chapter 18: equilibria in solutions of weak acids and bases all weak acids behave the same way in...

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

AOHOHHA 32

][

]][[

][

]][[ 3

HA

AH

HA

AOHKa

• 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

aa KK logp

• Weak bases behave in a similar manner in water

• For the “general” base B:

][

]][[

isconstant ionization base the2

B

OHHBK

OHHBOHB

b

• 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:

bb KK logp

w

ba

b

a

KOHOH

A

OHHA

HA

AOHKK

A

OHHAKOHHAOHA

HA

AOHKAOHOHHA

]][[

][

]][[

][

]][[

isproduct the

][

]][[

:base conjugate for the

][

]][[

:acid weak For the

3

3

2

332

• 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

C)25(at 00.14ppp

and o

wba

wba

KKK

KKK

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.

• 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

• This, and the pH, are often used or requested in equilibrium calculations

%100literper available moles

literper ionized molesionization percentage

• 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?

ANALYSIS: The reaction can be represented as:

At equilibrium, [OH-] = x = 10-pOH

x x x -0.010 E

x x x - Cx-0.010

x 0 0 0.010 I

][

]][[ )()(

2

2

B

OHBHKOHaqBHOHaqB b

SOLUTION: Use pOH = 14.00 – pH, substituting:

%3.1

%100010.0

ionization %

and 5.80,p so 106.1

)103.1010.0(

)103.1(

010.0

then ,103.1

10][

6

4

242

4

)10.1000.14(

x

K

x

xK

M

OH

b

b

Determining how to proceed in acid-base equilibrium problems. The nature of the solute species determines how the problem is approached. This flowchart can help get you started in the right direction.

• 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

• Ions can also be acids or bases• For example, NH4

+ is a weak acid and NO2- a weak

base

• In salts both anions and cations are present, either of which could affect the pH

• These ions can arise from a number of sources

• Properties of their aqueous solutions can be summarized

• 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

• There are four possibilities when a salt is added to pure water:

1) Neither the anion nor cation affects the pH and the solution remains neutral. For example: NaCl

2) Only the cation is acidic, so the solution becomes acidic. For example: NH4Cl

3) Only the anion is basic, so the solution becomes basic. For example: NaNO2

4) 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

• The quadratic equation can be used to solve equilibrium problems when simplifying assumptions are invalid

• Simplifying assumptions fail, for example, when the initial concentration of a weak acid or base in pure water is less that 400 times the ionization constant

• Solving the problem using the quadratic equation is more time consuming, so it is worth checking before it is used

• Example: Calculate the pH of a 0.0010 M solution of dimethylamine for which Kb=9.6x10-4.

ANALYSIS: 400* Kb>0.0010 M, so use of the quadratic equation is indicated.

SOLUTION: Set the problem up

0.0010 E

)0010.0( C

][

]][[ 0 0 0.0010 I

106.9 )()()(

2

42

xxx

x

xxxx

B

OHBH

KaqOHaqBHOHaqB-

b

Put in standard form

Solve for x and the equilibrium concentrations

7422

2

2

106.9106.90010.00

)0010.0(

)0010.0(

xxKxKx

xKx

x

xK

bb

b

b

MMxB

MxOHBH

M

x

4

4

4

7244

109.3)0010.0(][

101.6][][

and 101.6

)1(2

)106.9)(1(4)106.9(106.9

so allowed, are solutions positiveOnly

• The method of successive approximations can also be used:

The method of successive approximations. Following these steps leads to a solution of equilibrium problems when the usual simplifying assumptions fail.

• This can be applied to the last problem:

444

2

442

42

104.1or 106.9)108.90010.0(

:for solve ion,approximat Second

108.9or 106.9)0010.0(

:for solve ion,approximatFirst

106.9)0010.0(

:expression thefrom Starting

xx

x

xx

x

x

xKb

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

converged Stop, 101.6 101.6

102.6 100.6 102.6 100.6

102.6 109.5 103.6 108.5

105.6 106.5 107.6 104.5

100.7 109.4 105.7 101.4

102.8 100.3 101.9 104.1

108.9 :)in values(all guess Starting

422

421

420

419

418

417

416

415

414

413

412

411

410

49

48

47

46

45

44

43

42

41

40

xx

xxxx

xxxx

xxxx

xxxx

xxxx

xM

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.

• 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

• 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

OHaqAaqOHaqHA 2)()()(

)()()( aqHAaqHaqA

• 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 text)

• 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)

Example: What is the pH of a buffer made by adding 0.10 mol NH3 and 0.11 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:

) (from 055.0L 0.2

mol 11.0][

050.0L 0.2

mol 10.0][

][

]][[ )()()(

44

3

3

4423

ClNHMNH

MNH

NH

OHNHKaqOHaqNHOHaqNH b

SOLUTION: Solve for [OH-] and use this to calculate the pH

• There are two important factors that determine the pH of a buffer solution

9.30pH and 4.70pOH

100.2050.0

)055.0(108.1][

)055.0(

])[050.0(108.1

][

]][[

55

5

3

4

OH

OH

NH

OHNHKb

• 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

mol

mol][or

][

][][

gives grearrangin

][

]][[ )()()(

A

HAKH

A

HAKH

HA

AHKaqAaqHaqHA

aa

a

• 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

][

][logppH

or ][

][logppH

acid

saltK

HA

AK

a

initial

initiala

• Typically, the weak acid is selected so the the desired pH is within one unit of the pKa

• A buffer’s capacity is determined by the magnitudes of the molarities of its components

• Generally, the pH change in an experiment must be limited to about

1ppH:buffer useful aFor aK

unit pH 1.0

Example: A buffer made from 0.10 mol HA (pKa=7.20) and 0.15 mol NaA in 2.0 L has 0.02 mol of HCl added to it with no volume change. What is the pH change?

ANALYSIS: This buffer problem is “best” solved in terms of moles. HCl is a strong acid. The H+

it contributes to solution increases the amount of HA present at the expense of A-.

SOLUTION: The pH before addition of HCl was:

38.7mol 10.0

mol 15.0log20.7

][

][logppH

HA

AKa

After the HCl ionizes and reacts:

• The pH change was greater than –0.1 unit. The buffer effectively resisted the pH change, however, because if the HCl had been added to pure water, the pH change would have been much larger:

15.038.723.7pHpHpH

is change pH theand ,23.7mol 0.12

mol 0.13log7.20pH

:issolution new theof pH the

mol 0.12mol )02.010.0(][

mol 0.13mol )02.015.0(][

initialfinial

final

final

HA

A

00.500.70.2

02.0logpH

• 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 > Ka2 (> Ka3, if applicable)

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

13

24

34

334

24

8

42

24

22442

3

43

4214243

105.4

][

]][[ )()(

103.6

][

]][[ )()(

101.7

][

]][[ )()(

HPO

POHKPOaqHaqHPO

POH

HPOHKHPOaqHaqPOH

POH

POHHKPOHaqHaqPOH

a

a

a

– 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

stepfirst total

step thirdstep secondstepfirst

321

step thirdstep secondstepfirst total

][][

that so ][][][

that followsit ,

Since

][][][][

HH

HHH

KKK

HHHH

aaa

Example: What is the pH and [CO32-] in 0.10 M

carbonic acid (H2CO3)?

ANALYSIS: Carbonic acid is a diprotic acid. The pH will depend on the [H+] generated from the first ionization. Ionization constants can be obtained from Table 18.3

SOLUTION: Solve the first ionization first, then substitute the results into the second reaction.

7

32

31332

103.4

][

]][[ )()()(

COH

HCOHKaqHCOaqHaqCOH a

– Applying the usual procedures:

– The pH is 3.68. Substituting these results into the second ionization equation:

][][101.2 and 10.0

thatso 400

34

2

1

1

HCOHMxx

K

Kx

a

a

11232

3

3

23

2233

107.4][

that so ][][but

][

]][[ )()()(

COK

HCOH

HCO

COHKaqCOaqHaqHCO

a

a

• The salts of polyprotic acids can affect the pH of a solution– For simplicity, only salts containing nonacidic

cations will be considered

• These salts produce basic solutions

• The reactions involved are a generalization of the hydrolysis reactions associated with the anions of monoprotic acids

• Consider the addition of sodium carbonate to water

– Sodium carbonate is the sodium salt of carbonic acid

– It is soluble in water

2 1

8

1 3

3 22

3 2 2 3

4

223

31

3 223

constants acid for the Like

10 3. 2] [

] ][ [

) ( ) ( ) (

10 1. 2] [

] ][ [

) ( ) ( ) (

b b

a

wb

a

wb

K K

K

K

HCO

CO H OHK

aq OH aq CO H O H aq HCO

K

K

CO

HCO OHK

aq OH aq HCO O H aq CO

– Simplifying assumptions like those made for the acid ionization can be made:

• The overall procedure for an acid-base titration was discussed in Section 5.13

• The titration stops at the end point, as indicated by a color change of an acid-base indicator

stepfirst total

step secondstepfirst

21

step secondstepfirst total

][][

and ][][

that followsit Since

][][][

OHOH

OHOH

KK

OHOHOH

bb

• If the indicator was properly selected, the end point is very near the equivalence point (when stoichiometric amounts of acid and bases have been added) for the system

• If the pH of the solution is plotted versus volume of titrant added, a titration curve is obtained

• The pH of the solution can easily be obtained using a pH meter

• Some of the more important types of titrations will be considered

• 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).

• Titration of a weak acid by a strong base– This can be divided into four regions

1) Before the titration begins: this is simple a solution of weak acid

2) During the titration, but before the equivalence point: the solution is a buffer

3) At the equivalence point: the solution contains a salt of the weak acid, and hydrolysis can occur

4) 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

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).

• 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

1) Before the titration begins: this is a solution of a weak base in water

2) During the titration, but before the equivalence point: the solution is a buffer

3) At the equivalence point: the solution contains the salt of the weak base, and hydrolysis can occur

4) Past the equivalence point: excess added H+ determines the pH of the solution

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

+.

• 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.

• A few general comments about indicators can be made– Most dyes that are acid-base indicators are

weak acids, which can be represented as HIn– The color change can be represented as:

color)(another color) (one

form base form acid

][

]][[ )()()(

HIn

InHKaqInaqHaqHIn HIn

– The color change will “appear” to the human eye near the equivalence point of the indicator

– At the equivalence point, the concentration of the acid and base form are equal, so that

– The best indicators have intense color(s) so only a small amount will produce an intense color change that is “easy” to see and won’t consume too much of the titrant

HInKppH point eequivalenc at the