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1
Chapter 16. Acid Base Equilibria

• .

• .

2
Equilibria in Solutions of Weak Acids

• The dissociation of a weak acid is an equilibrium
  situation with an equilibrium constant, called
  the acid dissociation constant, Ka based on the
  equation
• HA (aq) H2O (l) ? H3O (aq) A- (aq)

3
Equilibria in Solutions of Weak Acids

• The acid dissociation constant, Ka is always
  based on the reaction of one mole of the weak
  acid with water.
• If you see the symbol Ka, it always refers to a
  balanced equation of the form
• HA (aq) H2O (l) ? H3O (aq) A- (aq)

4
Problem

• The pH of 0.10 mol/L HOCl is 4.23. Calculate Ka
  for hypochlorous acid.
• HOCl (aq) H2O (l) ? H3O (aq) ClO- (aq)

5
Calculating Equilibrium Concentrations in
Solutions of Weak Acids

• We can calculate equilibrium concentrations of
  reactants and products in weak acid dissociation
  reactions with known values for Ka.
• To do this, we will often use the ICE table
  technique we saw in the last chapter on
  equilibrium.

6
Calculating Equilibrium Concentrations in
Solutions of Weak Acids

• We need to figure out what is an acid and what is
  a base in our system.
• For example, if we start with 0.10 mol/L HCN,
  then HCN is an acid, and water is a base.
• HCN (aq) H2O (l) ? H3O (aq) CN- (aq)
• Ka 4.9 x 10-10

7

• Like our previous equilibrium problems, we then
  create a table of the initial concentrations of
  all chemicals, the change in their concentration,
  and their equilibrium concentrations in terms of
  known and unknown values.

8

• We can ALWAYS solve this equation using the
  quadratic formula and get the right answer, but
  it might be possible to do it more simply.

9

• Every time we do an weak acid equilibrium
  problem, divide the initial concentration of the
  acid by Ka.
• For this example
• 0.10 / 4.9 x 10-10 2 x 108

10

• 0.10 / 4.9 x 10-10 2 x 108
• Since this value is greater than 100, we can
  assume that the initial concentration of the acid
  and the equilibrium concentration of the acid are
  the same.
• This assumption will lead to answers with less
  than 5 error since this pre-check is greater
  than 100.

11

• The assumption we will make is that
• x ltlt HCNi so HCNeqm ? HCNI
• 4.9 x 10-10 x2 / 0.10
• x2 (4.9 x 10-10)(0.10)
• x ?4.9 x 10-11
• x ?7.0 x 10-6 mol/L

12

• Based on the assumption weve made, at
  equilibrium
• x H3Oeqm CN-eqm
• 7.0 x 10-6 mol/L
• (-ve value isnt physically possible)
• HCNeqm 0.10 mol/L.

13
.

• Any time we make an assumption,
• we MUST check it.
• We assumed x ltlt HCNi
• To check the assumption, we divide x by HCNi
  and express it as a percentage

14

• As long as the assumption check is less than 5,
  then
• the assumption is valid!
• If the assumption was not valid, we would have to
  go back and use the quadratic formula!

15
Remember!

• H2O (l) H2O (l) ? H3O (aq) OH- (aq)
• is always taking place in water whether or not we
  have added an acid or base.
• This reaction also contributes
• H3O (aq) and OH- (aq)
• to our system at equilibrium

16
Remember!

• Since at 25 ?C
• Kw H3O OH- 1.0 x 10-14
• it turns out that if our acid-base equilibrium
  were interested in gives a pH value between
  about 6.8 and 7.2 then the auto-dissociation of
  water contributes a significant amount of H3O
  and OH- to our system and the real pH would not
  be what we calculated in the problem.

17
Problem

• Acetic acid CH3COOH (or HAc) is the solute that
  gives vinegar its characteristic odour and sour
  taste. Calculate the pH and the concentration of
  all species present in
• a) 1.00 mol/L CH3COOH
• b) 0.00100 mol/L CH3COOH

18
Problem a)
Lets check the initial acid concentration / Ka
ratio. 1.00 / 1.8 x 10-5 ? 55000 is larger
than 100.
19
Problem a)
We can probably assume that x ltlt HAci so
HAceqm ? HAci 1.8 x 10-5 x2 / 1.00 x2
(1.8 x 10-5)(1.00) x ?1.8 x 10-5 x ?4.2 x
10-3 mol/L (but must be value since x H3O)
20
Problem a)

• So at equilibrium,
• H3O CH3COO- 4.2 x 10-3 mol/L
• CH3COOH 1.00 mol/L.
• The assumption was valid and so
• pH - log H3O
• pH - log 4.2 x 10-3
• pH 2.38

21
Problem b)
Lets check the initial acid concentration / Ka
ratio. 0.00100 / 1.8 x 10-5 ? 56 is smaller
than 100.
22
Problem b)
We can probably CAN NOT assume that x ltlt HAci
so HAceqm ? HAci
23
Problem b)
24
Problem b)
Since H3O x we must use the positive value,
so H3O CH3COO- 1.3 x 10-4
mol/L CH3COOH 0.00100 mol/L 1.3 x 10-4
mol/L 0.00087 mol/L.
25
Problem b)
Lets confirm that x ltlt HAci IS NOT TRUE pH
- log H3O pH - log 4.2 x 10-4 pH 3.38
26
Problem

• A vitamin C tablet containing 250 mg of ascorbic
  acid (C6H8O6 Ka 8.0 x 10-5 is dissolved in a
  250 mL glass of water to give a solution where
  C6H8O6 5.68 x 10-3 mol/L.
• What is the pH of the solution?

27
Problem

• Check the initial acid concentration / Ka ratio.
  
• 5.68 x 10-3/ 8.0 x 10-5 ? 71
• which is not larger than 100 so

28
Problem
29
Problem

• Since H3O x the answer must be the positive
  value
• H3O C6H7O6- 6.3 x 10-4 mol/L
• C6H8O6 (5.68 x 10-3 - 6.3 x 10-4) mol/L
• 5.05 x 10-3 mol/L
• pH - log H3O
• pH - log 6.3 x 10-4
• pH 3.20

30
Degree of ionization

• The pH of a solution of a weak acid like acetic
  acid will depend on the initial concentration of
  the weak acid and Ka. Therefore, we can define a
  second measure of the strength of a weak acid by
  looking of the
• degree (or percent) ionization of the acid.
• ionization HAionized / HAinitial x 100

31
Percent ionization

• In part a) of an earlier problem an acetic acid
  solution with initial concentration of 1.00 mol/L
  at equilibrium had
• H3Oeqm HAionized 4.2 x 10-3 mol/L
• ionization HAionized / HAinitial x 100
• ionization 4.2 x 10-3 mol/L / 1.00 mol/L x
  100
• ionized 0.42

32
Percent ionization

• In part b) of an earlier problem an acetic acid
  solution with initial concentration of 0.00100
  mol/L at equilibrium had
• H3O HAionized 1.3 x 10-4 mol/L
• ionization HAionized / HAinitial x 100
• ionization 1.3 x 10-4 mol/L / 0.00100 mol/L x
  100
• ionization 13

33
Figure
34
Equilibria in Solutions of Weak Bases

• The dissociation of a weak base is an equilibrium
  situation with an equilibrium constant, called
  the base dissociation constant, Kb based on the
  equation
• B (aq) H2O (l) ? BH (aq) OH- (aq)

35
Equilibria in Solutions of Weak Bases

• The base dissociation constant, Kb is always
  based on the reaction of one mole of the weak
  base with water.
• If you see the symbol Kb, it always refers to a
  balanced equation of the form
• B (aq) H2O (l) ? BH (aq) OH- (aq)

36
Equilibria in Solutions of Weak Bases

• Our approach to solving equilibria problems
  involving bases is exactly the same as for acids.
• 1. Set up the ICE table
• 2. Establish the equilibrium constant
  expression
• 3. Make a simplifying assumption when possible
• 4. Solve for x, and then for eqm amounts

37
Problem

• Strychnine (C21H22N2O2), a deadly poison used for
  killing rodents, is a weak base having Kb 1.8 x
  10-6. Calculate the pH if
• C21H22N2O2initial 4.8 x 10-4 mol/L

38
Problem
Check the initial base concentration / Kb ratio
4.8 x 10-4 / 1.8 x 10-6 ? 267 which is
greater than 100 We are probably good to make a
simplifying assumption that x ltlt C21H22N2O2i
39

• The assumption we will make is that
• x ltlt C21H22N2O2i so
• C21H22N2O2eqm ? C21H22N2O2I
• 1.8 x 10-6 x2 / 4.8 x 10-4
• x2 (1.8 x 10-6)(4.8 x 10-4)
• x ?8.64 x 10-10
• x ?2.94 x 10-5 mol/L

40
Problem

• Since x OH-, the answer must be the positive
  value,
• x C21H23N2O2 OH- 2.9 x 10-5 mol/L
• C21H22N2O2 4.8 x 10-4 mol/L 2.9 x 10-5
  mol/L
• 4.5 x 10-4 mol/L.
• We should check the assumption!

41
Problem

• In this case, the error is more than 5.
• I will leave it to you to go back and use the
  quadratic formula.
• Compare the two answers

42
Problem

• To continue towards the answer of the problem AS
  IF the assumption WERE VALID
• pOH - log OH-
• pOH - log 2.9 x 10-5
• pOH 4.54
• pH pOH 14.00
• pH 14.00 - pOH
• pH 14.00 - (4.54)
• pH 9.46

43
Relation Between Ka and Kb

• The strength of an acid in water is expressed
  through Ka, while the strength of a base can be
  expressed through Kb
• Since Brønsted-Lowry acid-base reactions involve
  conjugate acid-base pairs there should be a
  connection between the
• Ka value and the Kb value of a
• conjugate acid-base pair.

44
Relation Between Ka and Kb

• HA (aq) H2O (l) ? H3O (aq) A- (aq)
• A- (aq) H2O (l) ? OH- (aq) HA (aq)

45

• Since these reactions take place in the same
  beaker at the same time lets
• add them together

46

• The sum of the reactions is the dissociation of
  water reaction, which has the ion-product
  constant for water
• Kw H3O OH- 1.0 x 10-14 at 25 C
• Closer inspection shows us that

47

• As the strength of an acid increases (larger Ka)
  the strength of the conjugate base must decrease
  (smaller Kb) because their product must always be
  the dissociation constant for water Kw.

48

• Strong acids always have very weak conjugate
  bases. Strong bases always have very weak
  conjugate acids.
• Since Ka x Kb Kw
• then Ka Kw / Kb
• and Kb Kw / Ka

49
Problem

• a) Piperidine (C5H11N) is an amine found in
  black pepper. Find Kb for piperidine in Appendix
  C, and then calculate Ka for the C5H11NH cation.
• Kb 1.3 x 10-3
• b) Find Ka for HOCl in Appendix C, and then
  calculate Kb for OCl-.
• Ka 3.5 x 10-8

50
Acid-Base Properties of Salts

• When acids and bases react with each other,
• they form ionic compounds called salts.
• Salts, when dissolved in water, can lead to
  acidic, basic, or neutral solutions, depending on
  the relative strengths of the acid and base we
  derive them from.
• Strong acid Strong base ? Neutral salt solution
• Strong acid Weak base ? Acidic salt solution
• Weak acid Strong base ? Basic salt solution

51
Salts that Yield Neutral Solutions

• Strong acids and strong bases react to form
  neutral salt solutions. When the salt
  dissociates in water, the cation and anion do not
  appreciably react with water to form H3O or OH-.

52
Salts that Yield Neutral Solutions

• Strong base cations like the alkali metal cations
  (Li, Na, K) or alkaline earth cations (Ca2,
  Sr2, Ba2, but NOT Be2) and strong acid anions
  such as Cl-, Br-, I-, NO3-, and ClO4- will
  combine together to give neutral salt solutions
  with pH 7.

53
Salts that Yield Neutral Solutions

• Sodium chloride (NaCl) will dissociate into Na
  and Cl- in water.
• Cl- has no acidic or basic tendencies.
• Cl- (aq) H2O (l) ? no reaction
• Chloride ions DO NOT HAVE hydrolysis reactions
  with water since it is the conjugate of a
  strong acid, which makes it very, very weak.

54
Salts that Yield Neutral Solutions

• Na has no acidic or basic tendencies.
• Na (aq) H2O (l) ? no reaction
• Sodium ions DO NOT HAVE hydrolysis reactions with
  water since it is the conjugate of a strong
  base, which makes it very, very weak.

55
Salts that Yield Acidic Solutions

• The reaction of a strong acid with anions like
• Cl-, Br-, I-, NO3-, and ClO4-
• with a weak base will lead to an
• acidic salt solution.
• The solution is acidic because the anion shows no
  acidic or basic tendencies, but the cation does,
  as it is the conjugate acid of a weak base.

56
Salts that Yield Acidic Solutions

• Ammonium chloride (NH4Cl) will dissociate into
  NH4 and Cl- in water.
• Cl- has no acidic or basic tendencies.
• Cl- (aq) H2O (l) ? no reaction
• Chloride ions DO NOT HAVE hydrolysis reactions
  with water since it is the conjugate of a
  strong acid, which makes it very, very weak.

57
Salts that Yield Acidic Solutions

• NH4 has acidic tendencies.
• That is
• NH4 (aq) H2O (l)? NH3 (aq) H3O (aq)
• Ammonium ions hydrolyze in water because it is
  the conjugate acid of the weak base NH3, which
  means ammonium is a weak acid.

58
Salts that Yield Basic Solutions

• The reaction of a strong base with cations like
  Li, Na, K, Ca2, Sr2, and Ba2
• with a weak acid will lead to an
• basic salt solution.
• The solution is acidic because the cation shows
  no acidic or basic tendencies, but the anion
  does, as it is the conjugate base of a weak acid.

59
Salts that Yield Basic Solutions

• Sodium fluoride (NaF) will dissociate into Na
  and F- in water.
• Na (aq) H2O (l) ? no reaction
• Sodium ions DO NOT HAVE hydrolysis reactions with
  water since it is the conjugate of a strong
  base, which makes it very, very weak.

60
Salts that Yield Basic Solutions

• F- has basic tendencies.
• That is
• F- (aq) H2O (l)? HF (aq) OH- (aq)
• Fluoride ions hydrolyze in water because it is
  the conjugate base of the weak acid HF, which
  means fluoride is a weak base.

61
Problem

• Predict whether the following salt solution is
  neutral, acidic, or basic and calculate the pH.
• 0.25 mol/L NH4Br NH3 has a Kb value of 1.8 x
  10-5

62
Problem
Initial acid HA / Ka ratio is 0.25 / 5.56 x
10-10 ? 4.5 x 108 we can probably assume 0.25 gtgt
x 5.56 x 10-10 x2 / 0.25 x2 (5.56 x
10-10)(0.25) x2 1.39 x 10-10 x ?1.39 x
10-10 x 1.18 x 10-5 mol/L
63
Problem
Negative answer not physically possible
so therefore, H3O 1.18 x 10-5 mol/L Since
weve shown the assumption is valid pH -log
H3O - log 1.18 x 10-5 4.93.
64
Salts that Contain Acidic Cations and Basic
Anions

• If a salt is composed of an
• acidic cation
• and a
• basic anion,
• the acidity or basicity of the salt solution
• depends on the relative strengths of the acid and
  base.

65
Salts that Contain Acidic Cations and Basic
Anions

• If the acid cation is stronger than the base
  anion, it wins and the salt solution is acidic.
  
• If the base anion is stronger than the acid
  cation, it wins and the salt solution is basic.
  

66
Salts that Contain Acidic Cations and Basic
Anions

• Ka gt Kb
• the acid cation is stronger and the salt
  solution is acidic.
• Ka lt Kb
• the base anion is stronger and the salt
  solution is basic.
• Ka ? Kb
• the salt solution is close to neutral.

67
Problem

• Classify each of the following salts as acidic,
  basic, or neutral
• a) KBr
• b) NaNO2
• c) NH4Br
• d) NH4F

Ka for HF 6.6 x 10-4 Kb for NH3 1.8 x 10-5
68
The Common-Ion Effect

• Solutions consisting of both an acid and its
  conjugate base are very important because they
  are very resistant to changes in pH. Such buffer
  solutions regulate pH in a variety of biological
  systems.

69
The Common-Ion Effect

• Lets consider a solution made of 0.10 moles of
  acetic acid and 0.10 moles of sodium acetate with
  a total volume of 1.00 L, making the initial
  CH3COOH CH3COO- 0.10 mol/L.
• First we must identify all potential acids and
  bases in the system.
• CH3COOH CH3COO- Na
  H2O
• acid base
  neutral acid
• 
  or base

70
Point of view of the acid

• Our reaction will be
• CH3COOH (aq) H2O (l) ? H3O (aq) CH3COO- (aq)
• Ka 1.8 x 10-5

Note that the initial concentration of our
product CH3COO- is NOT ZERO!
71

• Lets check the
• initial acid concentration / Ka ratio.
• 0.10 / 1.8 x 10-5 ? 5500
• Its probably safe to assume that
• x ltlt HAci so HAceqm ? HAcI
• and x ltlt Ac-i so Ac-eqm ? Ac-i
• 1.8 x 10-5 x (0.10 x) / (0.10 x)
• 1.8 x 10-5 x (0.10) / (0.10)
• x 1.8 x 10-5 mol/L

72

• At equilibrium,
• H3O 1.8 x 10-5 mol/L
• CH3COO- 0.10 1.8 x 10-5 0.10 mol/L
• CH3COOH 0.10 - 1.8 x 10-5 0.10 mol/L
• Assumption was valid! Check for yourself!
• pH - log H3O
• pH - log 1.8 x 10-5
• pH 4.74

73
If we had started out with only 0.10 mol/L acetic
acid, the pH would be found from
74

• The initial acid concentration / Ka will still be
  the same, so we can assume
• x ltlt HAci so HAceqm ? HAci
• 1.8 x 10-5 x2 / (0.10 x)
• 1.8 x 10-5 x2/ (0.10)
• x ?1.8 x 10-6 mol/L
• x ?1.3 x 10-3 mol/L (cant be ve)
• pH - log H3O
• pH - log 1.3 x 10-3
• pH 2.89

75

• Without the acetate ion the pH of
• 0.10 M acetic acid is 2.89.
• With an equal concentration of acetate ion
  present, the pH of
• 0.10 M acetic acid 0.10 M acetate is 4.74
• The acetate ion makes a large difference on the
  equilibrium pH!

76
CH3COOH (aq) H2O (l) ? H3O (aq) CH3COO- (aq)

• Adding the conjugate base (a stress!) to the
  equilibrium system of an acid dissociation shows
  the common-ion effect, where the addition of a
  common ion causes the equilibrium to shift.
• This is an example of Le Chataliers Principle.

Addition of the weak base to the acid dissociation
77
Problem

• Calculate the concentrations of all species
  present, and the pH in a solution that is 0.025
  mol/L HCN and 0.010 mol/L NaCN.
• (Ka of HCN 4.9 x 10-10)

78
Problem
The initial base concentration / Ka ratio is
0.010 / 4.9 x 10-10 ? 2 x 107 Its probably safe
to assume that x ltlt HCNi so HCNeqm ?
HCNI and x ltlt CN-i so CN-eqm ? CN-i
79
Problem
4.9 x 10-10 x (0.010 x) / (0.025 x) 4.9 x
10-10 x (0.010) / (0.025) x 1.2 x 10-9
mol/L So at equilibrium, H3O 1.2 x 10-9
mol/L CN- 0.010 1.2 x 10-9 0.010 mol/L
HCN 0.025 - 1.2 x 10-9 0.025
mol/L. Assumption was valid! Check this for
yourself!
80
Problem
pH - log H3O pH - log 1.2 x 10-9 pH 8.91
81
Buffer Solutions

• Solutions that contain both a weak acid and its
  conjugate base are buffer solutions.
• These solutions are resistant to changes in pH.

82
Buffer Solutions

• If more acid (H3O) or base (OH-) is added to the
  system, the system has enough of the original
  acid and conjugate base molecules in the solution
  to react with the
• added acid or base, and so the new equilibrium
  mixture will be
• very close in composition to the original
  equilibrium mixture.

83
Buffer solutions

• A 0.10 mol?L-1 acetic acid 0.10 mol?L-1 acetate
  mixture has a pH of 4.74 and is a buffer
  solution!
• CH3COOH (aq) H2O (l) ? H3O (aq) CH3COO- (aq)

84
Buffer solutions

• If we rearrange the Ka expression to solve for
• H3O

85
Buffer solutions

• Assume x ltlt HAci so HAceqm ? HAcI
• and x ltlt Ac-i so Ac-eqm ? Ac-i,
• and we should see
• If CH3COOHi CH3COO-i,
• then H3O 1.8 x 10-5 M Ka
• and pH pKa 4.74

86
Buffer solutions

• What happens if we add 0.01 mol of NaOH (strong
  base) to 1.00 L of the acetic acid acetate
  buffer solution?
• CH3COOH (aq) OH- (aq) ? H2O (l) CH3COO- (aq)
• This reaction goes to completion and keeps
  occurring until we run out of the limiting
  reagent OH-

New CH3COOH 0.09 M and new CH3COO- 0.11 M
87
Buffer solutions

• With the assumption that x is much smaller than
  0.09 mol (an assumption we always need to check
  after calculations are done!), we find

Note weve made the assumption that x ltlt 0.09
M! pH - log H3O pH - log 1.5 x 10-5 pH
4.82
88
Buffer solutions

• Adding 0.01 mol of OH- to 1.00 L of water would
  have given us a pH of 12.0 because there is no
  significant amount of acid in water for the base
  to react with.
• Our buffer solution resisted this change in pH
  because there is a significant amount of acid
  (acetic acid) for the added base to react with.

89
Buffer solutions

• What happens if we add 0.01 mol of HCl (strong
  acid) to 1.00 L of the acetic acid acetate
  buffer solution?
• CH3COO- (aq) H3O (aq) ? H2O (l) CH3COOH (aq)
• This reaction goes to completion and keeps
  occurring until we run out of the limiting
  reagent H3O

New CH3COOH 0.11 M and new CH3COO- 0.09 M
90
Buffer solutions

• With the assumption that x is much smaller than
  0.09 mol (an assumption we always need to check
  after calculations are done!), we find

Note weve made the assumption that x ltlt 0.09
M! pH - log H3O pH - log 2.2 x 10-5 pH
4.66
91
Buffer solutions

• Adding 0.01 mol of H3O to 1.00 L of water would
  have given us a pH of 2.0 because there is no
  significant amount of acid in water for the base
  to react with.
• Our buffer solution resisted this change in pH
  because there is a significant amount of base
  (acetate) for the added acid to react with.

92
.
93
Buffer capacity

• Buffer capacity is the measure of the ability of
  a buffer to absorb acid or base without
  significant change in pH.
• Larger volumes of buffer solutions have a larger
  buffer capacity than smaller volumes with the
  same concentration.
• Buffer solutions of higher concentrations have a
  larger buffer capacity than a buffer solution of
  the same volume with smaller concentrations.

94
Problem

• Calculate the pH of a 0.100 L buffer solution
  that is 0.25 mol/L in HF and 0.50 mol/L in NaF.

With the assumption that x is much smaller than
0.25 mol (an assumption we always need to check
after calculations are done!), we find
Assume x ltlt HCNi so HCNeqm ? HCNi and x ltlt
CN-i so CN-eqm ? CN-i
95
Problem
pH - log H3O pH - log 1.75 x 10-4 pH 3.76
96
Problem

• a) What is the change in pH on addition of 0.002
  mol of HNO3?

New HF 0.27 M and new F- 0.48 M
97
Problem
Notice weve made the assumption that x ltlt 0.27
M. We should check this! pH - log H3O pH
- log 1.97 x 10-4 pH 3.71
98
Problem

• b) What is the change in pH on addition of 0.004
  mol of KOH?

New HF 0.21 M and new F- 0.54 M
99
Problem
Notice weve made the assumption that x ltlt 0.21
M. We should check this! pH - log H3O pH
- log 1.36 x 10-4 pH 3.87
100
The Henderson-Hasselbalch Equation

• Weve seen that, for buffer solutions containing
  members of a conjugate acid-base pair, that
• pH pKa log base / acid
• This is called the Henderson-Hasselbalch
  Equation.

101
The Henderson-Hasselbalch Equation

• If we have a buffer solution of a conjugate
  acid-base pair, then the pH of the solution will
  be close to the pKa of the acid.
• This pKa value is modified by the logarithm of
  ratio of the concentrations of the base and acid
  in the solution to give the actual pH.

102
Problem

• Use the Henderson-Hasselbalch Equation to
  calculate the pH of a buffer solution prepared by
  mixing equal volumes of 0.20 mol/L NaHCO3 and
  0.10 mol/L Na2CO3.
• We need the Ka and the concentrations of the acid
  (HCO3-) and the base (CO32-).
• Ka 5.6 x 10-11
• (we use the Ka for the second proton of H2CO3!).

103
Problem

• NOTE The concentrations we are given for the
  acid and the base are the concentrations
• before the mixing of equal volumes!

104
Problem

• If we mix equal volumes, the total volume is
  TWICE the volume for the original acid or base
  solutions.
• Since the number of moles of acid or base DONT
  CHANGE on mixing,
• the initial concentrations we use will be
• half the given values.
• pH pKa log base / acid
• pH (-log 5.6 x 10-11) log (0.05) / (0.10)
• pH 10.25 0.30
• pH 9.95