1. Chapter 16 Acid-Base Equilibria in Aqueous Solutions

2. Autoionization of Water
Self-ionization – when a substance reacts with itself to form ions
In pure water, a few molecules act as bases and a few act as acids.
H2O () H+ (aq) OH- (aq)
H2O () + H2O () H3O+ (aq) OH- (aq)
Water is amphoteric. It can act as an acid or as a base.

3. Ion-Product Constant for Water
H2O () H+ (aq) OH- (aq)
Kw = [H+] [OH-]
Kw is called the ion-product constant for water
The value of for the Kw of water is only at 25oC.
At other temperatures there are different values.
Remember – the concentrations of liquids such as water do not appear in the equilibrium-constant expression.
Kw is just another equilibrium constant
Kw = [H+] [OH-] = 1.0 x (at 25 C)

4. H2O () H+ (aq) + OH- (aq) In pure DI water at 25 C
[OH-] = 1.0 x M
[H+] = 1.0 x M
and
Since water dissociates into equal amounts of H+ and OH-, one can determine their concentrations each to be 10-7M.
DI water refers to deionized water – water from which the ions of dissolved salts have been removed.

5. Kw = [H+] [OH-] = 1.0 x 10-14 (at 25 C)
when [H+] = [OH-]
solution is neutral
when [H+] > [OH-]
solution is acidic
when [H+] < [OH-]
solution is basic
Kw = [H+] [OH-] = 1.0 x (at 25 C)
H2O auto-ionization occurs in all solutions.
When other ions are present
[H+] is usually NOT equal to [OH–]
But Kw = [H+][OH–] = 1.0 × 10–14
Kw expression is valid for ALL aqueous solutions (acidic, basic or neutral) at 25 C

6. Learning Check
In a sample of blood at 25 °C, [H+] = 4.6  10–8 M. Find [OH–] and determine if the solution is acidic, basic or neutral.
So 2.2 × 10–7 M > 4.6 × 10–8 M
[OH–] > [H3O+] so the solution is basic
Note that we compare the OH- concentration with the H+ concentration. Whichever is greater determines whether the solution is acidic or basic.

7. The pH Scale
Søren Sørensen 1868 – 1939
pH = -log [H+]

8. For Any Aqueous Solution at 25 C
When pH = 7 the solution is neutral
When pH < 7 the solution is acidic
When pH > 7 the solution is basic
pH scale range is from

9. Dr. Pepper Cherry (it’s amazingly smooth) has [H+] = 1. 2 x 10-3 M
Dr. Pepper Cherry (it’s amazingly smooth) has [H+] = 1.2 x 10-3 M. What is the pH of this pop?

10. Lemon juice has a pH = 2.41 Stomach juice has a pH = 2.09
vs.
pH = 2.09
pH = 2.41
Remember that the log scale is not a linear scale. That is, if something is twice as acidic as something else, the ratio of pH’s will not necessarily be two.
Which one is more acidic? Calculate, specifically by how much more acidic one is (vs. the other).

11. pH 4.00 solution has 10 times less H+ than pH 3.00 solution
Learning Check
What are [H+] and [OH–] of pH = 3.00 solution?
[H+] = 10–3.00 = 1.0  10–3 M
[OH–] = = 1.0  10–11 M
What are [H+] and [OH–] of pH = 4.00 solution?
pH = 4.00 [H+] = 1.0  10–4 M
[OH–] = = 1.0  10–10 M
pH 4.00 solution has 10 times less H+ than pH 3.00 solution
Note on significant figures:
The number of decimal places in the log term = number of significant digits in antilog term.

12. Can the pH of a solution be less than 0 or greater than 14?

13. pH

14. Blood has a [OH-] = 2.69 x 10-7 M at 25 C. What is the pH of blood ?

15. There is another scale analogous to pH
pOH = -log [OH-]
pH + pOH = (at 25 C)
Above expression is valid for ALL aqueous solutions (acidic, basic or neutral) at 25 C
There are other “p” scales –
The “p” in pH tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions).
Some similar examples are
pOH: log [OH]
pKw: log Kw
Because
[H3O+] [OH] = Kw = 1.0  1014
we know that
log [H3O+] + log [OH] = log Kw = 14.00
or, in other words,
pH + pOH = pKw = 14.00
Blood has a [OH-] = 2.69 x 10-7 M at 25 C. What is the pH of blood ?

16. Learning Check
Kw increases with increasing temperature. At 50 °C, Kw = ×10–14. What is the pH of a neutral solution at 50 °C ? A B C D
[H+] = [OH-]
Kw=[H+]x[OH-]
[H+]=Kw1/2
[H+]=(5.476x10-14)1/2
[H+]=2.34x10-7
pH=-log(2.34x10-7)

17. Calculations for strong acids/bases
Determine the pH of a 0.15 M HBr (aq) soln
Strong acids and bases are fully dissociated. Thus [H+] concentration = [HBr] concentration

18. Learning Check
What is the pH of 0.15 M HBr? HBr is a strong acid and dissociates completely, HBr(aq) + H2O  H+(aq) + Br–(aq) Int N/A 0 0 Final 0 N/A
pH = –log(0.15) = 0.82

19. 14. 2 g Ba(OH)2 (s) is dissolved in enough H2O to form 500. 0 mL soln
14.2 g Ba(OH)2 (s) is dissolved in enough H2O to form mL soln. What’s the resulting pH of the solution ?
How many hydroxyl groups does Barium hydroxide have? Is it fully dissociated?

20. Learning Check
What is the pH of M Ba(OH)2? A strong base and dissociates completely Ba(OH)2 (aq)  Ba2+ (aq) + 2 OH– (aq) Int Final
pOH = –log(0.166) = 0.780
pH = – pOH = – = 13.22

21. Calculations of Strong Acids and Bases
The auto-ionization of H2O will always add to [H+] and [OH–] of an acid or base. Does this have an effect on the last answer?
The previous problem had M [OH–] from the Ba(OH)2 but the [H+] must have come from water. If it came from water an equal amount of [OH–] comes from water and the total [OH–] is
[OH–]total = [OH–]from Ba(OH) [OH–]from H2O
[OH–]total = M × 10–14 M
= M (when properly rounded)
The autoionization of water is insignificant

22. Weak Acids CH3COOH (aq) CH3COO− (aq) + H+ (aq) or
H2O or
CH3COOH (aq) H+ (aq) CH3COO− (aq)
H2O
[ H+] [CH3COO−]
[CH3COOH]
Ka =
Technically water does have a concentration. It is 55.6 M, but it is constant so it is lumped in with Ka.
Ka is called the acid-dissociation constant
Ka is just an equilibrium constant that is specific for a weak acid

23. A 0. 050 M aqueous acetic acid solution has a pH = 3. 03
A M aqueous acetic acid solution has a pH = What is the Ka for acetic acid ?

24. Calculating Ka from the pH
The pH of a M solution of acetic acid, CH3COOH, at 25C is Calculate Ka for acetic acid at this temperature.
We know that
[H3O+] [HCOO]
[HCOOH]
Ka =

25. Calculating Ka from the pH
The pH of a M solution of acetic acid, CH3COOH, at 25C is Calculate Ka for acetic acid at this temperature.
To calculate Ka, we need the equilibrium concentrations of all three components.
We can find [H3O+], which is the same as [HCOO], from the pH.

26. Calculating Ka from the pH
pH = log [H3O+]
3.03 = log [H3O+]
3.03 = log [H3O+]
103.03 = 10log [H3O+] = [H3O+]
9.3  104 = [H3O+] = [HCOO]

27. Calculating Ka from pH Now we can set up an ICE table… [CH3COOH], M
[H3O+], M
[CH3COO], M
Initially
0.050
Change
At Equilibrium

28. Calculating Ka from pH Now we can set up a table… [CH3COOH], M
[H3O+], M
[CH3COO], M
Initially
0.050
Change
9.3  104
+9.3  104
At Equilibrium

29. Calculating Ka from pH Now we can set up a table… [CH3COOH], M
[H3O+], M
[CH3COO], M
Initially
0.050
Change
9.3  104
+9.3  104
At Equilibrium
0.050  9.3  104
= = 0.049
9.3  104

30. Calculating Ka from pH [9.3  104] [9.3  104] Ka = [0.049]
= 1.8  105

31. What is the pKa of acetic acid ?
pKa = -log Ka
What is the pKa of acetic acid ?
pKa = -log 1.8 x 10-5
= 4.7
What is the percentage of ionization of acetic acid ?

32. Calculating Percent Ionization
[H3O+]eq
[HA]initial
Percent ionization =  100
In this example,
[H3O+]eq = 9.3  104 M
[CH3COOH]initial = M
9.3  104
0.050
Percent ionization =  100
= 1.9%

33. The greater the value of Ka, the stronger is the acid.

34. What is the pH of a 0.60 M HNO2 (aq) ?

35. Calculating pH from Ka
Calculate the pH of a 0.60 M solution of nitrous acid, HNO2, at 25C.
HNO2(aq) + H2O(l) H3O+(aq) + NO2(aq)
Ka for nitrous acid at 25C is 4.5  104.

36. Calculating pH from Ka The equilibrium constant expression is
[H3O+] [NO2]
[HNO2]
Ka =

37. Calculating pH from Ka We next set up a table… [HNO2], M [H3O+], M
Initially
0.60
Change
At equilibrium

38. Calculating pH from Ka [HNO2], M [H3O+], M [NO2], M Initially 0.60
Change x +x
At equilibrium

39. Calculating pH from Ka
[HNO2], M
[H3O+], M
[NO2], M
Initially
0.60
Change x +x
At equilibrium
0.60  x  0.60 x
If we can ignore x for the equilibrium concentration of HNO2 we can avoid using the quadratic equation in solving for x and the math is much easier. We will run a check to see if we can legitimately ignore x after we have solved this problem.
We will assume that x is very small compared to 0.60 and can, therefore, be ignored.

40. Calculating pH from Ka Now, (x)2 4.5  104 = (0.60)

41. Calculating pH from Ka pH = log [H3O+] pH = log (1.6  102)

42. When can you legitimately disregard “X” in the denominator to avoid quadratic equation ?
Perform 5% check
1. disregard the “X” in the denominator
2. solve the equality with that approximation 3.
“value obtained for X”
[original]
x 100
In the nitrous acid problem we just did, x =
The original value for HONO was 0.60.
0.016/0.60 x 100 = 0.96 % which is less than 5% so the approximation was okay.
If value < 5%, approximation OK
If value > 5%, approximation is NOT OK and must go back and use quadratic equation 

43. Weak Bases Kb is called the base-dissociation constant
Kb is just an equilibrium constant that is specific for a weak base
Weak bases undergo hydrolysis
Hydrolysis – a substance reacts with H2O leaving either H+ or OH−

44. Weak Bases The equilibrium constant expression for this reaction is
B-(aq) + H2O(l)
HB(aq) + OH(aq)
[HB] [OH]
[B]
Kb =
where Kb is the base-dissociation constant.

45. Equilibrium favors the reactants (left)
NH3 (aq) + H2O ()
NH4+ (aq) + OH− (aq)
[ NH4+] [OH−]
[NH3]
Kb =
= 1.8 x 10−5
Equilibrium favors the reactants (left)

46. Kb can be used to find [OH] and, through it, pH.
Weak Bases
Kb can be used to find [OH] and, through it, pH.

47. (CH3)3N (aq) + H2O () (CH3)3NH+ (aq) + OH− (aq)
Trimethylamine, (CH3)3N has a lovely, putrefying smell (like rotting dead fish). What is the pH of a M aqueous solution of trimethylamine ?
Kb of trimethylamine is 7.4 x 10-5
hydrolysis
(CH3)3N (aq) + H2O ()
(CH3)3NH+ (aq) + OH− (aq)

48. pH of Basic Solutions Tabulate the data. [(CH3)3N], M [(CH3)3NH+], M
[OH], M
Initially
0.145
At equilibrium

49. pH of Basic Solutions Tabulate the data. [(CH3)3N], M [(CH3)3NH+], M
[OH], M
Initially
0.145
At equilibrium
0.145  x  0.145 x

50. pH of Basic Solutions (x)2 (0.145) 7.4  105 =
Was our assumption to ignore x in the denominator valid?

51. pH of Basic Solutions Therefore, [OH] = 1.04  103 M
pOH = log (1.04  103)
pOH = 2.98
pH =  2.98
pH = 11.02

52. pKb = -log Kb The pKb of methylamine is 3.38. What is the Kb ?
4.17 x 10-4

53. Ka and Kb Ka and Kb are related in this way: Ka  Kb = Kw
Therefore, if you know one of them, you can calculate the other.

54. [HNO2][OH−] [NO2− ] [HNO2]
NO2− H2O
HNO OH−
[HNO2][OH−]
[NO2− ]
Kb =
H2O
HNO2
H+ + NO2−
[ H+] [NO2 −]
[HNO2]
K = a

55. [H+] [NO2−] [HNO2]
Ka =
[HNO2] [OH−]
[NO2−]
Kb =
[H+] [NO2−]
[HNO2]
[HNO2] [OH−]
[NO2−]
(Ka)(Kb) =
(Ka)(Kb) = [H+] [OH−]
= Kw = 1.0 x 10−14
(at 25C)
for any conjugate acid/base pair at 25C
(Ka)(Kb) = 1.0 x 10−14

56. What is the Kb for NO2− ?
Ka (HNO2) = 4.5 x 10-4

57. Household bleach is a 0. 65 M NaOCl (aq) solution
Household bleach is a 0.65 M NaOCl (aq) solution. What is the pH of bleach?
Kb = 3.5 x 10-7 = [HOCl][OH-] / [OCl-]
Set up ICE Table

58. Equilibrium Calculations when Simplifications Fail
Weak acid, HA, ionizes in water to give ions
HA(aq) + H2O A–(aq) + H3O+(aq)
[HA]equilibrium = [HA]initial – x
where x = amount dissociated to ions = [A–] = [H+]
For the solutions we have looked at so far
[HA]equilibrium  [HA]initial

59. Simplifications work when
Simplifications fail when
In latter case, must solve using quadratic equation
Exact
Mathematically more complex

60. The Quadratic Equation
For a quadratic equation in the form
The values of x that satisfy this equation are
Take only the positive root for the answer
Negative [x] is meaningless
Always gives you the correct answer

61. The equilibrium law for this reaction is
What is the pH of a 0.15 M solution of dichloroacetic acid, CHCl2CO2H, in water? For dichloroacetic acid, Ka = 5.0 × 10–2.
CHCl2CO2H(aq) + H2O CHCl2CO2–(aq) + H3O+(aq)
The equilibrium law for this reaction is
[CHCl2CO2H ] (M)
[CHCl2CO2–] (M)
[H3O+] (M) I
0.15 C E
First, solve this problem assuming x is insignificant in its contribution to the denominator.
X is determined to be which is more than 5% of 0.15.
As a result, we cannot assume that x is insignificant and can be ignored.
As a result, we must use the quadratic equation.
– x
+ x
+ x
0.15 – x x x

62. Using Quadratic Equation
Or in terms of general quadratic equation
Where a = 1, b = 0.050, and c = –7.5 ×10–3
Now put into quadratic formula
The two roots are: x = M and x = – M

63. Using Quadratic Equation
Since only positive root has physical meaning, we use this answer
[H+] = x = M
[C2HCO2Cl2–] = x = M
[HC2HCO2Cl2] = 0.15 – = M
pH = –log(0.065) = 1.19

64. Buffers – solutions which resist changes in pH
1. A mixture of a weak acid + soluble salt of the conjugate base of that weak acid or
2. A mixture of a weak base + soluble salt of the conjugate acid of that weak base
Most important buffer:
In blood HCO3– and H2CO3
Absorb acids and bases produced by metabolism
Maintains a remarkably constant pH
Important as cells live only in a very narrow pH range.
Buffer Capacity – the amount of acid or base that can be added before the buffer is overwhelmed and pH dramatically changes

65. Buffers
If a small amount of hydroxide is added to an equimolar solution of HF and NaF, for example, the HF reacts with the OH to make F and water.
Similarly, if acid is added, the F reacts with it to form HF and water.

66. Buffer Calculations
Consider the equilibrium constant expression for the dissociation of a generic acid, HA:
HA + H2O
H3O+ + A
[H3O+] [A]
[HA]
Ka =

67. Buffer Calculations Rearranging slightly, this becomes [A] [HA] Ka =
[H3O+]
Taking the negative log of both side, we get
base
[A]
[HA]
log Ka =
log [H3O+] + log
pKa pH
acid

68. Buffer Calculations So [base] pKa = pH  log [acid]
Rearranging, this becomes
pH = pKa + log
[base]
[acid]
This is the Henderson–Hasselbalch Equation.

69. Henderson-Hasselbalch Equation
Lawrence Henderson 1878 – 1942
Karl Hasselbalch 1874 – 1962

70. Henderson-Hasselbalch Equation
[base]
[acid]
pH = pKa + log
pH = pH of the buffered solution
pKa = pKa of the weak acid
[base] and [acid] are initial [ ]’s of the conjugate acid/base pair

71. Steps to Working Any Buffer Problem
1. Determine how all species exist in solution
2. Write the equilibrium reaction describing the species
3. Leave out spectator ions such as Na+ and Cl- because they don’t affect the pH
Ka for formic acid = 1.8 X 10-4
Suppose g NaCHO2 (s) is dissolved into 305 mL of 0.45 M HCHO2 (aq). What is the pH of the resulting solution ?

72. Henderson–Hasselbalch Equation
What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4  104.
These are equilibrium concentrations which are entered directly into the Henderson-Hasselbalch Equation.

73. Henderson–Hasselbalch Equation
pH = pKa + log
[base]
[acid]
pH = log (1.4  104) + log
(0.10)
(0.12)
pH = (0.08)
pH = 3.77

74. pOH = pOH of the buffered solution
Henderson-Hasselbalch equation for base
[acid]
[base]
pOH = pKb + log
pOH = pOH of the buffered solution
pKb = pKb of the weak base
[acid] and [base] are initial [ ]’s of the conjugate acid/base pair

75. 1. Work this problem using the “base” form of Henderson-Hasselbalch
Determine the pH of a solution of 0.30 M NH3 (aq) and 0.18 M NH4Cl (aq). (Use data on Slide 53)
1. Work this problem using the “base” form of Henderson-Hasselbalch
2. Work this problem using the “acid” form of Henderson-Hasselbalch
pKa (NH4Cl) = 9.25
pKb (NH3) = 4.74
Do you get the same answers?

76. 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.
Ideally want a 1:1 mixture of salt and acid whose pKa is equal to the desired pH.
A suitable buffer is within pH = pKa  1 range.

77. A solution buffered at pH 3. 90 is needed for a reaction
A solution buffered at pH 3.90 is needed for a reaction. Would formic acid and its salt, sodium formate, make a good choice for this buffer? If so, what ratio of moles of the conjugate base anion (HCOO–) to the acid (HCOOH) is needed?
From table pKa of formic acid is 3.74.
This is within pH = pKa  1 range
2.90 to 4.90
So we can use this buffer system

78. Titration
In this technique, a known concentration of base (or acid) is slowly added to a solution of acid (or base).

79. 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.

80. 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.

81. Titration of a Strong Acid with a Strong Base
Just before (and after) the equivalence point, the pH increases rapidly.

82. 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.

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

84. Titration of a Weak Acid with a Strong Base
Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.
At the equivalence point the pH is >7.
Phenolphthalein is commonly used as an indicator in these titrations.

85. 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.

86. Which slide, this or the next slide, depicts the titration of the weaker acid?

87. Which slide, this or the previous slide, depicts the titration of the weaker acid?