1. Acid-Base Equilibria and Solubility Equilibria
Chapter 16
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2. CH3COONa (s) Na+ (aq) + CH3COO- (aq)
The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.
The presence of a common ion suppresses the ionization of a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).
CH3COONa (s) Na+ (aq) + CH3COO- (aq)
common ion
CH3COOH (aq) H+ (aq) + CH3COO- (aq)
16.2

3. Henderson-Hasselbalch equation
Consider mixture of salt NaA and weak acid HA.
NaA (s) Na+ (aq) + A- (aq)
Ka =
[H+][A-]
[HA]
HA (aq) H+ (aq) + A- (aq)
[H+] =
Ka [HA]
[A-]
Henderson-Hasselbalch equation
-log [H+] = -log Ka - log
[HA]
[A-]
pH = pKa + log
[conjugate base]
[acid]
-log [H+] = -log Ka + log
[A-]
[HA]
pH = pKa + log
[A-]
[HA]
pKa = -log Ka
16.2

4. What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?
Mixture of weak acid and conjugate base!
HCOOH (aq) H+ (aq) + HCOO- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.30
0.00
0.52 -x +x +x x x x
pH = pKa + log
[HCOO-]
[HCOOH]
Common ion effect
0.30 – x  0.30
pH = log
[0.52]
[0.30]
= 4.01
x  0.52
HCOOH pKa = 3.77
16.2

5. What is the pH of a buffer made by mixing 1. 00 L of 0
What is the pH of a buffer made by mixing 1.00 L of M benzoic acid, HC7H5O2, with 3.00 L of M sodium benzoate, NaC7H5O2? Ka for benzoic acid is 6.3 × 10-5.
This problem involves dilution first. Once we know the concentrations of benzoic acid and benzoate ion, we can use the acid equilibrium to solve for x.
We will use HBz to represent benzoic acid and Bz- to represent benzoate ion.

6. HBz(aq) +
H2O(l)
H3O+(aq) +
Bz-(aq)
Initial
0.0050
0.045
Change –x +x
Equilibrium
– x x x

9. Calculate the pH of a 0.10 M HF solution to which sufficient sodium fluoride has been added to make the concentration 0.20 M NaF. Ka for HF is 6.8 × 10-4.
We will use the acid equilibrium for HF.
NaF provides the conjugate base, F-, so [F-] = 0.20 M.

10. HF(aq) +
H2O(l)
H3O+(aq) +
F-(aq)
Initial
0.10
0.20
Change –x +x
Equilibrium
0.10 – x x x

12. A buffer solution is a solution of: A weak acid or a weak base and
The salt of the weak acid or weak base
Both must be present!
A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.
Consider an equal molar mixture of CH3COOH and CH3COONa
Add strong acid
H+ (aq) + CH3COO- (aq) CH3COOH (aq)
Add strong base
OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)
16.3

13. Henderson–Hasselbalch Equation
Buffers at a specific pH can be prepared using the Henderson-Hasselbalch Equation.
Buffers are prepared from a conjugate acid-base pair in which the ionization is approximately equal to the desired H3O+ concentration.
Consider a buffer made up of a weak acid HA and its conjugate base A-
The acid-ionization constant is 13

14. The preceding equation can be used to derive an equation for the pH of the buffer.
Take the negative logarithm of both sides of the equation.
The pKa of a weak acid is defined in a manner similar to pH and pOH
pKa = - log Ka

15. The equation can then be expressed as:
This equation is generally shown as:
This equation relates buffer pH for different concentrations of conjugate acid and base and is known as the Henderson-Hasselbalch equation.

16. Question: What is the [H3O+] for a buffer solution that is 0
Question: What is the [H3O+] for a buffer solution that is M in acid and M in the corresponding salt if the weak acid Ka = 5.80 x 10-7?
Use as example the following equation of a conjugate acid-base pair
pKa = - log(5.80 x 10-7) = -(-6.24) = 6.24
[H3O+] = pKa + log [A-]/[HA] = log 0.600/0.250
[H3O+] = = 2.40 x 10-7
[H3O+] = 2.40 x 10-7

17. HCl + CH3COO- CH3COOH + Cl-
HCl H+ + Cl-
HCl + CH3COO CH3COOH + Cl-
16.3

18. Which of the following are buffer systems? (a) KF/HF
(b) KBr/HBr, (c) Na2CO3/NaHCO3
(a) KF is a weak acid and F- is its conjugate base
buffer solution
(b) HBr is a strong acid
not a buffer solution
(c) CO32- is a weak base and HCO3- is its conjugate acid
buffer solution
16.3

19. NH4+ (aq) H+ (aq) + NH3 (aq)
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution?
NH4+ (aq) H+ (aq) + NH3 (aq)
pH = pKa + log
[NH3]
[NH4+]
pH = log
[0.30]
[0.36]
pKa = 9.25
= 9.17
start (moles)
0.029
0.001
0.024
NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)
end (moles)
0.028
0.0
0.025
final volume = 80.0 mL mL = 100 mL
[NH4+] =
0.028
0.10
[NH3] =
0.025
0.10
pH = log
[0.25]
[0.28]
= 9.20
16.3

20. Chemistry In Action: Maintaining the pH of Blood
16.3

21. Equivalence point – the point at which the reaction is complete
Titrations
In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
The indicator
changes color
(pink)
4.7

22. An acid–base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid).
Such curves are used to gain insight into the titration process. You can use the titration curve to choose an indicator that will show when the titration is complete.
I am not including the topic of adding an acid or base to a buffer solution.

23. This titration plot shows the titration of a strong acid with a strong base. Note that the pH at the equivalence point is 7.0.

24. The equivalence point is the point in a titration when a stoichiometric amount of reactant has been added.
The indicator must change color near the pH at the equivalence point.

25. When solving problems in which an acid reacts with a base, we first need to determine the limiting reactant. This requires that we use moles in the calculation.
The next step is to review what remains at the end of the reaction. In the case of a strong acid and a strong base, the solution is neutral.
When a weak acid or weak base is involved, the product is a salt. After finding the salt concentration, we use a hydrolysis equilibrium to find the pH.

26. Calculate the pOH and the pH of a solution in which 10. 0 mL of 0
Calculate the pOH and the pH of a solution in which 10.0 mL of M HCl is added to 25.0 mL of M NaOH.
First we calculate the moles of HCl and NaOH.
Then we determine what remains after the reaction.

27. The overall reaction is
HCl + NaOH  H2O + NaCl
Moles HCl = (0.100 M)(10.0 × 10-3 L)
= 1.00 × 10-3 mol
Moles NaOH = (0.100 M)(25.0 × 10-3 L)
= 2.50 × 10-3 mol
All of the HCl reacts, leaving 1.50 × 10-3 mol NaOH.
The new volume is 10.0 mL mL = 35.0 mL.

29. Calculate the [OH] and the pH at the equivalence point for the titration of 500. mL of 0.10 M propionic acid with M calcium hydroxide. (This can be used to prepare a preservative for bread.)
Ka = 1.3  105.
First we need to find the moles of propionic acid to find the volume of calcium hydroxide. Then we will examine the salt that is produced and use the hydrolysis equilibrium to find the pH.
We will use HPr to represent propionic acid and Pr- to represent propionate ion.

30. 2HPr + Ca(OH)2  Ca(Pr)2 + 2H2O
Moles propionic acid = (0.10 M)(500. × 10-3 L)
= mol HPr
To find the moles of calcium hydroxide, we need the balanced chemical equation:
2HPr + Ca(OH)2  Ca(Pr)2 + 2H2O
Moles calcium hydroxide =

31. Now we find the volume of Ca(OH)2 required to give 0.025 mol:
The new volume is 500. × 10-3 L L = 1.00 L
The concentration of Pr- is 2(0.025 M) = M

32. The hydrolysis reaction is Pr- + H2O  HPr + OH-
Pr-(aq) +
H2O(l)
HPr(aq) +
OH-(aq)
Initial
0.050
Change –x +x
Equilibrium
0.050 – x x

34. Strong Acid-Strong Base Titrations
NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)
OH- (aq) + H+ (aq) H2O (l)
16.4

35. Weak Acid-Strong Base Titrations
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)
CH3COOH (aq) + OH- (aq) CH3COO- (aq) + H2O (l)
At equivalence point (pH > 7):
CH3COO- (aq) + H2O (l) OH- (aq) + CH3COOH (aq)
16.4

36. Strong Acid-Weak Base Titrations
HCl (aq) + NH3 (aq) NH4Cl (aq)
H+ (aq) + NH3 (aq) NH4Cl (aq)
At equivalence point (pH < 7):
NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)
16.4

37. Exactly 100 mL of 0. 10 M HNO2 are titrated with a 0
Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ?
start (moles)
0.01
0.01
HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l)
end (moles)
0.0
0.0
0.01
[NO2-] =
0.01
0.200
= 0.05 M
Final volume = 200 mL
NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.05
0.00
0.00 -x +x +x x x x
Kb =
[OH-][HNO2]
[NO2-] = x2
0.05-x
pOH = 5.98
= 2.2 x 10-11
pH = 14 – pOH = 8.02
0.05 – x  0.05
x  1.05 x 10-6
= [OH-]

38. Acid-Base Indicators HIn (aq) H+ (aq) + In- (aq) [HIn]  10
Color of acid (HIn) predominates
 10
[HIn]
[In-]
Color of conjugate base (In-) predominates
16.5

39. pH
16.5

40. The titration curve of a strong acid with a strong base.
16.5

41. Which indicator(s) would you use for a titration of HNO2 with KOH ?
Weak acid titrated with strong base.
At equivalence point, will have conjugate base of weak acid.
At equivalence point, pH > 7
Use cresol red or phenolphthalein
16.5

42. To deal quantitatively with an equilibrium, you must know the equilibrium constant.
We will look at the equilibria of slightly soluble (or nearly insoluble) ionic compounds and show how you can determine their equilibrium constants.
Once you find these values for various ionic compounds, you can use them to answer questions about solubility or precipitation.

43. When an ionic compound is insoluble or slightly soluble, an equilibrium is established:
MX(s) M+(aq) + X-(aq)
The equilibrium constant for this type of reaction is called the solubility-product constant, Ksp.
For the above reaction,
Ksp = [M+][X-]

44. Solubility Equilibria
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-]
Ksp is the solubility product constant
MgF2 (s) Mg2+ (aq) + 2F- (aq)
Ksp = [Mg2+][F-]2
Ag2CO3 (s) Ag+ (aq) + CO32- (aq)
Ksp = [Ag+]2[CO32-]
Ca3(PO4)2 (s) Ca2+ (aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO43-]2
Dissolution of an ionic solid in aqueous solution:
Q < Ksp
Unsaturated solution
No precipitate
Q = Ksp
Saturated solution
Q > Ksp
Supersaturated solution
Precipitate will form
16.6

45. 16.6

46. Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution.
Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution.
16.6

47.  What is the solubility of silver chloride in g/L ?
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = 1.6 x 10-10
Initial (M)
Change (M)
Equilibrium (M)
0.00
0.00
Ksp = [Ag+][Cl-] +s +s
Ksp = s2
s = Ksp  s s
s = 1.3 x 10-5
[Ag+] = 1.3 x 10-5 M
[Cl-] = 1.3 x 10-5 M
1.3 x 10-5 mol AgCl
1 L soln
g AgCl
1 mol AgCl x
Solubility of AgCl =
= 1.9 x 10-3 g/L
16.6

48. AgBr(s) Ag+(aq) + Br-(aq)
Exactly mg of AgBr will dissolve in 1.00 L of water. What is the value of Ksp for AgBr?
Solubility equilibrium:
AgBr(s) Ag+(aq) + Br-(aq)
Solubility-product constant expression:
Ksp = [Ag+][Br-]

49. The solubility is given as 0.133 mg/1.00 L, but Ksp uses molarity:
AgBr(s)
Ag+(aq) +
Br-(aq)
Initial
Change +x
Equilibrium x

50. [Ag+] = [Cl-] = x = × 10-7 M
Ksp = (7.083 × 10-7)2
Ksp = 5.02 × 10-13

51. BaF2(s) Ba2+(aq) + 2F-(aq)
An experimenter finds that the solubility of barium fluoride is 1.1 g in 1.00 L of water at 25°C. What is the value of Ksp for barium fluoride, BaF2, at this temperature?
Solubility equilibrium:
BaF2(s) Ba2+(aq) + 2F-(aq)
Solubility-product constant expression:
Ksp = [Ba2+][F-]2

52. The solubility is given as 1.1 g/1.00 L, but Ksp uses molarity:
BaF2(s)
Ba2+(aq) +
2F-(aq)
Initial
Change +x
+2x
Equilibrium x 2x

53. [Ba2+] = x = 6.27 × 10-3 M
[F-] = 2x = 2(6.27 × 10-3) = 1.25 × 10-2 M
Ksp = (6.27 × 10-3)(1.25 × 10-2)2
Ksp = 9.8 × 10-7

54. When Ksp is known, we can find the molar solubility.

55. Hg2Cl2(s) Hg22+(aq) + 2Cl-(aq)
Calomel, whose chemical name is mercury(I) chloride, Hg2Cl2, was once used in medicine (as a laxative and diuretic). It has a Ksp equal to 1.3  1018. What is the solubility of Hg2Cl2 in grams per liter?
Solubility equilibrium:
Hg2Cl2(s) Hg22+(aq) + 2Cl-(aq)
Solubility-product constant expression:
Ksp = [Hg22+][Cl-]2

56. Ksp = x(2x)2 Ksp = x(4x2) Ksp = 4x3 1.3 × 10-18 = 4x3
Hg2Cl2(s)
Hg22+(aq) +
2Cl-(aq)
Initial
Change +x
+2x
Equilibrium x 2x
Ksp = x(2x)2
Ksp = x(4x2)
Ksp = 4x3
1.3 × = 4x3
x3 = 3.25 × 10-19
x = 6.88 × 10-7 M

57. The molar solubility is 6
The molar solubility is 6.9 × 10-7 M, but we also need the solubility in g/L:

58. 16.6

59. The ions present in solution are Na+, OH-, Ca2+, Cl-.
If 2.00 mL of M NaOH are added to 1.00 L of M CaCl2, will a precipitate form?
The ions present in solution are Na+, OH-, Ca2+, Cl-.
Only possible precipitate is Ca(OH)2 (solubility rules).
Is Q > Ksp for Ca(OH)2?
[Ca2+]0 = M
[OH-]0 = 4.0 x 10-4 M
Q = [Ca2+]0[OH-]0 2
= 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8
Ksp = [Ca2+][OH-]2 = 8.0 x 10-6
Q < Ksp
No precipitate will form
16.6

60. We can use the reaction quotient, Q, to determine whether precipitation will occur.

61. One form of kidney stones is calcium phosphate, Ca3(PO4)2, which has a Ksp of 1.0  1026. A sample of urine contains 1.0  103 M Ca2+ and 1.0  108 M PO43 ion.
Calculate Qc and predict whether Ca3(PO4)2 will precipitate.

62. Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+]3 [PO43-]2
Qc = (1.0 × 10-3)3 (1.0 × 10-8)2
Qc = 1.0 × 10-25
Ksp < Qc
A precipitate will form.

63. When a problem gives the amounts and concentrations of two samples that are then mixed, the first step in solving the problem is to calculate the new initial concentrations.

64. Exactly 0. 400 L of 0. 50 M Pb2+ and 1. 60 L of 2
Exactly L of 0.50 M Pb2+ and 1.60 L of 2.50  102 M Cl are mixed together to form 2.00 L of solution.
Calculate Qc and predict whether PbCl2 will precipitate. Ksp for PbCl2 is 1.6  105.

65. PbCl2(s) Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+] [Cl-]2 = 1.6 × 10-8
Ksp < Qc
A precipitate will form.
This slide is missing the Houghton Mifflin copyright line. Please add it.

66. AgBr (s) Ag+ (aq) + Br- (aq) Ksp = 7.7 x 10-13 Ksp = [Ag+][Br-]
What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br- and Cl- at a concentration of 0.02 M?
AgBr (s) Ag+ (aq) + Br- (aq)
Ksp = 7.7 x 10-13
Ksp = [Ag+][Br-]
[Ag+] =
Ksp
[Br-]
7.7 x 10-13
0.020 =
= 3.9 x M
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = 1.6 x 10-10
Ksp = [Ag+][Cl-]
[Ag+] =
Ksp
[Cl-]
1.6 x 10-10
0.020 =
= 8.0 x 10-9 M
3.9 x M < [Ag+] < 8.0 x 10-9 M
16.7

67. The Common Ion Effect and Solubility
The presence of a common ion decreases the solubility of the salt.
What is the molar solubility of AgBr in (a) pure water and (b) M NaBr?
NaBr (s) Na+ (aq) + Br- (aq)
AgBr (s) Ag+ (aq) + Br- (aq)
[Br-] = M
Ksp = 7.7 x 10-13
AgBr (s) Ag+ (aq) + Br- (aq)
s2 = Ksp
[Ag+] = s
s = 8.8 x 10-7
[Br-] = s 
Ksp = x s
s = 7.7 x 10-10
16.8

68. pH and Solubility Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)
The presence of a common ion decreases the solubility.
Insoluble bases dissolve in acidic solutions
Insoluble acids dissolve in basic solutions
add
remove
Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)
At pH less than 10.45
Ksp = [Mg2+][OH-]2 = 1.2 x 10-11
Lower [OH-]
Ksp = (s)(2s)2 = 4s3
OH- (aq) + H+ (aq) H2O (l)
4s3 = 1.2 x 10-11
Increase solubility of Mg(OH)2
s = 1.4 x 10-4 M
[OH-] = 2s = 2.8 x 10-4 M
At pH greater than 10.45
pOH = pH = 10.45
Raise [OH-]
Decrease solubility of Mg(OH)2
16.9

69. Complex Ion Equilibria and Solubility
A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions.
Co2+ (aq) + 4Cl- (aq) CoCl4 (aq) 2-
The formation constant or stability constant (Kf) is the equilibrium constant for the complex ion formation.
Kf =
[CoCl4 ]
[Co2+][Cl-]4 2-
Co(H2O)6 2+
CoCl4 2-
stability of complex Kf
16.10

70. Metal ions that form complex ions include Ag+, Cd2+, Cu2+, Fe2+, Fe3+, Ni2+, and Zn2+.
Complexing agents, called ligands, are Lewis bases. They include CN-, NH3, S2O32-, and OH-.
In each case, an equilibrium is established, called the complex-ion formation equilibrium.

71. Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
Zn2+(aq) + 4OH-(aq) Zn(OH)42-(aq)

72. 16.10

73. 16.11

74. Qualitative Analysis of Cations
16.11

75. Flame Test for Cations
lithium
sodium
potassium
copper
16.11

76. Chemistry In Action: How an Eggshell is Formed
Ca2+ (aq) + CO32- (aq) CaCO3 (s)
CO2 (g) + H2O (l) H2CO3 (aq)
carbonic
anhydrase
H2CO3 (aq) H+ (aq) + HCO3- (aq)
HCO3- (aq) H+ (aq) + CO32- (aq)