Chapter 33 - Stephen F. Austin State Universityfaculty.sfasu.edu/janusama/powerpoint/chapter33.pdf33.1.2 pH of Weak Acids •Once you know the value of K a, you can calculate the equilibrium

1

Chapter 33

Acid-Base Equilibria

Copyright (c) 2011 by Michael A. Janusa, PhD. All rights reserved.

2

Solutions of a Weak Acid or Base

• The simplest acid-base equilibria are

those in which a single acid or base

solute reacts with water.

– we will first look at solutions of pure weak

acids or bases.

– Next, we will also consider solutions of

salts, which can have acidic or basic

properties as a result of the reactions of

their ions with water. Salts can be neutral,

acidic, or basic.

3

33.1 Acid-Ionization Equilibria

• Acid ionization (or acid dissociation) is

the reaction of an acid with water to

produce hydronium ion (hydrogen ion) and

the conjugate base anion (hydrolysis).

– Because acetic acid is a weak electrolyte, it ionizes to a

small extent in water, hence double arrow; not 100% like

strong acid which wrote single arrow.

(aq)OHC(aq)OH 2323

)l(OH)aq(OHHC 2232 <<100%

K<<1

4

Acid-Ionization Equilibria

• For a weak acid, the equilibrium concentrations

of ions in solution are determined by the acid-

ionization constant (also called the acid-

dissociation constant, Ka).

– Consider the generic monoprotic acid, HA.

)aq(A)aq(OH )l(OH)aq(HA 32

5

Acid-Ionization Equilibria

)(A)(OH )()( 32 aqaqlOHaqHA

– The corresponding equilibrium expression is:

][

]][[ 3

HA

AOHK

– Recall that the concentrations of pure solids and pure

liquids are not included in the K expression.

– however, since we are referring to acid

hydrolysis (acid + H2O) the constant is known

as Ka , the acid-ionization constant.

aK

7

Experimental Determination of Ka

• Nicotinic acid is a weak monoprotic acid with

the formula HC6H4NO2. A 0.012 M solution of

nicotinic acid has a pH of 3.39 at 25 C.

Calculate the acid-ionization constant for this

acid at 25 C.

– It is important to realize that the solution was made

0.012 M in nicotinic acid, however, some molecules

ionize making the equilibrium concentration of

nicotinic acid less than 0.012 M, just a small amount

less.

8

A Problem To Consider

Initial[] 0.012 0 0 Change[]

Equilibrium[]

– Let x be the moles per liter of product formed.

)()(OH )()( 24632246 aqNOHCaqlOHaqNOHHC

][

]][[

246

2463

NOHHC

NOHCOHKa

- + + x x x

0.012 - x x x

9


– We can obtain the value of x from the given

pH.

pHpHantiOHx 10)log(][ 3

10


– Note that

012.001159.0)00041.0012.0()x012.0(

the concentration of unionized acid remains

virtually unchanged. Important point to

remember for later.

– Substitute this value of x in our equilibrium expression.

522

104.1)01159.0(

)00041.0(

)012.0(

x

xKa

HW 28

ionization <<100%

K<<1

][

]][[

246

2463

NOHHC

NOHCOHKa

11

33.1.1 Percent Ionzation

• The degree of ionization of a weak

electrolyte is the fraction of molecules

that react with water to give ions. – Electrical conductivity or some other colligative

property can be measured to determine the degree

of ionization.

– With weak acids, the pH can be used to determine

the equilibrium composition of ions in the solution

because it gives the concentration of hydronium ions

at equil. pH of solution [H3O+]eq amount ionized

12

– To obtain the degree of ionization:

%4.3%1000.012

0.00041 ionization %

100% x ][

][ ionization %

x

acid

H

o

eq

– weak acids typically less than 5%. The lower the %ionized, less H+

formed, the lower Ka, weaker the acid. Ka is measure of strength of

acid or Kb if base. Strength based on ionization not pH directly.

Lower pH more acidic, does not mean stronger acid

note: % is unit

13

Ka is a measure of strength of acid. It indicates the amount ionized to form

hydronium ions. The more hydronium ions formed, naturally the more

acidic and lower pH. However, strength of acid is measure of amount

ionized, not conc of acid. Conc can vary which ultimately will vary the pH

of solution. To determine the strength of an acid you must compare Ka

values not pH of solution. pH is dependent on Ka as well as conc. of

solution. Must compare apples and apples.

10M HC2H3O2 Ka = 1.8 x 10-5 pH = 1.87

0.1M HF Ka = 6.9x10-4 pH = 2.08

Which stronger acid?

0.1M HC2H3O2 Ka = 1.8 x 10-5 pH = 2.87

0.1M HF Ka = 6.9x10-4 pH = 2.08

HF, larger Ka

14

33.1.2 pH of Weak Acids

• Once you know the value of Ka, you can

calculate the equilibrium

concentrations of species HA, A-,

and H3O+ for solutions of different

molarities. – The general method for doing this is same way we

did Kc or Kp except we can simplify the math

because of percent ionize is so small compared to

initial conc of acid; remember 0.012-x=0.012. This

wasn't the case for Kc or Kp problems so can't

neglect in those type problems

15

Calculations With Ka

• Note that in our previous example, the degree of

ionization was very small as compared to the

concentration of nicotinic acid.

• It is the small value of the degree of ionization that

allows us to ignore the subtracted x in the

denominator of our equilibrium expressions to

simplify our calculations.

• The degree of ionization of a weak acid depends on

both the Ka and the concentration of the acid solution

16

Calculations With Ka • How do you know when you can use

this simplifying assumption?

– then this simplifying assumption of ignoring the

subtracted x gives an acceptable error of less than 5%

otherwise must do quadratic formula. Another way

look at it factor of two - three difference in power of 10

is needed between Ka and conc of acid.

– It can be shown that if the acid concentration, Ca,

divided by the Ka exceeds 100, that is,

100[acid] aK

if

17

ex. Calculate the pH in a 1.00 M solution of nitrous acid, HNO2,

for which Ka = 6.0 x 10-4 at 25oC.

HNO2 + H2O H3O+ + NO2

-

[HNO2] [H3O+] [NO2

-] Initial, [ ]o 1.00 0 0 Change, D[ ]

Equilibrium, [ ]eq

- + + x x x

1.00 - x x x

=

acid, base, or salt? acid strong or weak? weak

Does answer make sense?

Yes, acid pH

18


• Really two equil: Kw as well but neglect most

of the time as long as [H+] is greater than

10-6 M from the acid.

• Really [H+]tot = [H+]acid +[H+]H2O (water so

small neglect)


-

H2O + H2O H3O+ + OH-

[H+]tot = [H+]HNO2 +[H+]H2O

[H+]tot = 0.024 M + <10-7M = 0.024 M

19

[HNO2] [H3O+] [NO2

-]

Initial, [ ]o 0.100 0 0

Change, D[ ]

Equilibrium, [ ]eq

- + + x x x

0.100 - x x x

acid, base, or salt? acid

strong or weak?

weak


Yes, acid pH

ex. Calculate the pH of 10.00 mL of 1.00 M solution of nitrous acid , HNO2, diluted

to 100.0 mL with water for which Ka = 6.0 x 10-4 at 25oC.


-

20


• What is the pH at 25 C of a 3.6 x 10-3 M

solution of acetyl salicylic acid (aspirin),

HC9H7O4? The acid is monoprotic and

• Ka=3.3 x 10-4 at 25 C.

acid, base, or salt? acid

strong or weak? weak

21


Initial, [ ]o 3.6 x 10-3 0 0

Change, D[ ] Equilibrium, [ ]eq

)()(OH )()( 47932479 aqOHCaqlOHaqOHHC

x x x

3.6 x 10-3 - x x x

- + +

22


– Note that

which is less than 100 or 10-3 vs 10-4 not factor 3

difference; therefore, won't work, so we must

solve the equilibrium equation exactly .

23

– If we substitute the equilibrium concentrations and the

Ka into the equilibrium constant expression, we get

Initial, [ ]o 3.6 x 10-3 0 0

Change, D[ ] -x +x +x

Equilibrium, [ ]eq 3.6 x 10-3 – x x x

)()(OH )()( 47932479 aqOHCaqlOHaqOHHC

24

– Rearranging the equation to put it in the form

ax2 + bx + c = 0

0)102.1(x)103.3(x642

– You can solve this equation exactly by using the

quadratic formula.

a2

ac4bbx

2

)1(2

)102.1)(1(4)103.3()103.3( 6244 x

Mx 4104.9

Mx 3103.1

25

– Taking the positive value and neglect water


MOHx 4

3 104.9][

– Now we can calculate the pH.

03.3)104.9log(pH4

HW 29

Does answer make sense? Yes, acid pH

][ 479

OHC

26

Calculate the pH of a solution that has a concentration of 1.0 x 10-8 M NaOH.

NaOH Na+ + OH-

1.0 x 10-8 M

pOH = - log [OH-] = - log 1.0 x 10-8 = 8.00

pH = 14.00 – pOH = 14.00 – 8.00 = 6.00

Does the answer make sense?

No, it is a base - what’s wrong?

The principle reaction is the water not the NaOH; therefore,

you can’t neglect the water. Must rework problem with

water equilibrium and include the additional base from

NaOH (pH = 7.02).

1.0 x 10-8 M

27

33.1.3 pH of Polyprotic Acids

• Some acids have two or more protons

(hydrogen ions) to donate in aqueous solution.

These are referred to as polyprotic acids.

– Sulfuric acid, for example, can lose two protons in

aqueous solution.

)aq(HSO)aq(OH)l(OH)aq(SOH 43242

)aq(SO)aq(OH )l(OH)aq(HSO2

4324

)aq(OH)aq(OH )l(OH)l(OH 322

[H3O+]tot = [H3O

+]H2SO4 + [H3O+]HSO4- + [H3O

+]H2O

Typically worry about principle rxn only

28

Polyprotic Acids

– For a weak diprotic acid like carbonic acid, H2CO3,

several simultaneous equilibria must be

considered.

)aq(HCO)aq(OH )l(OH)aq(COH 33232

)aq(CO)aq(OH )l(OH)aq(HCO2

3323

)aq(OH)aq(OH )l(OH)l(OH 322

[H3O+]tot = [H3O

+]H2CO3 + [H3O+]HCO3- + [H3O

+]H2O

< 5%

<< 5%

Once again, typically worry about principle rxn only.

i.e. NaHCO3 would use the second equation.

29

– Each equilibrium has an associated acid-ionization

constant.

Polyprotic Acids

– For the loss of the first proton

7

32

331a 103.4

]COH[

]HCO][OH[K

– For the loss of the second proton

11

3

2

332a 108.4

]HCO[

]CO][OH[K

)aq(HCO)aq(OH )l(OH)aq(COH 33232

)aq(CO)aq(OH )l(OH)aq(HCO2

3323

30

– In the case of a triprotic acid, such as H3PO4, the

third ionization constant, Ka3, is smaller than the

second one, Ka2.

Polyprotic Acids

– In general, the second ionization constant, Ka2, for

a polyprotic acid is smaller than the first ionization

constant, Ka1. Harder to pull proton off charged

species. Total H+ is sum of all equil H+ but usually

neglect all but the principal rxn.

Ka3 Ka2 Ka1

HPO42- H2PO4

- H3PO4

<< <<

31

– However, reasonable assumptions can be made

that simplify these calculations as we show in the

next example.

Polyprotic Acids

– When several equilibria occur at once, it might

appear complicated to calculate equilibrium

compositions.

32

– For diprotic acids, Ka2 is so much smaller than Ka1

that the smaller amount of hydronium ion produced in

the second reaction can be neglected.


• Ascorbic acid (vitamin C) is a diprotic acid,

H2C6H6O6. What is the pH of a 0.100 M

solution?

Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

(aq)(aq)OH )()( 666326662

OHHClOHaqOHCH

(aq)(aq)OH )()(2

66632666

OHClOHaqOHHC

33


(aq)(aq)OH )()( 666326662

OHHClOHaqOHCH

Initial, [ ]o 0.100 0 0

Change, D[ ] Equilibrium, [ ]eq

x x x

0.100 - x x x

- + +

][

]][[

6662

66631

OHCH

OHHCOHKa

34


52

109.7)100.0(

x

x

652 109.7)100.0()109.7( x

– Assuming that x is much smaller than 0.10

(1265>100), you get

55.2)0028.0log(pH

][

]][[

6662

66631

OHCH

OHHCOHKa

][][0028.0108.2 6663

3 OHHCOHMMx

35

– The ascorbate ion, C6H6O62-, which we will call y, is

produced only in the second ionization of H2C6H6O6.

A Problem To Consider • Ascorbic acid (vitamin C) is a diprotic acid,

H2C6H6O6. What is the pH of a 0.10 M

solution? What is the concentration of the

ascorbate ion, C6H6O62- in previous example?

The acid ionization constants are Ka1 = 7.9 x

10-5 and Ka2 = 1.6 x 10-12.

(aq)(aq)OH )()(2

66632666

OHClOHaqOHHC

from 1st equil some from 1st eq none to start

36

– Assume the starting concentrations for HC6H6O6- and

H3O+ to be those from the first equilibrium.

Initial, [ ]o 0.0028 0.0028 0

Change, D[ ] -y +y +y

Equilibrium, [ ]eq 0.0028-y 0.0028+y y

(aq)(aq)OH )()(2

66632666

OHClOHaqOHHC

][

]][[

666

2

66632

OHHC

OHCOHKa

(aq)(aq)OH )()( 666326662

OHHClOHaqOHCH0.0028 M 0.0028 M

from 1st equil

37

– Substituting into the equilibrium expression

12106.1

)0028.0(

)y)(0028.0(

12106.1

)y0028.0(

)y)(y0028.0(

– Assuming y is much smaller than 0.0028 (1.75 x 109 >

100), the equation simplifies to

– Hence, My 122

666 106.1]OHC[

– typically, the concentration of the anion ion

equals Ka2 if starting with reactants of first equil. HW 30

[H3O+]tot = 0.0028 MKa1 + 1.6 x 10-12 Ka2 + <1 x10-7 H2O = 0.0028 M

38

33.2 Base-Ionization Equilibria

• Equilibria involving weak bases are

treated similarly to those for weak acids.

– Ammonia, for example, ionizes in water as

follows.

)aq(OH)aq(NH )l(OH)aq(NH 423

][

]][[

3

4

NH

OHNHK

39

Base-Ionization Equilibria

– The corresponding equilibrium expression is:

– Recall that the concentrations of pure solids and pure

liquids are not included in the K expression.

– however, since we are referring to base

hydrolysis (base + H2O) the constant is known

as Kb , the base-ionization constant.

bK


40

Base-Ionization Equilibria • Equilibria involving weak bases are treated similarly to those for

weak acids. Larger Kb , the stronger base, ionize more OH-,

more basic.

– In general, a weak base B with the base ionization

)aq(OH)aq(HB )l(OH)aq(B 2

has a base ionization constant equal to

]B[

]OH][HB[Kb

HW 31

41

A Problem To Consider • What is the pH of a 0.20 M solution of

pyridine, C5H5N, in aqueous solution?

The Kb for pyridine is 1.4 x 10-9.

1. Write the equation and make a table of

concentrations (ICE).

2. Set up the equilibrium constant expression.

3. Solve for x = [OH-].

– As before, we will follow the three steps in

solving an equilibrium.

acid, base, or salt? base strong or weak? weak

42


)aq(OH)aq(NHHC )l(OH)aq(NHC 55255

[C5H5N] [C5H5NH+] [OH-]

Initial, [ ]o 0.20 0 0

Change, D[ ]

Equilibrium, [ ]eq x x x

0.20 - x x x

- + +

43



[C5H5N] [C5H5NH+] [OH-]

Initial, [ ]o 0.20 0 0

Change, D[ ]


0.20 - x x x

- + +

44


– Note that

which is much greater than 100 (also factor of 8

difference between the powers of 10 of Kb [2.0 x 10-1]

and concentration of base [1.4 x 10-9]), so we may use

the simplifying assumption that (0.20-x) (0.20).

89 104.1

104.120.0

b

b

KC

45



[C5H5N] [C5H5NH+] [OH-]

Initial, [ ]o 0.20 0 0

Change, D[ ]


0.20 - x x x

- + +

46


– Solving for pOH

77.4)107.1log(]log[ 5 OHpOH

23.977.400.1400.14 pOHpH

– Since pH + pOH = 14.00

HW 32

Does answer make sense? Yes, base pH

or combine equations into one:

47

33.3 Acid-Base Properties of Ions and

Salts • One of the successes of the Brønsted-

Lowry concept of acids and bases was

in pointing out that some ions can act

as acids or bases.

– A 0.1 M solution has a pH of 11.1 and is therefore

fairly basic.

– Consider a solution of sodium cyanide, NaCN.

)aq(CN)aq(Na)s(NaCN

OH 2

Cations: neutral or acidic Anion: neutral or basic

48

– Sodium ion, Na+, is unreactive with water,

but the cyanide ion, CN-, reacts to produce

HCN and OH- making it a basic solution

)aq(OH)aq(HCN )l(OH)aq(CN 2

Acid-Base Properties of a Salt Solution

– From the Brønsted-Lowry point of view, the

CN- ion acts as a base, because it accepts

a proton from H2O.

49

– You can also see that OH- ion is a product, so you

would expect the solution to have a basic pH. This

explains why NaCN solutions are basic.


– The reaction of the CN- ion with water is referred to

as the hydrolysis of CN-.


NaCN is a basic salt.

50

• The hydrolysis of an ion is the reaction of an

ion with water to produce the conjugate acid

and hydroxide ion or the conjugate base

and hydronium ion.

– The CN- ion hydrolyzes to give the

conjugate acid and hydroxide.



51

– The hydrolysis reaction for CN- has the form of a base

ionization so you write the Kb expression for it.



][

]][[

CN

OHHCNK b

Note: need Kb of CN- but probably given Ka HCN (ionization

constant tables contain molecular species), will need to convert.

52

– The NH4+ ion hydrolyzes to the conjugate

base (NH3) and hydronium ion.


Acid-Base Properties of a Salt

Solution • An example of an cation that undergoes hydrolysis is

the ammonium ion

)()( 44 aqClaqNHClNH

– A 0.1 M solution has a pH of 5.1 and is therefore acidic.

][

]][[

4

33

NH

OHNHKa

53

33.3.1 Salts: Acidic, Basic, or Neutral

• How can you predict whether a

particular salt will be acidic, basic, or

neutral and undergoes hydrolysis?

– The Brønsted-Lowry concept illustrates the

inverse relationship in the strengths of conjugate

acid-base pairs.

– Consequently, the anions of weak acids are

good proton acceptors.

– Anions of weak acids are basic.

54

Predicting Whether a Salt is Acidic,

Basic, or Neutral – On the other hand, the anions of strong acids (good

proton donors) have virtually no basic character, that is,

they do not hydrolyze. Therefore, conjugate base of strong

acids have do not undergo hydrolysis - neutral.

– For example, the Cl- ion, which is conjugate to the strong

acid HCl, shows no appreciable reaction with water. If did,

strong acid breaks up 100% and hydroxide produced

cancels with hydronium of strong acid and make neutral

solution

OHHCllOHaqCl

CllOHaqHCl

)()(

OH)()(

2

32

OH3Cl

reaction no)()(

,

2 lOHaqCl

essencein

55


Basic, or Neutral – Small highly charged (attract O more weakens O-H bond)

cations have an attraction for the O on water and cause the

production of H+ while large low charged cations do not.

– For example, reaction no)l(OH)aq(Na 2

In summary,

If cation is that of the strong base (Na+, K+, Li+, Sr2+, Ba2+, Ca2+),

it will have neutral hydrolysis; otherwise, cation will have acidic

hydrolysis

If anion of monoprotic strong acid (NO3-, Cl-, Br-, I-, ClO4

-), it

will have neutral hydrolysis; otherwise, anion will have basic

hydrolysis

Low charged large ions are group I and II metals (cations of

strong bases) and do not undergo hydrolysis - neutral.

56

• Salts may be acidic/basic/neutral and are composed of cations (positive ions) and anions (negative ions)

cation anion

Acidic or neutral basic or neutral

Cations of strong bases Anions of monoprotic strong

are neutral; otherwise, acids are neutral; otherwise,

cation contributes acidity anion contributes basicity

Na+, K+, Li+, Ca2+, Sr2+, Ba2+ Cl-, Br-, I-, NO3-, ClO4

- add

add neutral; rest of cations add neutral; rest of anions add

acidity to salt basicity to salt

57


Basic, or Neutral

• To predict the acidity or basicity of a

salt, you must examine the acidity or

basicity of the ions composing the salt.

Depending on the two components (cation/anion) the overall salt will be acidic/neutral/basic :

nn n

an a

nb b

ab depends on which is larger Ka or Kb

58


Basic, or Neutral

)()( )( 44 aqClaqNHsClNH

acid neutral acidic salt

pH < 7

)()()( aqCNaqNasNaCN neutral basic basic salt

pH > 7

KC2H3O2(s) K+(aq) + C2H3O2-(aq)

neutral basic basic salt

pH > 7

59

Predicting Whether a Salt is Acidic, Basic, or Neutral • AlCl 3

• Zn(NO3)2

• KClO4

• Na3PO4

• LiF

• NH4F (Ka = 5.6x10-10 > Kb = 1.4x10-11)

• NH4ClO (Ka = 5.6x10-10 < Kb = 3.6x10-7)

• NH4C2H3O2 (Ka = 5.6x10-10 = Kb = 5.6x10-10)

HW 33

acid neutral acidic

acid neutral acidic

neutral neutral neutral

neutral basic basic

neutral basic basic

acid basic acidic

acid basic basic

acid basic neutral

60

33.3.2 pH of Salt Solution • To calculate the pH of a salt solution

would require the Ka of the acidic cation

or the Kb of the basic anion.

– The ionization constants of ions are not listed

directly in tables because the values are easily

related to their conjugate species.

– Thus the Kb for CN- is related to the Ka for HCN.

61

The pH of a Salt Solution

• To see the relationship between Ka and Kb for

conjugate acid-base pairs, consider the acid

ionization of HCN and the base ionization of

CN-.

)aq(CN)aq(OH )l(OH)aq(HCN 32


Ka

Kb

)()( )()( 322 aqOHaqOHlOHlOH Kw

62

)aq(CN)aq(OH )l(OH)aq(HCN 32


Ka

Kb

– Therefore,

)()( )O( )( 322 aqOHaqOHlHlOH Kw

CxKKK o

wba [email protected] 14

HW 34

– When two reactions are added, their equilibrium

constants are multiplied.

Calculate the Ka of NH4+ given the Kb of NH3 is 1.8 x 10-5 at 25oC

63

The pH of a Salt Solution

• For a solution of a salt in which only

one ion hydrolyzes, the calculation of

equilibrium composition follows that of

weak acids and bases.

– The only difference is first must obtain the

Ka or Kb for the ion that hydrolyzes.

64

Attack of hydrolysis problem

When attacking pH (hydrolysis problems) always ask yourself

1) reaction or single solute

2) acid, base, or salt

3) if acid or base, is it strong (straight forward) or weak (Ka or Kb

problem)

if salt, write dissociation eq and examine cation/anions involved to

determine if acid or base hydrolysis is available then Ka or Kb

problem.

single solute

reaction neutralization

acid

base

salt

strong

strong

weak

100%

Ka

Kb

100%

weak

soluble

insoluble

cation

anion

cation SB

neutral

otherwise Ka

Anion of monoprotic SA

otherwise Kb

neutral

Ksp

[CN-] [HCN] [OH-]

Initial, [ ]o 0.10 0 0


Equilibrium, [ ]eq 0.10 - x x x

65

A Problem To Consider • What is the pH of a 0.10 M NaCN solution

at 25 C? The Ka for HCN is 4.9 x 10-10. acid, base, or salt? basic salt soluble or insoluble? soluble

NaCN(s) Na+(aq) + CN-(aq)

0.10 M 0.10 M

CN- (aq) + H2O(l) HCN(aq) + OH-(aq)

66



Yes, basic salt

67

example: What is the pH of 0.10 M NH4I solution?

Kb NH3 = 1.8 x 10-5

HW 35

acid, base, or salt? acidic salt

soluble or insoluble? soluble

NH4I(s) NH4+(aq) + I-(aq)

0.10 M 0.10 M

NH4+ (aq) + H2O(l) H3O

+(aq) + NH3 (aq)

[NH4+] [H3O

+] [NH3]

Initial, [ ]o 0.10 0 0



Yes, acidic salt


68

33.4 The Common Ion Effect

• The common-ion effect is the shift in

an ionic equilibrium caused by the

addition of a solute that provides an ion

common to the equilibrium.

(aq)OHC(aq)OH 2323

)l(OH)aq(OHHC 2232 Initial: 0 0

pH of acetic acid is lower than the pH of acetic

acid/sodium acetate because of the common ion effect.

Add NaC2H3O2

pH will increase

69

The Common Ion Effect

– The equilibrium composition would shift to the left

and the degree of ionization of the acetic acid is

decreased, meaning less hydronium formed and

higher pH of solution as compared to pure acid.

(aq)OHC(aq)OH 2323

)l(OH)aq(OHHC 2232

– This repression of the ionization of acetic acid by

sodium acetate is an example of the common-ion

effect. Another source of ion coming from an

additional solute added to the solution.

)()( 232232 aqOHCaqNaOHNaC

pH x x

y M

+ y M

70

33.4.1 Buffers • When common ion effect involves weak acid/conjugate base (or

vice versa), it is referred to as a buffer. Buffer is a solution

characterized by the ability to resist changes in pH when limited

amounts of acid or base are added to it.

– Buffers contain either a weak acid and its conjugate base or a

weak base and its conjugate acid. Typically conj is in form of

salt. The components differ by one proton.

– Thus, a buffer contains both an acid species and a base species in

equilibrium which allows it to absorb any added acid or base and

keep pH of solution relatively constant

HC2H3O2 / NaC2H3O2

HCN / NaCN

H2CO3 / KHCO3

NH4Cl / NH3

H3PO4 / Na2HPO4 buffer?

HCl / NaCl buffer?

No, differ by 2 protons

No, strong acid

71

Buffers – Consider a buffer with equal molar amounts of HA

and its conjugate base A-.

)()( 32 (aq)A(aq)OHlOHaqHA

)()( aqAaqNaNaA

If add small amount acid – react with acid or base component?

)( )( 23 aqHAlOH(aq)OH(aq)A

neutralization

base

replace consumed A- and maintain pH

(slight decrease in pH due to additional hydronium)

If add small amount of base – reacts with acid component , produce A-, and

shifts to left replace consumed HA and maintain pH

(slight increase in pH due to decrease in hydronium)

72

An aqueous solution is 0.25 M in formic acid, HCHO2 and 0.18 M in

sodium formate, NaCHO2. What is the pH of the solution. The Ka

for formic acid is 1.7 x 10-4.

NaCHO2(s) Na+(aq) + CHO2-(aq)

0.18 M 0.18 M

HCHO2 (aq) + H2O(l) H3O

+(aq) + CHO2- (aq)

neutralization or buffer? buffer

[HCHO2] [H3O+] [CHO2

-]

Initial, [ ]o 0.25 0 0.18


Equilibrium, [ ]eq 0.25 - x x 0.18 +x

buffer, similar conc

can neglect x

Note: 0.25 M HCHO2

pH = 2.18

0.25 M HCHO2/0.18 M NaCHO2

pH = 3.63

73

A solution is prepared to be 0.15 M HC2H3O2 and 0.20 M NaC2H3O2.

What is the pH of this solution at 25oC? Ka HC2H3O2 @ 25oC = 1.8 x 10-5

neutralization or buffer? buffer NaC2H3O2 (s) Na+(aq) + C2H3O2

-(aq)

0.20 M 0.20 M

HC2H3O2 (aq) + H2O(l) H3O

+(aq) + C2H3O2 - (aq)

[HC2H3O2 ] [H3O+] [C2H3O2

-]

Initial, [ ]o 0.15 0 0.20



Note: 0.15 M HC2H3O2

pH = 2.78

0.15 M HC2H3O2 /0.18 M NaC2H3O2

pH = 4.87

74

Calculate the pH of 75 mL of a buffer solution (0.15 M HC2H3O2 / 0.20 M

NaC2H3O2 pH = 4.87) to which 9.5 mL of 0.10 M HCl is added. Ka HC2H3O2

@ 25oC = 1.8 x 10-5

Step 1: Calculate the initial mmols of each component

Step 2: Neutralization reaction

NaC2H3O2(aq) + HCl(aq ) HC2H3O2(aq) + NaCl (aq)

mmolo 15.00 0.95 11.25 0

Dmmol

mmolneu

Step 3: New concentrations for species affecting pH

basic salt weak acid neutral salt

-0.95 -0.95 +0.95 +0.95

14.05 0.00 12.20 0.95

75


Step 4: Hydrolysis

NaC2H3O2 (s) Na+(aq) + C2H3O2

-(aq)

0.166 M 0.166 M


+(aq) + C2H3O2 - (aq)

[HC2H3O2] [H3O+] [C2H3O2

-]

Initial, [ ]o 0.144 0 0.166



76

What was the change in pH to the buffer by additional HCl?

Calculate the pH if we added the same 9.5 mL of 0.10 M HCl to 75 mL of pure H2O.

HCl (aq) H+(aq) + Cl -(aq)

0.0112 M 0.0112 M

buffer holds pH relatively

constant when small amounts of

acid or base or added

HW 36

77

Buffers – Two important characteristics of a buffer are its

buffer capacity and its pH.

– Buffer capacity depends on the amount of acid

and conjugate base present in the solution;

typically, 1 pH unit.

– The next topic illustrates how to calculate the pH

of a buffer by special eq but suggest you do it just

like we did to keep with thought process and not

memorize a equation.

78

The Henderson-Hasselbalch Equation • How do you prepare a buffer of given pH?

– A buffer must be prepared from selecting a conjugate

acid-base pair in which the Ka of the acid is

approximately equal to the desired H3O+

concentration.

– Note: you can add extra conjugate base, higher pH, or

less base, lower pH, to get to exact pH wanted.

79

The Henderson-Hasselbalch Equation

– The acid ionization constant is:

– By rearranging, you get an equation for the H3O+

concentration.

]A[

]HA[K]O[H a3

]HA[

]A][OH[K 3

a

)aq(A)aq(OH )l(OH)aq(HA 32

– Consider a buffer of a weak acid HA and its

conjugate base A-.

80

The Henderson-Hasselbalch Equation – Taking the negative logarithm of both sides of the

equation we obtain:

)][

][log()Klog(]Olog[H-

)][

][Klog(]Olog[H-

a3

a3

A

HA

A

HA

– The previous equation can be rewritten

]HA[

]A[logKppH a

Previous example: An aqueous solution is 0.25 M in formic acid,

HCHO2 and 0.18 M in sodium formate, NaCHO2 , had pH 3.63. The

Ka for formic acid is 1.7 x 10-4.

81

The Henderson-Hasselbalch Equation – More generally, you can write

– This equation relates the pH of a buffer to the

concentrations of the conjugate acid and base. It is known

as the Henderson-Hasselbalch equation (for buffers

only). Also way to work previous buffer problem. Note

pH = pKa when conc base=acid

],[

],[logKpH a

HAacid

Abasep

82

• Let’s prepare a buffer with pH of 4.90 – So to prepare a buffer of a given pH (for example, pH 4.90)

we need a conjugate acid-base pair with a pKa close to the

desired pH or 10-4.90 to find Ka close to 1.3 x 10-5 (desired

hydronium conc. equivalent to pH desired)

HC2H3O2 Ka = 1.8 x 10-5

H2CO3 Ka = 4.3 x 10-7

HNO2 Ka = 4.5 x 10-4

– The Ka for acetic acid is 1.8 x 10-5, and its pH/pKa

is 4.74 (calculated using 1.8 x 10-5 as [H3O+]) if

equal conc of acid/salt are used.

– You could get a buffer of pH 4.90 by increasing the

ratio of [base]/[acid], meaning more basic salt.

want Ka close to [H3O+]

83

What if I wanted to use 0.100 M HC2H3O2 (Ka = 1.8 x 10-5), what

concentration of NaC2H3O2 would I need to obtain a buffer with pH =4.90.

][][14.0100.045.1

][100.010

100.0

][10

100.0

][log16.074.490.4

100.0

][log74.490.4

],[

],[logKpH

232232

232

16.0

23216.0

232

232

a

OHNaCOHCM

OHC

OHC

OHC

OHC

HAacid

Abasep

HW 37


+(aq) + C2H3O2 - (aq)

pKa = -log 1.8 x 10-5 = 4.74

84

33.5 Acid-Base Titrations or Mixtures • 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 titration curves to choose an

appropriate indicator that will show when the

titration is complete.

3232 NaHCOOHNaOHCOH

3223 CONaOHNaOHNaHCO

32232 2 2 CONaOHNaOHCOH

___________________________ HCl + NaOH NaCl + H2O

titrant

Figure: Curve for the titration of a

strong acid by a strong base.

85

end pt

end pt

Ebbing, D. D.; Gammon, S. D. General

Chemistry, 8th ed., Houghton Mifflin,

New York, NY, 2005.

86

• Figure shows a curve for the titration of HCl with NaOH.

– Note that the pH changes slowly until the titration

approaches the equivalence point.

– The equivalence point is the point in a titration when

a stoichiometric amount of reactant has been added;

contrary to end point (observable point).

– At this equivalence point of SA and SB, the pH of the

solution is 7.0 because it contains a neutral salt, NaCl,

that does not hydrolyze. NaOH + HCl NaCl + H2O

– To detect the end point, you need an acid-base

indicator that changes color beyond equivalence

point (titration error).

– Phenolphthalein can be used because it changes

color in the pH range 8.2-10.

87

Previous example equivalent pt was at 7 but not all neutralization reactions

eq pts are at 7. During a titration or just mixing an acid and base together to

make a solution, the resulting solution will contain a salt and water

Acid + Base salt + water

One of the chemical properties of acids and bases is that they neutralize one

another; however, that doesn’t mean that the product will be neutral.

The misconception is that if the acid and base are in stoich proportions that

the resulting solution is neutral. This is not true. The salt formed may be a

acidic, basic, or neutral salt and will dictate the pH of the solution. Common

sense can help you predict the pH of a mixture.

SA + SB

WA + SB

SA + WB

neutral salt only true neutralization pH =7

basic salt original acid and base neutralize but

product has basic properties and basic pH

acidic salt original acid and base neutralize but

product has acidic properties and acidic pH

88

When working acid-base titration problems or mixtures treat

them all the same but must be able to examine the components

remaining after neutralization to determine what affects pH or not

(strong, weak, salts). They are all done the same. The problem will

dictate where you go. We can derive titration curve by doing a

calculation at different points and draw curve or measure with pH

meter to measure and draw curve.

1.) Calculate mmols initially

2.) write the neutralization reaction and find mmols of

components left after neutralization

3.) find new conc. of species that affect pH

4.) do hydrolysis of main species that affect pH (like earlier in chapter)

89

A Problem To Consider • Calculate the pH of a solution in which 50.00 mL of 1.000 M

NaOH is added to 50.00 mL of 1.000 M HCl.


HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)

Step 2: neutralization reaction

mmolo 50.00 50.00 0

Dmmol -50.00 -50.00 +50.00

mmolneu 0.00 0.00 50.00


neutral salt

Sodium Chloride is a neutral salt; therefore, no further calculations are needed,

pH = 7.00. Note: this only happened because we are mixing the stoichiometric

amounts of a strong acid and a strong base. If we mix a weak and a strong species,

the equivalence point will not be at 7 as we will see in later examples.

mmolo 50.00 49.99 0

Dmmol -49.99 -49.99 +49.99

mmolneu 0.01 0.00 49.99

90

• Calculate the pH of a solution in which 49.99 mL of 1.000 M






neutral salt Strong acid

Step 4: Hydrolysis

HCl (aq) + H2O (l) H3O

+ (aq) + Cl- (aq)

mmolo 50.00 50.01 0

Dmmol -50.00 -50.00 +50.00

mmolneu 0.00 0.01 50.00

91

• Calculate the pH of a solution in which 50.01 mL of 1.000 M


HW 38





neutral salt Strong base

Step 4: Hydrolysis

NaOH(aq) Na+ (aq) + OH- (aq)

mmolo 50.00 20.00 0

Dmmol -20.00 -20.00 +20.00

mmolneu 30.00 0.00 20.00

92

• Calculate the pH of a solution in which 20.00 mL of 1.000 M NaOH is

added to 50.00 mL of 1.000 M HC2H3O2. Ka HC2H3O2= 1.8x10-5




Basic salt Weak acid

HC2H3O2 (aq) + NaOH (aq) NaC2H3O2 (aq) + H2O (l)

BUFFER

93

Step 4: Hydrolysis

NaC2H3O2 (s) Na+ (aq) + C2H3O2- (aq)

0.2857 M 0.2857 M

HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2

- (aq)

[HC2H3O2] [H3O+] [C2H3O2

-]

Initial, [ ]o 0.4286 0 0.2857



mmolo 50.00 50.00 0

Dmmol -50.00 -50.00 +50.00

mmolneu 0.00 0.00 50.00

94






Basic salt


95

Step 4: Hydrolysis

NaC2H3O2 (s) Na+ (aq) + C2H3O2- (aq)

0.5000 M 0.5000 M

C2H3O2- (aq) + H2O (l) HC2H3O2(aq) + OH- (aq)

[C2H3O2-] [HC2H3O2] [OH-]

Initial, [ ]o 0.5000 0 0



Notice at eq pt and pH is not 7

because weak/strong

mmolo 50.0 51.0 0

Dmmol -50.0 -50.0 +50.0

mmolneu 0.00 1.0 50.0

96






Basic salt


Strong base [OH-] from basic salt is extremely small as

compared to strong base ; therefore, neglect

WA/conj basic salt – buffer

conj acidic salt/WB - buffer

SA/WA

SB/WB

SA/SA – add [H3O+]

SB/SB - add [OH-]

acid/base -neutralization rxn

97

HW 39

Step 4: Hydrolysis

NaOH (aq) Na+ (aq) + OH- (aq)

Notice the pH - 9.22 and [OH-] - 1.7 x 10-5 M changed tremendously compared to the

previous example. This indicates that neglecting the contribution of NaC2H3O2 would

not affect the overall answer.

Documents

Chapter 33 - Stephen F. Austin State Universityfaculty.sfasu.edu/janusama/powerpoint/chapter33.pdf33.1.2 pH of Weak Acids •Once you know the value of K a, you can calculate the equilibrium