Ch18' Acids and Bases: Equilibria

1
Ch18. Acids and Bases Equilibria

• Brady Senese, 4th

2
Chapter 18 Equilibria in Solutions of Weak Acids
and Bases

• All weak acids behave the same way in aqueous
  solution they partially ionize
• In terms of the general weak acid HA, this can
  be written as
• Following the procedures in Chapter 16

3

• Ka is called the acid ionization constant
• These are often reported as the pKa
• Table 18.1 and Appendix C list the Ka and pKa for
  a number of acids
• A large pKa,means a small value of Ka and
  only a small fraction of the acid molecules
  ionize
• A small pKa,means a large value of Ka and a
  large fraction of the acid molecules ionize

4

• Weak bases behave in a similar manner in water
• For the general base B

5

• Values of Kb and pKb for a number of weak bases
  are listed in Table 18.2 and in Appendix C
• Where, like for acids
• There is an interesting relations ship between
  the acid and base ionizations constants for a
  conjugate acid-base pair
• Using the general weak acid HA

6

7

• Thus, for any conjugate acid-base pair
• Most tables of ionization constants only give
  values for the molecular member of the conjugate
  acid-base pair
• The ionization constant of the ion member of the
  conjugate acid-base pair is then calculated as
  needed

8
For Conjugate Pairs KwKa x Kb
Relative strengths of conjugate acid-base pairs.
The stronger the acid is, the weaker the
conjugate base. The weaker the acid, the stronger
the conjugate base. Very strong acids ionize 100
and their conjugate bases do not react to any
measurable extent.

9
Learning Check

• What is the value of Kb for the following weak
  conjugate base?
• NaF
• KwKaKb Ka for HF 6.810-4
• 1.4710-11
• What is the value of Ka for the following weak
  conjugate acid?
• NH4Cl
• KwKaKb Kb for NH3 1.810-5
• 5.5610-10

10
Determining the pH Of Aqueous Weak Acid Solutions

• Dominant equilibrium is Ka reaction
• write the net ionic equation
• look up the Ka value for the acid
• set up ICE table
• solve for x
• Calculate pH from the hydronium concentration at
  equilibrium

11

• The primary goal is usually to determine the
  equilibrium concentration for all species in the
  mass action expression
• The percentage ionization of the acid or base is
  defined as

12

• Example Morphine is very effective at relieving
  intense pain and is a weak base. What is the Kb,
  pKb, and percentage ionization of morphine if a
  0.010 M solution has a pH of 10.10?
• The reaction can be represented as
• At equilibrium, OH- x 10-pOH

13

• SOLUTION Use pOH 14.00 pH, substituting

14

• As in Chapter 16, if the extent of reaction (x)
  is small it is possible to simplify the mass
  action expression
• For the case of a weak acid or base added to pure
  water
• If the initial solute concentration is at least
  400 times larger than the ionization constant,
  the initial concentration of the solute can be
  used as though they were the equilibrium
  concentration
• If the solute concentration is too small, or the
  equilibrium constant too large, then the
  quadratic equation must be used

15
Learning Check

• Determine the pH of 0.1M solutions of
• HC2H3O2
• Ka1.810-5
• HCN
• Ka6.210-10

0.1M
N/A
0
0
-x
x
x
-x
(0.1-x)0.1
N/A
x
x
pH2.87
0.1M
N/A
0
0
-x
x
x
-x
(0.1-x) 0.1
N/A
x
x
pH4.60
16
Determining The pH Of Base Solutions

• Dominant equilibrium is Kb reaction
• write the net ionic equation
• set up ICE table for starting quantities
• solve for x
• Calculate pOH from the OH- at equilibrium, and
  convert to pH

17
Learning Check

• Determine the pH of 0.1M solutions of
• N2H4
• Kb1.710-6
• NH3
• Kb1.810-5

0.1M
N/A
0
0
-x
x
x
-x
(0.1-x) 0.1
N/A
x
x
pOH3.38
pH10.62
0.1M
N/A
0
0
-x
x
x
-x
(0.1-x) 0.1
N/A
x
x
pH11.13
pOH2.87
18
Learning Check

• Determine the ionization of 0.2M solution of
  HC2H3O2
• Ka1.810-5

0.2M
N/A
0
0
-x
x
x
-x
(0.2-x) 0.2
N/A
x
x
x1.9010-3M
0.95 ionized
19
Learning Check

• Determine the ionization of 0.1M solution of
  HC2H3O2
• Ka1.810-5

0.1M
N/A
0
0
-x
x
x
-x
(0.1-x) 0.1
N/A
x
x
x1.34 x 10-3M
1.3 ionized
20

• Aqueous cations
• Cations that are conjugate acids of weak
  molecular bases are weak acids
• Metal cations with a high charge density (like
  Al3, Fe3, and Cr3) yield aqueous solutions
  that are acidic
• Aqueous anions
• The anion of a strong acid is too weak a base to
  influence the pH of a solution
• Anions of weak acids tend to make solution basic

21
Predicting AcidBase Properties Of A Salt

• There are four possibilities when a salt is added
  to pure water
• Neither the anion nor cation affects the pH and
  the solution remains neutral. For example NaCl
• Only the cation is acidic, so the solution
  becomes acidic. For example NH4Cl
• Only the anion is basic, so the solution becomes
  basic. For example NaNO2
• The anion is basic and the cation is acidic, the
  pH of the solution will be determined by the
  relative strengths of the acid and base. For
  example NH4NO2 produces an acidic solution and
  NH4OCl produces a basic solution

22
Learning Check

• 0.1M solutions of the following are acid/ base/
  neutral or amphoteric?
• HCl
• NaCl
• NaCN
• HCN
• Na2S
• Na3PO4
• NH4Cl

• acid
• neutral
• base
• acid
• base
• base
• acid

23
Learning Check

• Determine the pH of a 0.1M solution of the salt
  NaF
• basic due to F-
• KbKw/Ka

0.1M
N/A
0
0
-x
x
x
-x
(0.1-x) 0.1
N/A
x
x
pOH5.92
pH8.08
24
Using The Quadratic Equation

• Dropping x in binomial terms usually works when
  the concentration of the binomial is gt400 K
• If the value of x gt 5 of the initial
  concentration, we are often left with a
  complicated expression
• If the expression can be reduced to the form
• ax2 bx c0, it may be solved explicitly using
  the quadratic equation.

25

• Example Calculate the pH of a 0.0010 M solution
  of dimethylamine for which Kb9.6x10-4.
• ANALYSIS 400 Kbgt0.0010 M, so use of the
  quadratic equation is indicated.
• SOLUTION Set the problem up

26

• Put in standard form
• Solve for x and the equilibrium concentrations

27
Solving by Successive Approximations

• When the equation cannot be reduced to a
  quadratic, successive approximation may be used
  to estimate the answer

28

• This can be applied to the last problem

29

• Using a subscript to keep track of the iterations
  the method gives

Comment This method works best when the initial
guess is close to the final answer. At
equilibrium, the base is 61 ionized, so
starting with a guess of 0 ionized is far
from the final solution.
30
Buffers

• When a small amount of strong acid or base is
  added to certain solutions, only a small change
  in pH is observed
• These solutions are called buffers
• Buffer solution usually contains two solutes, one
  providing a weak acid and the other a weak base
• If the weak acid is molecular, then the conjugate
  base can be supplied as a soluble salt of the acid

31

• Buffers work because the weak acid can react
  with added base and the weak base can react with
  added acid
• Consider the general buffer made so that both HA
  and A- are present in solution
• When base (OH-) is added
• When acid (H) is added
• Net result small changes in pH

32
Learning Check

• what is the pH of a buffer made by combining 0.1M
  HC2H3O2 (Ka 1.810-5) and 0.2M NaC2H3O2
• inhibition limits
• the value of x

0.1M
N/A
0
0.2
-x
x
x
-x
(0.1-x) 0.1
N/A
x
(0.2x)0.2
x9.00 x 10-6 M
pH5.05
33

• Because of these reactions, calculations
  involving buffer solutions can be greatly
  simplified
• For buffer solutions, the initial concentration
  of both the weak acid and its conjugate base can
  be used as though they were equilibrium values
  (more complicated buffer systems where this
  assumption is not valid will not be encountered
  in this class)
• For buffer solutions only, either molar
  concentrations or moles can be used in the Ka (or
  Kb) expression to express the amounts of the
  members of the conjugate acid-base pair (the same
  units must be used for both members of the pair)

34

• Example What is the pH of a buffer made by
  adding 0.11 mol NH3 and 0.10 mol NH4Cl to 2.0 L
  of solution? The Kb for ammonia is 1.8x10-5
• ANALYSIS This is a buffer, initial
  concentrations can be used as equilibrium values

35

• SOLUTION Solve for OH- and use this to
  calculate the pH
• There are two important factors that determine
  the pH of a buffer solution

36

• For the general weak acid HA
• Thus both the value of Ka and the ratio of the
  molarities (or the ratio of moles) affect the pH
• These last two relations are often expressed in
  logarithmic form

37

• The first is called the Henderson-Hasselbalch
  equation, and is frequently encountered in
  biology courses
• When preparing a buffer, the concentration ratio
  is usually near 1, so the pH is mostly determined
  by the pKa of the acid

38
Learning Check

• What is the pH of a buffer made from 0.5M HF
  (Ka6.810-4) and 0.2M NaF?

pKa-log(6.810-4)
pH2.77
39
Buffer Characteristics

• Buffer capacity is the ability to compensate for
  added acid or base, defined for each component
• acid buffer capacitymoles of conjugate base
  present
• base buffer capacity moles of acid present.
• If acid is added, remove from base and add to the
  acid of the pair
• If base is added, remove from acid of the pair
  and add to the acid of the pair.

40
Buffer Characteristics

• Consider 500 mL of a sodium acetate/acetic acid
  buffer in which the concentration of sodium
  acetate is 0.5M and that of acetic acid is 0.1M.
  What is the pH after 10 mL of 0.1M NaOH is added?
  (Ka acetic acid 1.810-5)
• HC2H3O2 is the acid of the buffer, thus it reacts
  the NaOH
• HC2H3O2(aq) OH-(aq) ? H2O(l) C2H3O2-(aq)
• 0.5L 0.1M 0.1L 0.1M N/A 0.5L 0.5M
• 0.05mol 0.01mol N/A 0.25mol start

pH5.56
41
Selecting A Buffer

• To choose buffer system, try to find a system
  whose pKa is close to desired system pH
• Then, calculate necessary ratio of components to
  obtain desired pH.

• Typically, the weak acid is selected so the the
  desired pH is within one unit of the pKa
• Generally, the pH change in an experiment must be
  limited to about

42
Learning Check

• Choose a buffer system to maintain a pH of 5.5.
  Describe its composition completely.
• select acid with pKa5.5. thus, look for Ka
  3.210-6
• H2C8H4O4, phthalic acid has Ka23.910-6. We
  could use NaHC8H4O4 and Na2C8H4O4.

43
Mixtures Of Acids

• Acids will not react with one another, but they
  each have a Ka reaction with water
• The most efficient reaction (higher Ka) occurs
  1st.
• Carry the common ion (H3O) concentration as a
  starting quantity in subsequent reactions
• Determine the pH at the end.
• Bases will not react with one another, but they
  each react with water according to their Kb
• The most efficient reaction (higher Kb) occurs
  1st.
• The common ion (OH-) concentration is a starting
  quantity in subsequent reactions
• Use the OH- to find pOH at the end, then find pH

Mixtures Of Bases
44
Learning Check

• Determine the pH of a mixture of 0.05M NH3
  (Kb1.810-5) and 0.2M NaF (Kb1.4710-11)

NH3 NaF ?NR
NH3 H2O(l) ? OH-(aq)
NH4(aq) 0.05 N/A 0 0 I -x N/A x x C 0.05
-x N/A x x E F-(aq) H2O(l) ?
OH-(aq) HF (aq) 0.2 N/A 9.4010-4 0 I -y -
y y y C (0.2-y)0.2 N/A (9.4010-4 y) y E
x9.4010-4
pOH3.03 pH10.97
y3.1310-9
45

• Acids that can donate more than one H to
  solution are called polyprotic acids
• Table 18.3 (and Appendix C) list the ionization
  constants of a number of polyprotic acids
• The ionization constants for these acids are
  numbered to keep tract of the degree of
  ionization
• Note that, for a given polyprotic acid, the
  magnitudes of the ionization constants are
  always Ka1 gt Ka2 (gt Ka3, if applicable)

46

• Suppose some H3PO4 is added to water, using the
  constants in Table 18.3

47

• The total H is then
• This is generally true when any polyprotic acid
  is added to water
• This greatly simplifies the determination of the
  pH in solutions of polyprotic acids

48
Learning Check

• What is the pH of a 0.1M solution of H3PO4?
  Ka17.110-3 Ka26.310-8 Ka34.510-13
• H3PO4(aq) H2O(l) ? H3O(aq) H2PO4-(aq)
• 0.1M N/A 0 0 I
• -x -x x x C
• (0.1-x) N/A x x E

x0.0233M
Since Ka1 gtgt Ka2, the amount of H3O contributed
by step 2 is negligible. pH1.63
49
Learning Check

• What is the concentration of all ions present in
  a 0.1M H2CO3 at equilibrium? Ka14.510-7
  Ka24.710-11
• H2CO3(aq) H2O(l) ? H3O(aq) HCO3-(aq)
• 0.1 N/A 0 0
• -x -x x x
• 0.1-x N/A x x
• 0.09978 N/A 2.1210-4 2.1210-4

x0.000212 M
50
Learning Check (Cont.)

• What is the concentration of all ions present in
  a 0.1M H2CO3 at equilibrium? Ka14.510-7
  Ka24.710-11
• HCO3- H2O(l) ? H3O(aq)
  CO32-(aq)
• 2.1210-4 N/A 2.1210-4
  0 I
• -y -y y y C
• 2.1210-4 -y N/A (2.1210-4 y) y E
• 2.1210-4 N/A 2.1210-4 4.710-11
• H2CO3 0.09978 HCO3-H3O 2.1210-4
  CO32-4.710-11

y4.710-11 M
51
Salts Of Polyprotic Acids

• Are amphoteric, and most are basic in water.
• To predict the behavior in water, compare the Ka
  to its Kb.
• Ex. Is an aqueous solution of NaHCO3 likely to be
  basic or acidic?
• NaHCO3 is a salt of neutral Na and amphoteric
  HCO3-.
• Ka4.710-11 Kb10-14/4.510-72.22 10-8.
• Since Kbgt Ka, the salt is basic in water

52
Learning Check

• What is the pH of a 0.1M solution of NaHCO3?
• Ka4.710-11 Kb10-14/4.510-72.22 10-8.
• HCO3- H2O(l) ? OH-(aq)
  H2CO3(aq)
• 0.1 N/A 0 0 I
• -x -x x x C
• 0.1 -x N/A x x E

8
x4.7 x 10-5
pOH4.33 pH9.67
53
Titration

• The controlled addition of one substance
  (titrant) to a known quantity of another
  substance (titrate) until the stoichiometric
  requirements are met
• Equivalence point - the volume of titrant needed
  to achieve an equimolar concentration of titrant
  and of titrate
• Endpoint - the volume of titrant necessary to
  achieve the stoichiometric ratio of reactants to
  change the indicators color
• The endpoint may be the last of several
  equivalence points in the case of polyprotic acids

54

• Titration of a strong acid by a strong base

Titration curve for the titration of 25.00 mL of
0.2000 M HCl (a strong acid) with the 0.2000 M
NaOH (a strong base). The equivalence point
occurs at 25.00 mL added base with a pH of 7.0
(data from Table 18.4).
55
Learning Check

• 25 mL of 2.5M HCl are titrated with 2.2M NaOH.
  What is the pH at 0mL?
• H3O1(aq) OH-(aq) ? 2H2O(l)

(25/25)2.5
(0/25)2.2
2.5
0
Neutral. N/A
None- no base yet. Treat as aqueous acid
solution. Here is strong acid. pH-log(2.5M)-.398
56
Learning Check (Cont.)

• 25 mL of 2.5M HCl are titrated with 2.2M NaOH.
  What is the pH at 15 mL?
• H3O1(aq) OH-(aq) ? 2H2O(l)

(25/40)2.5
(15/40)2.2
1.56
.825
-.825
-.825
.735
0
Here again, excess strong acid. pH-log(.735).134
57
Learning Check ( Cont. )

• 25 mL of 2.5M HCl are titrated with 2.2M NaOH.
  What is the pH at 30 mL? H3O1(aq) OH-(aq) ?
  2H2O(l)

(25/55)2.5
(30/55)2.2
1.14
1.2
-1.14
-1.14
.06
0
Here , excess strong base. pOH-log(.06)1.22 pH1
4-pOH12.78
58

• Titration of a weak acid by a strong base
• This can be divided into four regions
• Before the titration begins this is simple a
  solution of weak acid
• During the titration, but before the equivalence
  point the solution is a buffer
• At the equivalence point the solution contains a
  salt of the weak acid, and hydrolysis can occur
• Past the equivalence point the excess added OH-
  is used to determine the pH of the solution
• Data for the titration of acetic acid with sodium
  hydroxide is tabulated in Table 18.5

59

The titration curve for the titration of 25.00 mL
of 0.200 M acetic acid with 0.200 M sodium
hydroxide. Due to hydrolysis, the pH at the
equivalence point higher than 7.00 (data from
Table 18.5).
60

• Titration of a weak base by a strong acid
• This is similar to the titration of a weak acid
  by strong base
• Again dividing into four regions
• Before the titration begins this is a solution
  of a weak base in water
• During the titration, but before the equivalence
  point the solution is a buffer
• At the equivalence point the solution contains
  the salt of the weak base, and hydrolysis can
  occur
• Past the equivalence point excess added H
  determines the pH of the solution

61
Learning Check

• 25 mL of 2.5M HC2H3O2 are titrated with 3.2M
  NaOH. What is the pH at 0 mL?
• HC2H3O2(aq) OH-(aq) ? H2O(l) C2H3O2-(aq)

(25/25)2.5
(0/25)3.2
250
2.5
0
0
None- excess weak acid
62
Learning Check (Cont.)

• 25 mL of 2.5M HC2H3O2 are titrated with 3.2M
  NaOH. What is the pH at 0mL?(con)
• Write the Ka reaction for the weak acid
• HC2H3O2(aq) H2O(l) ? H3O(aq) C2H3O2-(aq)

0
0
2.5M
N/A
-x
x
-x
x
N/A
2.5-x
x
x
63
Learning Check (Cont.)

• 25 mL of 2.5M HC2H3O2 are titrated with 3.2M
  NaOH. What is the pH at 15 mL?
• HC2H3O2(aq) OH-(aq) ? H2O(l) C2H3O2-(aq)

(25/40)2.5
(15/40)3.2
2515
1.56
1.20
-1.20
-1.20
1.20
0
.36
1.20
We have a buffer mix, so use buffer equation
64
Learning Check (Cont.)

• 25 mL of 2.5M HC2H3O2 are titrated with 3.2M
  NaOH. What is the pH at 30 mL?
• HC2H3O2(aq) OH-(aq) ? H2O(l) C2H3O2-(aq)

(25/55)2.5
(30/55)3.2
2530
1.745
1.136
1.136
-1.136
-1.136
0
.609
1.136
We have excess strong base, so only consider
OH- pOH-log(.609). pH 14-(.2154)13.784
65

Titration curve for the titration of 25.00 mL of
0.200 M NH3 with 0.200 M HCl. The pH at the
equivalence point is below 7.00 because of the
hydrolysis of NH4.
66

• Titration curves for diprotic acids
• The features are similar to those for monoprotic
  acids, but two equivalence points are reached

The titration of the diprotic acid H2A by a
strong base. As each equivalence point is
reached, the pH rises sharply.
67
Reactions Of Acids And Bases

• Produce conjugate acid and bases
• May not create neutral products
• strong acids strong bases ? neutral salts
• weak acids strong bases ? basic salts
• strong acids weak bases ? acid salts
• weak acids weak bases ? amphoteric salts check
  Ka vs. Kb for dominance (we will not do titration
  calculations based on this type of reaction)
• Occur in order of strongest acid to weakest acid
  and strongest base to weakest base.

68
Predicting Effects Of Ions On pH

• Most metal cations are neutral or extremely weak
  acids
• Conjugate acids of weak bases are acidic
• Conjugate bases of weak acids are basic
• Conjugate bases of the strong acids are neutral
  (except HSO4- which is still fairly acidic)

69
Polyprotic Acids

• Have more than one ionizable hydrogen
• Each successive ionization has a specific
  ionization constant (Ka)
• Function as a mixture of acids the first
  ionization inhibits the second ionization, etc
• For H3PO4
• H3PO4(aq) H2O(l) ?H3O(aq) H2PO4-(aq) Ka,1
• H2PO4-(aq) H2O(l) ?H3O(aq) HPO42-(aq) Ka,2
• HPO42-(aq) H2O(l) ?H3O(aq) PO43-(aq) Ka,3