendix D; addition of NaCH3COO does not change the value of this constant. Substituting the equilibrium constant concentrations from our table into the equilibrium expression gives:
Because Ka is small, we assume that x is small compared to the original concentrations of CH3COOH and CH3COO– (0.30 M each). Thus, we can ignore the very small x relative to 0.30 M, giving: 8

9. The resulting value of x is indeed small relative to 0
The resulting value of x is indeed small relative to 0.30, justifying the approximation made in simplifying the problem.
Finally, we calculate the pH from the equilibrium concentration of H+(aq):
x = 1.8  105 M = [H+]
pH = –log(1.8  105) = 4.74
Comment In Section 16.6 we calculated that a 0.30 M solution of CH3COOH has a pH of 2.64, corresponding to [H+] = 2.3  103. Thus, the addition of NaCH3COO has substantially decreased [H+], as we expect from Le Châtelier’s principle.
Practice Exercise
Calculate the pH of a solution containing M nitrous acid (HNO2, Ka = 4.5  104) and 0.10 M potassium nitrite (KNO2).
Answer:
3.42 9

10. Sample Exercise 17.2 Calculating Ion Concentrations When a Common Ion is Involved
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.
Solution
Analyze We are asked to determine the concentration of F– and the pH in a solution containing the weak acid HF and the strong acid HCl.
In this case the common ion is H+.
Plan We can again use the four steps outlined in Sample Exercise 17.1. 10

11. Again, notice where the “initial” concentrations come from:
The weak electrolyte, 0.20 M HF starts entirely as HF molecules so the initial concentrations of HF, H+ and F- for the HF equilibrium,
HF  F- + H+
are,
[HF] = 0.20 M
[F-] = 0 M
[H+] = 0 M
The strong electrolyte 0.10 M HCl dissociates “immediately” (HCl(aq)  H+(aq) + Cl-(aq))
so its concentrations are:
[H+] = 0.10 M
*[Cl-] = 0.10 M
* Cl - is a spectator of course so it won’t be needed – of course!.

12. Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.
Solve Because HF is a weak acid and HCl is a strong acid, the major species in solution are HF, H+, and Cl–. The Cl–, which is the conjugate base of a strong acid, is merely a spectator ion in any acid–base chemistry.
The problem asks for [F–], which is formed by ionization of HF. Thus, the important equilibrium is
The common ion in this problem is the hydrogen (or hydronium) ion. Now we can tabulate the initial and equilibrium concentrations of each species involved in this equilibrium. Since we started with 0.10 M acid, let x be the F- and in this case the increase in H+ due to the HF contribution:
Again notice starting conc. of hydrogen comes from the immediate dissociation of the strong electrolyte HCl

13. Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.
The equilibrium constant for the ionization of HF, from Appendix D, is 6.8  104. Substituting the equilibrium-constant concentrations into the equilibrium expression gives
If we assume that x is small relative to 0.10 or 0.20 M, this expression simplifies to
This F– concentration is substantially smaller than it would be in a 0.20 M solution of HF with no added HCl. The common ion, H+, shifts the equilibrium to the left, and suppresses the ionization of HF.
The concentration of H+(aq) is
Thus,
pH = 1.00 13

14. Comment Notice that for all practical purposes, the hydrogen ion concentration is due entirely to the HCl; the HF makes a negligible contribution by comparison.

15. Strategy! In a problem involving mixing strong electrolytes with weak:
1) Assume that strong electrolytes added dissociate immediately and completely
2) Assume that weak electrolytes dissociate next, and the extent will depend on the presence of the strong electrolyte already present
3) The equilibrium concentrations will depend on the weak electrolyte shift that occurs.

16. Strategy!
Repeat: The final concentration depends on the shift that occurs in the weak electrolyte
1) Assume that strong electrolytes added dissociate immediately and completely
2) Assume that weak electrolytes dissociate next, and the extent will depend on the presence of the strong electrolyte already present
3) The equilibrium concentrations will depend on the weak electrolyte shift that occurs.

17. Strategy!
Example: suppose you wish to find the final concentration of acetate ions (C2H3O2-) in a solution of acetic acid (HC2H3O2) when a strong acid like HCl is added.

18. Strategy! Example: adding HCl to a solution of HC2H3O2
(strong acid) (weak acid)
1) Assume HCl completely dissociates giving off all of its H+ ions.

19. Strategy! Example: adding HCl to a solution of HC2H3O2
(strong acid) (weak acid)
2) Assume HC2H3O2 will next begin to ionize.
(The assumption is that acetic acid is waiting to ionize until after the strong acid is added. Silly I suppose but it works!)

20. Strategy! Example: adding HCl to a solution of HC2H3O2
(strong acid) (weak acid)
The final pH will depend on the
weak electrolyte acetic acid (not on the HCl)

21. Strategy! Example: adding HCl to a solution of HC2H3O2
(strong acid) (weak acid)
You will need to set up and ICE table and Ka expression to calculate the equilibrium concentration of H+ that remains after the acetic acid shift!

22. Strategy!
Try keeping this strategy in mind as you work problems involving strong and weak electrolyte mixtures.

23. Practice Exercise
Calculate the formate ion concentration and pH of a solution that is M in formic acid (HCOOH, Ka = 1.8  104) and 0.10 M in HNO3.
Answer:
[HCOO–] = 9.0  105, pH = 1.00

24. © 2012 Pearson Education, Inc.
17.2 Buffers
Buffers are solutions of a weak conjugate acid–base pair.
They are particularly resistant to pH changes, even when strong acid or base is added.
© 2012 Pearson Education, Inc.

25. Buffers crash course

26. © 2012 Pearson Education, Inc.
Buffers
If a small amount of hydroxide is added to an equimolar solution of HF in NaF, the HF reacts with the OH to make F and water HF + OH-  F- + H2O
© 2012 Pearson Education, Inc.

27. © 2012 Pearson Education, Inc.
Buffers
Similarly, if acid is added, the F reacts with it to form HF and water. H3O+ + F-  H2O + HF
© 2012 Pearson Education, Inc.

28. Sample Exercise 17.3 Calculating the pH of a Buffer
What is the pH of a buffer that is 0.12 M in lactic acid [HC3H5O3] and 0.10 M in sodium lactate [NaC3H5O3]? For lactic acid, Ka = 1.4  104.
Solution
Analyze We are asked to calculate the pH of a buffer containing lactic acid (HC3H5O3) and its conjugate base, the lactate ion (C3H5O3–).
Plan We will first determine the pH using the method described in Section 17.1.
Because HC3H5O3 is a weak electrolyte and NaC3H5O3 is a strong electrolyte, the major species in solution are HC3H5O3, Na+, and C3H5O3–. The Na+ ion is a spectator ion.
The HC3H5O3–C3H5O3– conjugate acid–base pair determines [H+] and, thus, pH;
[H+] can be determined using the acid-dissociation equilibrium of lactic acid. 28

29. Notice where the “initial” concentrations come from:
The weak electrolyte, 0.12 M lactic acid starts entirely as HC3H5O3 molecules so the initial concentrations of HC3H5O3, H+ and C3H5O3- for the HC3H5O3 equilibrium,
HC3H5O3  C3H5O3- + H+
are,
[HC3H5O3] = 0.12 M
[C3H5O3 -] = 0 M
[H+] = 0 M
The strong electrolyte 0.10 M sodium lactate dissociates “immediately” (NaC3H5O3(aq)  Na+(aq) + C3H5O3 -(aq))
so its concentrations are:
*[Na+] = 0.10 M
[C3H5O3 -] = 0.10 M
Notice: This will give you all of the “I” values for your ICE table
* Na+ is a spectator of course so it won’t be needed – of course!.

30. Solve The initial and equilibrium concentrations of the species involved in this equilibrium are
The equilibrium concentrations are governed by the equilibrium expression:
Because Ka is small and a common ion is present, we expect x to be small relative to either 0.12 or 0.10 M. Thus, our equation can be simplified to give

31. Solving for x gives a value that justifies our approximation:
Practice Exercise
Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. Appendix D: benzoic acid HC7H5O2 ka=6.3 x 10 -5
Answer:
4.42 31

32. © 2012 Pearson Education, Inc.
pH Range
The pH range is the range of pH values over which a buffer system works effectively.
It is best to choose an acid with a pKa close to the desired pH.
© 2012 Pearson Education, Inc.

33. Sample Exercise 17.4 Preparing a Buffer
How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00?
(Assume that the addition of NH4Cl does not change the volume of the solution.)
Solution
Analyze We are asked to determine the amount of NH4+ ion required to prepare a buffer of a specific pH.
Plan The major species in the solution will be NH4+, Cl–, and NH3. Of these, the Cl– ion is a spectator (it is the conjugate base of a strong acid).
Thus, the NH4+–NH3 conjugate acid–base pair will determine the pH of the buffer. The equilibrium relationship between NH4+ and NH3 is given by the base-dissociation reaction for NH3:
The key to this exercise is to use this Kb expression to calculate [NH4+]. 33

34. Solve We obtain [OH] from the given pH:
and so
Because Kb is small and the common ion [NH4+] is present, the equilibrium concentration of NH3 essentially equals its initial concentration:
pOH = – pH = – 9.00 = 5.00
[OH–] = 1.0  105 M
Reaction
NH3
H2O
NH4+
OH-
Initial
0.10 M - x
Change
- 10-5
+ 10-5
Equilibrium
x+ 10-5
10-5 M
[NH3] = 0.10 M

35. We now use the expression for Kb to calculate [NH4+]:
Thus, for the solution to have pH = 9.00, [NH4+] must equal 0.18 M. The number of moles of NH4Cl needed to produce this concentration is given by the product of the volume of the solution and its molarity:
(2.0 L)(0.18 mol NH4Cl/L) = 0.36 mol NH4Cl

36. Practice Exercise
Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (C6H5COOH) to produce a pH of 4.00.
Answer:
Notice: sodium benzoate (strong electrolyte with the conjugate base) with benzoic acid (weak conjugate acid)
0.13 M

37. pKa – a useful concept
Ex: A buffer solution made of 0.10 moles of acetic acid HC2H3O2 and 0.10 moles of NaC2H3O2 in 1.0 liters.
The weak electrolytes: HC2H3O2 determines the equilibrium pH:
HC2H3O2  H C2H3O2-
Ka = [H+ ] [C2H3O2-]
[HC2H3O2]
= X (0.10 M) = [H+ ]
(0.10 M)
Recall that we assume the equilibrium concentrations of HC2H3O2 and C2H3O2- to be essentially the same as initial
So 0.10 M – X = 0.10 M and 0.10 M + X = 0.10 M

38. pKa – a useful concept
Ex: A buffer solution made of 0.10 moles of acetic acid HC2H3O2 and 0.10 moles of NaC2H3O2 in 1.0 liters.
The weak electrolytes: HC2H3O2 determines the equilibrium pH:
HC2H3O2  H C2H3O2-
Ka = [H+ ] [C2H3O2-]
[HC2H3O2]
= X (0.10 M) = [H+ ]
(0.10 M)
Notice when concentrations of the acid and its conjugate base are equal, they cancel out.

39. pKa
A buffer solution made of 0.10 moles of acetic acid HC2H3O2 and 0.10 moles of NaC2H3O2 in 1.0 liters.
The weak electrolytes: HC2H3O2 determine the equilibrium pH:
HC2H3O2  H C2H3O2-
Ka = [H+ ] [C2H3O2-]
[HC2H3O2]
= X (0.10 M) = [H+ ]
(0.10 M)
So, Ka = x = X = [H+ ]
And pKa = - log Ka = - log (1.8 x 10-5 ) = pH

40. pKa
A buffer solution made of 0.10 moles of acetic acid HC2H3O2 and 0.10 moles of NaC2H3O2 in 1.0 liters.
The weak electrolytes: HC2H3O2 determine the equilibrium pH:
HC2H3O2  H C2H3O2-
Ka = [H+ ] [C2H3O2-]
[HC2H3O2]
= X (0.10 M) = [H+ ]
(0.10 M)
Get it? Ka = [H+ ] at equilibrium
And pKa = pH
Only when [HA] = [A- ]

41. Henderson Hasselbalch – another useful relationship
pH = pKa + log [A-]
[HA]
pH = - log (1.8 x 10-5 ) + log (0.10)
(0.10)
pH = log (1)
pH =
In a buffer mix, when [HA] = [A- ]
The pH = the pKa
pH = 4.74

42. Henderson Hasselbalch
pH = pKa + log [A-]
[HA]
pH = - log (1.8 x 10-5 ) + log (0.10)
(0.10)
pH = log (1)
pH =
Lets look at what happens when [HA] ≠ [A- ]
And the pH ≠ the pKa
pH = 4.74

43. Henderson Hasselbalch
pH = pKa + log [A-]
[HA]
Suppose the buffer solution made of 0.12 moles of acetic acid HC2H3O2 and 0.10 moles of NaC2H3O2 in 1.0 liters.
pH = - log (1.8 x 10-5 ) + log (0.10)
(0.12)
pH = - log (1.8 x 10-5 ) + log 0.83
pH = (-0.081)
Notice: As the acid concentration increases, the pH decreases
pH = 4.66

44. Adding acid or base to a buffer
Buffered aspirin reduces the affect of extra acid in people with excess stomach acid
A buffer absorbs acid or base added to it, so as to resist changes in pH.

45. When Strong Acids or Bases Are Added to a Buffer
When strong acids or bases are added to a buffer, it is safe to assume that all of the strong acid or base is consumed in the reaction.
© 2012 Pearson Education, Inc.

46. Addition of Strong Acid or Base to a Buffer
Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.
Use traditional Ka methods to determine the final pH

47. Sample Exercise 17.5 Calculating pH Changes in Buffers
A buffer is made by adding mol CH3COOH and mol CH3COONa to enough water to make L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1).
(a) Calculate the pH of this solution after 5.0 mL of 4.0 M NaOH(aq) solution is added. (b) For comparison, calculate the pH of a solution made by adding 5.0 mL of 4.0 M NaOH(aq) solution to L of pure water. 47

48. Then we use the resultant buffer composition and
A buffer is made by adding mol CH3COOH and mol CH3COONa to enough water to make L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1).
(a) Calculate the pH of this solution after 5.0 mL of 4.0 M NaOH(aq) solution is added.
Solution
Analyze We are asked to determine the pH of a buffer after addition of a small amount of strong base and to compare the pH change with the pH that would result if we were to add the same amount of strong base to pure water.
Plan Solving this problem involves the two steps outlined in Figure First we do a stoichiometry calculation to determine how the added OH– affects the buffer composition.
Then we use the resultant buffer composition and
the equilibrium-constant expression
for the buffer to determine the pH.

49. Solve (a)
Stoichiometry: Prior to this neutralization reaction, there are mol each of CH3COOH and CH3OO–. The amount of base added is L  4.0 mol/L = mol. The OH– provided by NaOH reacts with CH3COOH, the weak acid component of the buffer. Neutralizing the mol OH– requires mol of CH3COOH.
Consequently, the amount of CH3COOH decreases by mol, and the amount of the product of the neutralization, CH3COO–, increases by mol.
0.020 moles
CH3 COOH +
0.020 moles
OH-
0.020 moles
CH3 COO-
forms
This table summarizes the changes occurring during the neutralization: I C F
-0.020
+0.020 49

50. Converting moles to Molarity:
Notice that all of the added OH- has been neutralized. Also, our quantities in this table are in moles, not Molarity. The final volume will be the combined volumes of the two solutions, which will determine final molarities
Equilibrium Calculation: We now turn our attention to the equilibrium for the ionization of acetic acid, the relationship that determines the buffer pH:
[H+][CH3COO-]
Ka =
[CH3COOH]
Converting moles to Molarity:
[CH3COO-] = moles/1.005 L = M
[CH3COOH] = moles/1.005 L = M

51. [H+][0.318] pH = - log [H+ ] [0.279] = -log (1.58 x 10-5 ) = 4.80
Reaction
CH3COOH H+
CH3COO-
Initial
0.279 M
0.318
Change
- x
+ x +x
Equilibrium
0.279 M -x x x
“Since acetic acid is a weak acid (Ka is a low 10-5 ) we assume that x is insignificant compared to our equilibrium concentrations of CH3COOH
and CH3COO- so M –x = M and –x = M”
And substituting into our eq expression:
[H+][0.318]
pH = - log [H+ ]
= -log (1.58 x )
= 4.80
[0.279]
[H+] = 1.58 x 10-5 M

52. Method 2: Using the quantities of CH3COOH– and CH3COO remaining in the buffer, we determine the pH using the Henderson–Hasselbalch equation.
[CH3COO-]
[CH3COOH]
pH = - log(1.8 x 10-5) + log
[0.318]
[0.279]
pH = log
pH = = 4.80
Does this seem easier? We must do most of the same steps to get to this point?

53. pOH = –log[OH–] = –log(0.020) = +1.70 pH = 14 – (+1.70) = 12.30
(b) To determine the pH of a solution made by adding mol of NaOH to L of pure water, we first determine the concentration of OH– ions in solution,
[OH–] = mol/(1.005 L) = M
We use this value in Equation to calculate pOH and then use our calculated pOH value in Equation to obtain pH:
pOH = –log[OH–] = –log(0.020) = +1.70
pH = 14 – (+1.70) = 12.30
Comment Note that the small amount of added NaOH changes the pH of water significantly. In contrast, the pH of the buffer changes very little when the NaOH is added, as summarized in Figure 17.4. 53

54. Sample Exercise 17.5 Calculating pH Changes in Buffers
Continued
Practice Exercise
Determine (a) the pH of the original buffer described in Sample Exercise 17.5 after the addition of mol HCl
and (b) the pH of the solution that would result from the addition of mol HCl to L of pure water.
Answers :
(a) 4.68, (b) 1.70 54

55. © 2012 Pearson Education, Inc.
17.3 Titration
In this technique, a known concentration of base (or acid) is slowly added to a solution of acid (or base).
© 2012 Pearson Education, Inc.

58. © 2012 Pearson Education, Inc.
Titration
A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.
© 2012 Pearson Education, Inc.

59. Titration of a Strong Acid with a Strong Base

60. Titration of a Strong Acid with a Strong Base
From the start of the titration to near the equivalence point, the pH goes up slowly.

61. Titration of a Strong Acid with a Strong Base
Just before (and after) the equivalence point, the pH increases rapidly.
© 2012 Pearson Education, Inc.

62. Titration of a Strong Acid with a Strong Base
At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

63. Titration of a Strong Acid with a Strong Base
As more base is added, the increase in pH again levels off.

64. Sample Exercise 17.6 Calculating pH for a Strong Acid–Strong Base Titration
Calculate the pH when (a) 49.0 mL and (b) 51.0 mL of M NaOH solution have been added to 50.0 mL of M HCl solution.
Solution
Analyze We are asked to calculate the pH at two points in the titration of a strong acid with a strong base. The first point is just before the equivalence point, so we expect the pH to be determined by the small amount of strong acid that has not yet been neutralized. The second point is just after the equivalence point, so we expect this pH to be determined by the small amount of excess strong base.
Plan (a) As the NaOH solution is added to the HCl solution, H+(aq) reacts with OH–(aq) to form H2O. Both Na+ and Cl– are spectator ions, having negligible effect on the pH. To determine the pH of the solution, we must first determine how many moles of H+ were originally present and how many moles of were added. We can then calculate how many moles of each ion remain after the neutralization reaction.
To calculate [H+], and hence pH, we must also remember that the volume of the solution increases as we add titrant, thus diluting the concentration of all solutes present. 64

65. Sample Exercise 17.6 Calculating pH for a Strong Acid–Strong Base Titration
Continued
Solve The number of moles of H+ in the original HCl solution is given by the product of the volume of the solution and its molarity:
Likewise, the number of moles of OH– in 49.0 mL of M NaOH is 65

66. Because we have not reached the equivalence point, there are more moles of H+ present than OH–. Each mole of reacts OH– with 1 mol of H+. Using the convention introduced in Sample Exercise 17.5, we have the neutralization reaction:
The volume of the reaction mixture increases as the NaOH solution is added to the HCl solution. Thus, at this point in the titration, the volume in the titration flask is
50.0 mL mL = 99.0 mL = L
Thus, the concentration of H+(aq) in the flask is
The corresponding pH is
–log(1.0  103) = 3.00

67. Sample Exercise 17.6 Calculating pH for a Strong Acid–Strong Base Titration
Continued
Plan (b) We proceed in the same way as we did in part (a) except we are now past the equivalence point and have more OH– in the solution than H+. As before, the initial number of moles of each reactant is determined from their volumes and concentrations. The reactant present in smaller stoichiometric amount (the limiting reactant) is consumed completely, leaving an excess of hydroxide ion. 67

68. Solve
In this case the volume in the titration flask is
Hence, the concentration of OH–(aq) in the flask is
and we have
50.0 mL mL = mL = L
pOH = –log(1.0  103) = 3.00
pH = – pOH = – 3.00 = 11.00

69. Sample Exercise 17.6 Calculating pH for a Strong Acid–Strong Base Titration
Comment Note that the pH increased
by only two pH units, from 1.00 (Figure 17.7)
to 3.00, after the first 49.0 mL of NaOH solution
was added, but jumped by eight pH units,
from 3.00 to 11.00, as 2.0 mL of base solution
was added near the equivalence point.
Such a rapid rise in pH near the equivalence point
is a characteristic of titrations
involving strong acids and strong bases.
Practice Exercise
Calculate the pH when (a) 24.9 mL and (b) 25.1 mL of M HNO3 have been added to 25.0 mL of M KOH solution.
Answers:
(a) 10.30, (b) 3.70 69

70. Titration of a Weak Acid with a Strong Base

71. Unlike in the previous case, the stronger conjugate base of the weak acid affects the pH when it is formed (hydrolysis)

72. At the equivalence point the pH is >7.

73. Unlike in the previous case, the stronger conjugate base of the weak acid affects the pH when it is formed (hydrolysis)

74. Titration of a Weak Acid with a Strong Base
At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.
© 2012 Pearson Education, Inc.

75. Titration of a Weak Acid with a Strong Base
With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.
© 2012 Pearson Education, Inc.

76. Sample Exercise 17.7 Calculating pH for a Weak Acid–Strong Base Titration
Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH3COOH (Ka = 1.8  105).
Solution
Analyze We are asked to calculate the pH before the equivalence point of the titration of a weak acid with a strong base.
Plan We first must determine the number of moles of CH3COOH and CH3COO– present after the neutralization reaction. We then calculate pH using Ka, [CH3COOH], and [CH3COO–].

77. Solve Stoichiometry Calculation: The product of the volume and concentration of each solution gives the number of moles of each reactant present before the neutralization:
The 4.50  103 of NaOH consumes 4.50  103 of CH3COOH:
The total volume of the solution is
45.0 mL mL = 95.0 mL = L 77

78. The resulting molarities of CH3COOH and CH3COO– after the reaction are therefore
Equilibrium Calculation: The equilibrium between CH3COOH and CH3COO– must obey the equilibrium-constant expression for CH3COOH:
Reaction
CH3COOH H+
CH3COO-
Initial
0.053 M
0.0474
Change
- x
+ x +x
Equilibrium
0.053 M -x x x 78

79. 0.053 M –x = 0.053
Making similar assumptions as before:
and x =
Comment We could have solved for pH equally well using the Henderson–Hasselbalch equation – once the post neutralization concentrations have been calculated!
Practice Exercise
(a) Calculate the pH in the solution formed by adding 10.0 mL of M NaOH to 40.0 mL of M benzoic acid (C6H5COOH, Ka = 6.3  105. (b) Calculate the pH in the solution formed by adding 10.0 mL of M HCl to 20.0 mL of M NH3.
Answers:
(a) 4.20, (b) 9.26

80. Sample Exercise 17.8 Calculating the pH at the Equivalence Point
Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH3COOH with M NaOH.
Solution
Analyze We are asked to determine the pH at the equivalence point of the titration of a weak acid with a strong base. Because the neutralization of a weak acid produces its anion, which is a weak base, we expect the pH at the equivalence point to be greater than 7. (The conjugate base of the weak acid will hydrolyze the water)
Plan The initial number of moles of acetic acid equals the number of moles of acetate ion at the equivalence point. We use the volume of the solution at the equivalence point to calculate the concentration of acetate ion. Because the acetate ion is a weak base, we can calculate the pH using Kb and [CH3COO–].

81. Moles = M  L = (0.100 mol)/L(0.0500 L) = 5.00  103mol CH3COOH
Solve The number of moles of acetic acid in the initial solution is obtained from the volume and molarity of the solution:
Moles = M  L = (0.100 mol)/L( L) = 5.00  103mol CH3COOH
Hence, 5.00  103 mol of [CH3COO–] is formed. It will take 50.0 mL of NaOH to reach the equivalence point (Figure 17.9). … since they are both M solutions
The volume of this salt solution at the equivalence point is the sum of the volumes of the acid and base, 50.0 mL mL = mL = L.
Thus, the concentration of CH3COO– is 81

82. Practice Exercise The CH3COO– ion is a weak base:
The Kb for CH3COO– can be calculated from the Ka value of its conjugate acid, Kb = Kw/Ka = (1.0  1014)/(1.8  105) = 5.6  1010. Using the Kb expression, we have
Making the approximation that and then solving for x, we have x = [OH–] = 5.3  106 M, which gives pOH = 5.28 pH = 8.72.
Check The pH is above 7, as expected for the salt of a weak acid and strong base.
Practice Exercise
Calculate the pH at the equivalence point when (a) 40.0 mL of M benzoic acid (C6H5COOH, Ka = 6.3  105) is titrated with M NaOH; (b) 40.0 mL of M NH3 is titrated with M HCl.
Answers: (a) 8.21, (b) 5.28 82

83. Titration of a Weak Base with a Strong Acid
The pH at the equivalence point in these titrations is <7, so using phenolphthalein would not be a good idea.
Methyl red is the indicator of choice.
© 2012 Pearson Education, Inc.

84. Buffer point
Recall a buffer is a solution of a weak acid plus its conjugate base
Notice the point when half the acid has been titrated
The resulting conjugate base creates a buffer solution
pKa = pH
at ½ equivalence point

85. Buffer point Example: acetic acid titrated with sodium hydroxide
HC2H3O OH-  C2H3O H2O
pKa = pH
at ½ equivalence point

86. Buffer point Example: acetic acid titrated with sodium hydroxide
HC2H3O OH-  C2H3O H2O
½ the acid
remains
½ is left as
conjugate base
At the buffer point:
[HC2H3O2] = [C2H3O2-]
Ka = [H+] [C2H3O2-]
[HC2H3O2]
pKa = pH
at ½ equivalence point
Ka = [H+]
And pKa = pH

87. Ex: for HC2H3O2: HC2H3O2 + OH-  H2O + C2H3O2-
100 mL of
0.100 M =
0.010 moles
of H+
50 mL of
0.100 M =
0.005 moles
of OH+
Only 0.005
Moles react;
0.005 moles
HC2H3O2
Left over
0.005
Moles OH-
react;
0.005
Moles C2H3O2-
produced
pH depends on leftover HC2H3O2 in water
HC2H3O2  H+ + C2H3O2-
Ka = [H+ ][C2H3O2- ]
[HC2H3O2 ]
Ka = [H+ ](.005 Mol/vol)
(0.005 Mol/vol)
pKa = pH
Remaining solution
Contains equal quantities of
HC2H3O2 and C2H3O2-

88. Titrations of Polyprotic Acids
When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation.
Ex: For H2CO3
© 2012 Pearson Education, Inc.

90. 17.4 Solubility Products BaSO4(s) Ba2+(aq) + SO42(aq)
In a solution of a slightly soluble salt, an equilibrium will exist between the excess solid and the small concentration of dissolved ions.
For example, Barium Sulfate; a mostly insoluble salt
BaSO4(s)
Ba2+(aq) + SO42(aq)

92. Solubility Products *BaSO4(s) Ba2+(aq) + SO42(aq)
The equilibrium constant expression for this equilibrium can be written:
Ksp = [Ba2+] [SO42]
where the equilibrium constant, Ksp, is called the solubility product constant.
*Recall that solids (and liquids) are left out of the expression, so the Ksp only includes the products.

93. Sample Exercise 17.9 Writing Solubility-Product (Ksp) Expressions
Write the expression for the solubility-product constant for CaF2,
Appendix D gives 3.9  1011 for this Ksp.
Solution
Analyze We are asked to write an equilibrium-constant expression for the process by which CaF2 dissolves in water.
Plan We apply the general rules for writing an equilibrium-constant expression, excluding the solid reactant from the expression. We assume that the compound dissociates completely into its component ions:
Solve The expression for Ksp is
Ksp = [Ca2+][F–]2 93

94. Practice Exercise
Give the solubility-product-constant expressions and Ksp values (from Appendix D) for (a) barium carbonate,
(b) silver sulfate.
Answers:
(a) Ksp = [Ba2+][CO32–] = 5.0  109,
(b) Ksp = [Ag+]2[SO42–] = 1.5  105

95. © 2012 Pearson Education, Inc.
Solubility Products
Ksp is not the same as solubility.
Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).
© 2012 Pearson Education, Inc.

96. Sample Exercise 17.10 Calculating Ksp from Solubility
Solid silver chromate is added to pure water at 25 C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag2CrO4(s) and the solution.
Analysis of the equilibrated solution shows that its silver ion concentration is 1.3  104 M. Assuming that Ag2CrO4 dissociates completely in water and that there are no other important equilibria involving Ag+ or CrO42– ions in the solution, calculate Ksp for this compound.

97. Solution Analyze We are given the equilibrium concentration of Ag+ in a saturated solution of Ag2CrO4 and asked to determine the value of Ksp for Ag2CrO4.
Plan The equilibrium equation and the expression for Ksp are
To calculate Ksp, we need the equilibrium concentrations of Ag+ and CrO42– .We know that at equilibrium [Ag+] = 1.3  104 M. All the Ag+ and CrO42– ions in the solution come from the Ag2CrO4 that dissolves. Thus, we can use [Ag+] to calculate [CrO42–].

98. Ksp = [Ag+]2[CrO42–] = (1.3  104)2(6.5  105) = 1.1  1012
Solve From the chemical formula of silver chromate, we know that there must be 2 Ag+ ions in solution for each CrO42– ion in solution. Consequently, the concentration of CrO42– is half the concentration of Ag+:
and Ksp is
Ksp = [Ag+]2[CrO42–] = (1.3  104)2(6.5  105) = 1.1  1012
Check We obtain a small value, as expected for a slightly soluble salt. Furthermore, the calculated value agrees well with the one given in Appendix D, 1.2  1012 98

99. Practice Exercise
A saturated solution of Mg(OH)2 in contact with undissolved Mg(OH)2(s) is prepared at 25 C. The pH of the solution is found to be Assuming that Mg(OH)2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg2+ or OH– ions, calculate Ksp for this compound.
Answer:
1.6  1012

100. Sample Exercise 17.11 Calculating Solubility from Ksp
The Ksp for CaF2 is 3.9  1011 at 25 C . Assuming that CaF2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF2 in grams per liter.

101. Solution Analyze We are given Ksp for CaF2 and asked to determine solubility. Recall that the solubility of a substance is the quantity that can dissolve in solvent, whereas the solubility-product constant, Ksp, is an equilibrium constant.
Plan To go from Ksp to solubility, we follow the steps indicated by the red arrows in Figure We first write the chemical equation for the dissolution and set up a table of initial and equilibrium concentrations. We then use the equilibrium-constant expression. In this case we know Ksp, and so we solve for the concentrations of the ions in solution. Once we know these concentrations, we use the formula weight to determine solubility in g/L.

102. That makes the concentration of Ca2+ ions be X, and F- 2X
Solve Assume that initially no salt has dissolved, and then allow x mol L of CaF2 to dissociate completely when equilibrium is achieved:
That makes the concentration of Ca2+ ions be X,
and F- 2X
102

103. The stoichiometry of the equilibrium dictates that 2x mol L- of F– are produced for each x mol L- of CaF2 that dissolve. We now use the expression for Ksp and substitute, and simplify the equilibrium concentrations to solve for the value of x:

104. (Remember that To calculate the cube root of a number, use the yx function on your calculator, with ) Thus, the molar solubility of CaF2 is 2.1  104 mol L.
The mass of CaF2 that dissolves in water to form 1 L of solution is

105. Check We expect a small number for the solubility of a slightly soluble salt.
If we reverse the calculation, we should be able to recalculate the solubility product:
Ksp = (2.1  104)(4.2  104)2 = 3.7  1011
close to the value given in the problem statement, 3.9  1011.
Comment Because F– is the anion of a weak acid, you might expect hydrolysis of the ion to affect the solubility of CaF2. The basicity of F– is so small ( Kb = 1.5  1011), however, that the hydrolysis occurs to only a slight extent and does not significantly influence the solubility. The reported solubility is g/L at 25 C, in good agreement with our calculation.
Practice Exercise
The Ksp for LaF3 is 2  1019. What is the solubility of LaF3 in water in moles per liter?
Answer:
9  106 mol/L
105

106. 17.5 Factors Affecting Solubility
The Common-Ion Effect
If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease:
BaSO4(s)
Ba2+(aq) + SO42(aq)
Example: adding sulfuric acid H2SO4 (aq) to a solution of Barium sulfate, will increase the concentration of SO4 2- ions. The equilibrium above shifts left to remove the excess sulfate and BaSO4 precipitates out.

107. Notice: as concentrated acid H2 SO4 is added, white BaSO4 precipitates
BaSO4 (s)  Ba2+ + SO42-

108. Calculate the molar solubility of CaF2 at 25 C in a solution that is
Sample Exercise Calculating the Effect of a Common Ion on Solubility
Calculate the molar solubility of CaF2 at 25 C in a solution that is
(a) M in Ca(NO3)2, (b) M in NaF.
Solution Analyze We are asked to determine the solubility of CaF2 in the presence of two strong electrolytes, each containing an ion common to CaF2.
In (a) the common ion is Ca2+, and NO3– is a spectator ion.
In (b) the common ion is F–, and Na+ is a spectator ion.
108

109. Plan Because the slightly soluble compound is CaF2, we need to use Ksp for this compound, which Appendix D gives as 3.9  1011.
The value of Ksp is unchanged by the presence of additional solutes. Because of the common-ion effect, however, the solubility of the salt decreases in the presence of common ions. We use our standard equilibrium techniques of starting with the equation for CaF2 dissolution, setting up a table of initial and equilibrium concentrations, and using the Ksp expression to determine the concentration of the ion that comes only from CaF2.

110. Substituting into the solubility-product expression gives
Solve
(a) The initial concentration of Ca2+ is M because of the dissolved Ca(NO3)2:
Substituting into the solubility-product expression gives
Ksp =  1011 = [Ca2+][F–]2
= ( x)(2x)2

111. This would be a messy problem to solve exactly, but fortunately it is possible to simplify matters. Even without the common-ion effect, the solubility of CaF2 is very small (2.1  104 M). Thus, we assume that the M concentration of Ca2+ from Ca(NO3)2 is very much greater than the small additional concentration resulting from the solubility of CaF2; that is, x is much smaller than M, and
We then have

112. (b) The common ion is F–, and at equilibrium we have
Assuming that 2x is much smaller than M
(that is, ),
we have
Thus, 3.9  107 mol of solid CaF2 should dissolve per liter of M NaF solution.
[Ca2+] = x and [F–] = x
112

113. Comment The molar solubility of CaF2 in water is 2
Comment The molar solubility of CaF2 in water is 2.1  104 M (Sample Exercise 17.11). By comparison, our calculations here give a CaF2 solubility of 3.1  105M in the presence of M Ca2+ and 3.9  107 M in the presence of M F– ion. Thus, the addition of either Ca2+ or F– to a solution of CaF2 decreases the solubility.
However, the effect of F– on the solubility is more pronounced than that of Ca2+ because [F–] appears to the second power in the Ksp expression for CaF2, whereas [Ca2+] appears to the first power.

114. Sample Exercise 17.12 Calculating the Effect of a Common Ion on Solubility
Continued
Practice Exercise
For manganese(II) hydroxide,Mn(OH)2, Ksp = 1.6  1013. Calculate the molar solubility of Mn(OH)2 in a solution that contains M NaOH.
Answer:
4.0  1010 M
114

115. Factors Affecting SolubilitypH
If a substance has a basic anion, it will be more soluble in an acidic solution.
Substances with acidic cations are more soluble in basic solutions.
© 2012 Pearson Education, Inc.

116. Sample Exercise 17.13 Predicting the Effect of Acid on Solubility
Which of these substances are more soluble in acidic solution than in basic solution:
(a) Ni(OH)2(s), (b) CaCO3(s), (c) BaF2(s), (d) AgCl(s)?
Solution
Analyze The problem lists four sparingly soluble salts, and we are asked to determine which are more soluble at low pH than at high pH.
Plan Ionic compounds that dissociate to produce a basic anion are more soluble in acid solution.
Solve
(a) Ni(OH)2(s) is more soluble in acidic solution because of the basicity of OH–; the H+ reacts with the OH– ion, forming water:
116

117. What would be observed as the CaCO3 dissolves?
Solution
(b) CaCO3(s) ; Similarly, CaCO3(s) dissolves in acid solutions because CO32– is a basic anion:
The reaction between CO32– and H+ occurs in steps, with HCO3– forming first and H2CO3 forming in appreciable amounts only when [H+] is sufficiently high.
What would be observed as the CaCO3 dissolves?
Bubbles! …of CO2

118. Sample Exercise 17.13 Predicting the Effect of Acid on Solubility
Continued
(c) BaF2(s)?, The solubility of BaF2 is enhanced by lowering the pH because F– is a basic anion:
(d) AgCl(s)? The solubility of AgCl is unaffected by changes in pH because Cl– is the anion of a strong acid and therefore has negligible basicity.
118

119. Practice Exercise
Write the net ionic equation for the reaction between an acid and (a) CuS, (b) Cu(N3)2.
Answers: (a)
(b)

120. Factors Affecting Solubility
Complex Ions
Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.
© 2012 Pearson Education, Inc.

121. Factors Affecting Solubility
Complex Ions
The formation of these complex ions increases the solubility of these salts.

122. Sample Exercise 17.14 Evaluating an Equilibrium Involving a Complex Ion
Calculate the concentration of Ag+ present in solution at equilibrium when concentrated ammonia is added to a M solution of AgNO3 to give an equilibrium concentration of [NH3] = 0.20 M. Neglect the small volume change that occurs when NH3 is added.
Solution
Analyze Addition of NH3(aq) to Ag+ (aq) forms Ag(NH3)2+, as shown in Equation We are asked to determine what concentration of Ag+(aq) remains uncombined when the NH3 concentration is brought to 0.20 M in a solution originally M in AgNO3.
122

123. Plan We assume that the AgNO3 is completely dissociated, giving 0
Plan We assume that the AgNO3 is completely dissociated, giving M Ag+. Because Kf for the formation of Ag(NH3)2+ is quite large, we assume that essentially all the Ag+ is converted to Ag(NH3)2+ and approach the problem as though we are concerned with the dissociation of Ag(NH3)2+ rather than its formation. To facilitate this approach, we need to reverse Equation and make the corresponding change to the equilibrium constant:

124. Solve If [Ag+] is 0. 010 M initially, [Ag(NH3)2+] will be 0
Solve If [Ag+] is M initially, [Ag(NH3)2+] will be M following addition of the NH3.We construct a table to solve this equilibrium problem. Note that the NH3 concentration given in the problem is an equilibrium concentration rather than an initial concentration.
Because [Ag+] is very small, we can ignore x, so that Substituting these values into the equilibrium-constant expression for the dissociation of Ag(NH3)2+, we obtain
Formation of the Ag(NH3)2+ complex drastically reduces the concentration of free Ag+ ion in solution.
124

125. Practice Exercise
Calculate [Cr3+] in equilibrium with Cr(OH)4 when mol of Cr(NO3)3 is dissolved in 1 L of solution buffered at pH 10.0.
Answer:
1  1016 M
125

126. Factors Affecting Solubility
Amphoterism
Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.
Examples of such cations are Al3+, Zn2+, and Sn2+.

127. 17.6 Will a Precipitate Form?
In a solution,
If Q = Ksp, the system is at equilibrium and the solution is saturated.
If Q < Ksp, more solid can dissolve until Q = Ksp.
If Q > Ksp, the salt will precipitate until Q = Ksp.
© 2012 Pearson Education, Inc.

128. Sample Exercise 17.15 Predicting Whether a Precipitate Forms
Does a precipitate form when 0.10 L of 8.0  103 M Pb(NO3)2 is added to 0.40 L of 5.0  103 M Na2SO4?
Solution
Analyze The problem asks us to determine whether a precipitate forms when two salt solutions are combined.
Plan We should determine the concentrations of all ions just after the solutions are mixed and compare the value of Q with Ksp for any potentially insoluble product. The possible metathesis products are PbSO4 and NaNO3. Like all sodium salts is soluble, but PbSO4 has a Ksp of 6.3  107 (Appendix D) and will precipitate if the Pb+2 and SO42 concentrations are high enough for Q to exceed Ksp.
128

129. Solve When the two solutions are mixed, the volume is 0. 10 L + 0
Solve When the two solutions are mixed, the volume is 0.10 L L = 0.50 L. The number of moles of Pb2+ in 0.10 L of 8.0  103 M Pb(NO3)2 is
The concentration of Pb2+ in the 0.50-L mixture is therefore
The number of moles of SO42 in 0.40 L of 5.0  103 M Na2SO4 is
Therefore
and
Q = [Pb2+][SO42] = (1.6  103)(4.0  103) = 6.4  106
Because Q > Ksp (6.3  107) , PbSO4 precipitates.

130. Sample Exercise 17.15 Predicting Whether a Precipitate Forms
Continued
Practice Exercise
Does a precipitate form when L of 2.0  102 M NaF is mixed with L of 1.0  102 M Ca(NO3)2?
Answer:
Yes, CaF2 precipitates because Q = 4.6  108 is larger than Ksp = 3.9  10 11
130

131. Sample Exercise 17.16 Calculating Ion Concentrations for Precipitation
A solution contains 1.0  102 M Ag+ and 2.0  102 M Pb2+. When Cl is added, both AgCl (Ksp = 1.8  1010) and PbCl2 (Ksp = 1.7  105 M) can precipitate. What concentration of Cl is necessary to begin the precipitation of each salt? Which salt precipitates first?
Solution
Plan We are given Ksp values for the two precipitates. Using these and the metal ion concentrations, we can calculate what Cl concentration is necessary to precipitate each salt. The salt requiring the lower Cl ion concentration precipitates first.
Solve For AgCl we have Ksp = [Ag+][Cl] = 1.8  1010
Because [Ag+] = 1.0  102 M, the greatest concentration of Cl that can be present without causing precipitation of AgCl can be calculated from the Ksp expression:
Any Cl in excess of this very small concentration will cause AgCl to precipitate from solution.
131

132. Proceeding similarly for PbCl2, we have
Thus, a concentration of Cl in excess of causes PbCl2 to precipitate. Comparing the concentration required to precipitate each salt, we see that as Cl is added, AgCl precipitates first because it requires a much smaller concentration of Cl. Thus, Ag+ can be separated from by Pb2+ slowly adding Cl so that the chloride ion concentration remains between 1.8  1010 M and 2.9  102 M.
Comment Precipitation of AgCl will keep the Cl concentration low until the number of moles of Cl added exceeds the number of moles of Ag+ in the solution. Once past this point, [Cl] rises sharply and PbCl2 will soon begin to precipitate.
132

133. Sample Exercise 17.16 Calculating Ion Concentrations for Precipitation
Continued
Practice Exercise
A solution consists of M Mg2+ and Cu2+. Which ion precipitates first as OH is added? What concentration of OH is necessary to begin the precipitation of each cation? [Ksp = 1.8  1011 for Mg(OH)2, and Ksp = 4.8  1020 for Cu(OH)2.]
Answer:
Cu(OH)2 precipitates first, beginning when [OH] > 9.8  1010 M.
Mg(OH)2 begins to precipitate when [OH] > 1.9  105 M.
133

134. 17.7 Selective Precipitation of Ions
One can use differences in solubilities of salts to separate ions in a mixture.
© 2012 Pearson Education, Inc.