1422 chapt 18 acids bases

8/10/2019 1422 Chapt 18 Acids Bases

1/74

Chapters 18 & 19

Acids&BasesH2O + HF(aq) H3O

+(aq) + F(aq)


2/74

Acids & Bases 2

Acid andBaseDefinitions

Arrhenius

Acid increases H+ concentration

Base increases OH-concentration

Brnsted-Lowry (1923)

Acid donates a H+

Base combines with or accepts H+

Lewis

Acid electron acceptor

Base electron donor


3/74

Acids & Bases 3

Water itself has some ionic character:

H2O(l)+H2O(l) H3O+(aq)+ OH(aq)

Shorthand:

H2O(l) H

+

(aq)+ OH

(aq)This is calledself-ionization. Although most chemists simplywrite H+, it is important to realize that H+by itself representsa naked proton. In water the H+is hydrated, that is, it formsionic-diople interactions with other water molecules. A

common way of more accurately representing the fact that it

is hydrated (interacting with waters) is to write H3O+.


4/74

Acids & Bases 4

H2O(l) H+(aq) + OH(aq)

+

2

eqOHH[ [

KO

] ]

[H ]

=

But H2Ois a pure liquid and its concentration isconstant ([H2O] = 55.6M, activity = 1), so it is notincluded in the equilibrium expression:

14

+

@ 20 C

K [ ] [ ]

K 1.0 10

H

HO =

=

w

w

[H+][OH]= 11014

[H+]=[OH]= 1107M

So the [H+] and [OH] concentration in pure water

is 1

10

7MWe use Kwto indicate the water self-or auto-

ionization. This is still a Keqor Kc. Chemists use

many subscripts on the equilibrium constant K to

indicate specific types of equilibria:

Ka= acid equilibria Kb= base equilibriaKsp= slightly soluble (solubility product)equilibria

Because of the often very small nature of H+concentrations, chemists (and others) have devised


5/74

Acids & Bases 5

a logarithmic scaleto simplify expressing these

values:

pH = log [H+]

The negative sign in front of the log makes surethat most small concentrations of acid are given by

positivevalues.

[H+] = 1 107 M

pH= log(1 107) = 7.0

The greater the [H+]concentration, theLOWERthepHvalue!!!

pH < 7.0 Acidic

pH = 7.0 NeutralpH > 7.0 Basic (Alkaline)


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Acids & Bases 6

Substance pH

10 M HCl1.0

1 M HCl 0.0

Stomach Acid (HCl) 1.4

Lemon Juice 2.1

Orange Juice 2.8Wine 3.5

Black Coffee 5.0

Urine 6.0

Pure Water 7.0

Blood 7.4

Baking Soda Solution 8.5

Ammonia Solution 11.9

1 M NaOH 14.0

10 M NaOH 15.0


7/74

Acids & Bases 7

We can also define:

pOH = log [OH]

Although most chemists mainly usepH, pOHcanbe useful in base equilibrium numerical problems

that we will run into later.

Another definition we use is:

pKw =

log Kw = 14So, for a given water solution:

pOH+pH = pKw= 14or:

pOH= 14 pH

pH= 14 pOH


8/74

Acids & Bases 8

Example: ThepHof wine is 3.5. What is the[H+]? What is the [OH]? What is thepOH?

pH =

log[H+]= 3.5[H+]= antilog(3.5) = 103.5= 3.16 104M

HOH

K[ ]

1411

4

1 103.1 10

3.16 10

= = =

+[ ]

w

pOH =

log[OH

]=

log(3.1

10

11) = 10.5---- or ----

pOH = pKw pH = 14 3.5 = 10.5

Problem: ThepHin your stomach is around 1.What is the [H+]? What is the [OH]? What is thepOH?


9/74

Acids & Bases 9

Dissociation Equilibrium Constants

An acidis a compound that will ionizein solution

(usually water) to form a H+(aq) and a counter-anion. This can be writen in a general fashion as:

HA(aq) H+(aq) + A(aq)

The equilibrium expression for this reaction is:

[ ] [A ]HK H[ ]A

=

+

eq

The equilibrium constant, Keq, is often given a

special name for acids: Kaor the acid dissociation

constant. The largerthe Kavalue the moreH+is

being produced (the lowerthe correspondingpH),therefore, the strongerthe acid!!!!

Just as withpHwe can also definepKaas:

pKa = log Ka

The smaller(and more negative) thepKathestrongerthe acid!!!! This can be confusing, but is

very important: pKas are commonly used in

biology & chemistry!


10/74


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Acids & Bases 11

Strong Acids

These are acids that are essentially completely

dissociated in solution (usually water).HCl(aq) H+(aq) + Cl(aq)

C[ ] [ ]K

[ ]H

l

Cl

H 810

=

+

a

In general, Ka> 1 for a strong acid(although there

is no firm dividing line!)

The common strong acidsthat you are expected to

know are:

HCl, HBr, HI (the hydrohalic acids)

H2SO4(sulfuric acid) HNO3(nitric acid)

HClO4(perchloric acid)HCl, H2SO4,andHNO3are often referred to asmineral acids.

A strong acid completely dissociates the first H+, so

the H+concentration is the same as the given acidconcentration. thepHof a strong acid is just

log of the acid concentration (= log[H+]).


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Acids & Bases 12

Problem: What are thepH's of the followingsolutions?

a) 0.001 M HCl

b) 0.1 M HNO3

c) 1 105M H2SO4

d) 10 M HBr

e) 1 M HI

f) 0.1 M HF

g) 0.01 M HCl

h) 1 104M HNO3

i) 1 M acetic acidj) 1 1014M HCl


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Acids & Bases 13

Explanation forj)on the last page:

Theactualproton concentration for any acid

dissolved in water is more precisely defined as:

[H+]total = [H+]acid + [H+]waterNormally, [H+]wateris 1 10

7Mand is much less

than [H+]acid, so that we usually ignore it.

BUT, in this example, [H+]acidturns out to beonly 1 1014M, which ismuch, much lessthan

the [H+]water.

So in this example we can actually ignore the tiny

amount of H+contributed from the strong acid andonly consider the H+naturally present in water:

This will work for acid concentrations of 1 108Mand

lower. It gets complicated mathematically right around1 107M(not dealt with in this course).


14/74

Acids & Bases 14

Lewis Dot Structures: Wheres the Proton?

When we write the formula for HNO3, it doesNOT

mean that the proton (H+) is attached to the

nitrogen atom. The proton always binds to, or isassociated with a lone pair of electrons, usually on

one of the outer negatively charged atoms of a

polyatomic anion. Consider the Lewis Dot

structuresfor NO3, SO4

2, and ClO4:

Cl can have

mor e t h an 8 v e-

O

NO O

w hen N h as 4 b on ds,

i t i s f orm al l y cat i oni c

O

nit rate anion

SO

O

O

S can have

mo r e t han 8 ve-

sulfate dianion

2O

ClO

O

O

perchlorate anion

Resonance, of course, will spread out the negative

charges and bonding over all the O atoms. H+in

each of these cases binds to the oxygen atom(s) that

is (are) negatively charged:

O

NO O

nitr ic acid

S

O

O

O

O

O

sulfuric acid

ClO O

O

H H H

H

perchloric acid


15/74

Acids & Bases 15

Problem: Draw the Lewis Dot structure that bestminimizes the formal charges for the following

acids (some strong, some weak). Cleary show

where the H+

is coordinated.a) HBF4 (fluoroboric acid, strong)

b) H2CO3 (carbonic acid, weak)

c) H3PO4 (phosphoric acid, weak)

d) CF3SO3H (triflic acid, strong)

e) HCO2H (formic acid, weak)


16/74

Acids & Bases 16

Strong & Weak vs. pH

A common mistakeis to confusepHwith strong or

weak acids/bases. For example, if I tell you asolution has apHof 5.0, most of you would

incorrectly assume that it is a weak acid. Maybe,

maybe not.

ApHof 5.0 tells you that it is a weakly acidic

solution. This solution might have been made

from a medium to considerable amount of a weakacid, or a very small amountof a strong acid.

Without knowing the concentration (or amount

and nature) of the acid used (not just the resulting

H+concentration orpH), you cant tell whether it is

composed of a strong or weak acid.It is true that solutions with very lowpHs(for

example, 1.0) can pretty much only be composed

of strong acids.

So you need to be very careful with your language

in dealing with strong & weak acids and solutionsthat are strongly or weakly acidic.


17/74

Acids & Bases 17

Polyprotic Acids

Sulfuric Acid:

H2SO4(aq) H+

(aq)+ HSO4

(aq) Ka1

108

HSO4

(aq) H+(aq)+ SO42(aq) Ka210

-2

Phosphoric Acid:H3PO4(aq) H

+(aq)+ H2PO4

(aq) Ka110-3

H2PO4

(aq) H+(aq)+ HPO42(aq) Ka210

-8

HPO42(aq) H+(aq)+ PO4

3(aq) Ka310-13

Carbonic Acid:

H2CO3(aq) H+

(aq)+ HCO3

(aq) Ka1

10-7

HCO3

(aq) H+(aq)+ CO32(aq) Ka210

-11


18/74

Acids & Bases 18

Note that for polyproticacidsonly the first

dissociation(the firstH+) is likely to occur. Thesecond (or third) dissociation process is far less

likely, so the contribution of these subsequentdissociations to the overall [H+]total.

So, generally we only have to worry about the

firstproton and thefirst dissociation constant

unless one is doing a titration or other acid-

base reaction!!


19/74

Acids & Bases 19

Strong Bases

These are usually (and most commonly) alkali

metal hydroxidesthat dissociate completely insolution:

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

O[ ] [ ]K

[ ]

N

a

Ha

OHN

810

=

+

b

The common strong basesthat you are expected to

know are:

LiOH, NaOH, KOH, RbOH, & CsOH

The alkaline earth hydroxidesCa(OH)2, Sr(OH)2,Ba(OH)2are medium strong bases.

Be(OH)2& Mg(OH)2are considered to be weak

basessince they only partially dissociate in water.

The smaller ionic radius of Be2+and Mg2+cations

polarize coordinated H2O enough to promotehydrolysis(that is, loss of H+ from H2O).


20/74

Acids & Bases 20

It is very importantto remember that

there are almost always two stepsin

converting from [OH]topH:

1) convert [OH]to [H+]:

K[ ]

[ ]H

OH

141 10

[ ]

OH

= =

+ w

2) then convert [H+]topH.

---- or ----

1) convert [OH]topOH

2) then convertpOHtopH:

pH = 14 pOH


21/74

Acids & Bases 21

Problem: What are thepH's of the followingsolutions? What are thepOH's?

a) 0.001 M NaOH

b) 0.1 M CsOH

c) 1 105M KOH

d) 10 M RbOH

e) 1 M NaOH

f) 0.1 M Be(OH)2

g) 0.01 M LiOH

h) 1 104M NaOH

i) 1 M ammoniaj) 1 1014M KOH


22/74

Acids & Bases 22

Explaination for j) on the last page:

Theactualhydroxideconcentration for any base

dissolved in water is more precisely defined as:

[OH

]total = [OH

]base + [OH

]waterNormally, [OH]wateris 1 10

7Mand is much

less than [OH]base, thus we can usually ignore it.

BUT, in this example, [OH]baseturns out to beonly 1 1014M, much less than the [OH]water.

So in this example we can actually ignore the tiny

amount of OHcontributed from the base and onlyconsider the OHnaturally present in water:

This will work for base concentrations of 1 108Mand

lower. It gets complicated mathematically right around

1

107M(not dealt with in this course).


23/74

Acids & Bases 23

Conjugate Acid-Base Pairs

Bronsted-LowryDefinition of a Base: a substance

that combines or accepts a H+

. Consider thedissociation of an acid:

H2O + HF(aq) H3O+(aq) + F(aq)

Because this is an equilibrium, F(aq) is back reacting withH+(aq) to produce undissociated acid HF(aq). Therefore,F(aq) is acting like a base! Since it was originally part of theacid (HF), theres a special name for it: conjugate base.

Strong acidshave weakconjugatebases.Weakacidshavestrongerconjugatebases

(but usually not as strong as OH).

To indicate that an equilibrium favors one side of a

rxn chemists sometimes use the double equilibrium

arrows where one arrow is shorter. For a strongacid (weak conjugate base) one could write:

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


24/74

Acids & Bases 24

Bronsted-LowryDefinition of a Acid: a substance

that donates a H+.

Consider the reaction of a weak basewith water:

NH3(aq) + H2O NH4+(aq) + OH(aq)

Because this is an equilibrium, NH4+(aq) donates a H+(aq)

that reacts with OH(aq) to produce the original NH3(andwater). Therefore, NH4

+(aq) is acting like an acid! Since itstarted as a weak base(NH3), we call NH4

+a conjugate acid.

Strong baseshave weakconjugateacids.

Weakbaseshavestrongerconjugateacids.In the two reactions shown on this page and the previous,

water is acting as either an acidor a base. Any chemical that

can act as either an acidor baseis called amphoteric.

In the following reaction, Na+is more accurately called

a conjugateLewis Acid(not a proton donor).NaOH(aq) Na+(aq) + OH(aq)


25/74

Acids & Bases 25

Weak Acids

These are acids that only dissociate to produce a

small amount of H+in solution:

HF(aq) H+(aq) + F(aq)H

H

F

F

[ ] [ ]

K [ ]4

3.5 10

= =

+

a

Most of a weak acid is dissolved in solution in its

undissociated, neutral form. An acid is weak

because its conjugatebase(counter-anion) is a

good base and likes to bind to H+.

A weak acid typically has Ka< 1 103

Most weak acids are organic acids based on the

carboxylic group:

Some of the more common weakacidsthat we run

into on a daily basis include:


26/74

Acids & Bases 26

Acetic Acid Carbonic Acid Phosphoric A

H3C

C

O

OH HO

C

O

OH

O

P

HO OHOH

O

C

CC

C

CC C

OH

OOH3C

HOCH2

H

H

H

H

O

OHHO

O

H

HOH

cid

Acetylsalicylic Acid Ascorbic Acid

active ingredientin vinegar

formed when CO2dissolves in water

used in soda(Coke, Pepsi)

Vitamin CAspirin Note that the OH groups in all these examples are not

hydroxides!!The oxygen atom of the OH is strongly bonded

to the atom they are attached to and will not fall off as OH

.Instead they dissociate H+due to the Os electronegativityand ability to stabilize negative charge(s).


27/74

Acids & Bases 27

Example: What is thepHof a 0.1Msolution of

acetic acid? Ka1 105

Abbreviation for acetic acid: HOAc

Initial Cond: 0.1M 0M 0M

HOAc(aq) H+(aq) + OAc(aq)

@ Equilib: 0.1 -x x x

Substitute ourxvalues into the equilibrium

expression and solve forx:

+[ ] [ Ac ]K

[

H

H

O

OAc]5

5

1 10

( )( ) 1 10(0.1 )

x xx

= =

=

a

But, this will be a quadratic expressionand we'll

have to use the quadratic formulato solve forx

(ugh!).

There is, however, a very good approximationwe

can make todramatically simplify the algebra.


28/74

Acids & Bases 28

Because Kais fairly small (105) we know that

acetic acid is a weak acidand that we are only

going to make small quantitiesof H+. That means

thatxshould be a rather small number as well.This in turn means that 0.1 x should be 0.1.

Our assumption, therefore, will be thatxwill be

much lessthan 0.1 M (initial concentration of

acid). This really simplifies the algebra:

}

5

5

2 6

( )( ) 1 10(0.1 )

( )( )1 10

(0.1)1 10

x xx

x x

x

=

=

=

assume that x


29/74

Acids & Bases 29

When can I dropx??

When is it a good approximation??

When Keqis 1 103

or smaller andxworks out tobe at least an order of magnitude smallerthan the

initial concentration that one is subtractingxfrom.

This also works when one is addingxto the initial

concentration (e.g., common ion problems).

Problem: What is thepHof a 10Msolution ofhydrogen sulfide? Ka 1 10

7

H2S(aq) H+(aq) + HS(aq)


30/74

Acids & Bases 30

Problem: Prof. Stanley makes a new organic acid.He prepares a 0.01Msolution and finds that the

pHis 4.0. What is the Kafor this new acid?

HA(aq) H+(aq) + A(aq)

Problem: What is thepHof a 0.01Msolution ofHCN? Ka 4 10

10


31/74

Acids & Bases 31

Effect of Structure on Acid/Base Behavior

Hydrohalic Acids

A combination of factors affects the acid strengthof hydrohalic acids:

1) Polarity of the H-Xbond

2) Strength of the H-Xbond

3) Stability of theconjugatebase, X

Electrostatic

attraction!!

Thus, HFis a weak acidbecause the rather smallfluoride ion (F) has a concentrated negative

charge that very effectively and strongly attracts

the H+cation,notallowing it to dissociate and

become a strong acid (like HCl, HBr, or HI).

HF HCl HBr HI


32/74

Acids & Bases 32

Oxyacids

Oxyacids are those acids in which the central atom

(most commonly N, Cl, Sor P) is bonded to at least

one, and usually more, oxygenatoms. Theresulting negative charge on this unit is balanced

by the proper # of H+that associate with the

oxygen atoms (one per oxygen atom).

A list of common oxyacidsand their names:

HNO2 HNO3Nitrous Nitric

HClO HClO2 HClO3 HClO4

Hypochlorous Chlorous Chloric Percloric

H2SO3 H2SO4

Sulfurous Sulfuric

H3PO2 H3PO3 H3PO4

Hypophosphorous Phosphorus Phosphoric


33/74

Acids & Bases 33

The strength of oxyacids increasesfor a series of

compounds as follows:

1) given the same central atom, the more oxygens

present the stronger theacidReasoning: the more electronegative oxygen atoms

present, the more the negative charge on the anion is

spread out over a larger volume. This means that the

negative charge will be less concentrated on any single

oxygen. Thus, there will be a lower electrostatic attraction

to the H+cations, allowing them to dissociate more easily.

2) for the same # of oxygens, the more electro-negative the central atom, the stronger theacid

Reasoning: the more electronegative the central atom, the

more the negative charge on the anion will be pulled

towards the central atom and away from the outlying

oxygens. Thus, there will be a lower electrostaticattraction to H+cations, allowing easier dissociation.

3) For almost all acids, the higher the negative

charge on the anion(conjugate base), thelower

the acidity of theacid.

Reasoning: the higher the negative charge on the anion(mono- or polyatomic) the stronger the electrostatic

attraction to the H+cations. This makes it harder for the

H+cation to dissociate.


34/74

Acids & Bases 34

4) For almost all acids, the more electronegative

atoms present(like O or F)the higher the

acidity of thatacid.

Reasoning: the more electronegative atoms present, themore the negative charge on the anion will be pulled

towards these atoms and away from the atom that the H+is

associated with.

Example:


Hypochlorous Chlorous Chloric Percloric

Ka= 3 x108 Ka= 1 x10

2 Ka= 5 x102

Ka 1 x1010

Electrostatic charge potential (ECP) surface plots for ClO

through ClO4 anions (no H+). The red color (dark) indicates

more negative charge and a stronger electrostatic attractionto the H+cation (weaker acid). Positive charge is indicatedby the blue color (darker color on center atom).


35/74

Acids & Bases 35

Here are the ECP surface plots for the acids:


Note that the blue area around the H atom indicating positivecharge is increasing as O atoms are added.

Problem: Which is the stronger acid of the pair?

a) H2SO4 or H3PO4

b) HNO2 or HNO3

c) HOBr or HOI

d) H2SeO4 or H2SeO3

e) HI or HF

f) H2SO3 or H2SO4

g)

CC

O

OH

O

OH

CC

orH

H

H

F

F

F


36/74

Acids & Bases 36

Weak Bases

These are bases that only react to a relatively small

extent with H+in solution:

:NH3(aq) + H+(aq) NH4

+(aq)

The convention, however, for writing equilibria for

weak bases is to react the base with water, which

generates a small quantity of OH-:

:NH3(aq) + H2O NH4+(aq) + OH

(aq)This is the equilibrium we use to define our base

equilibrium constant, Kb:

4

3

[ ] [ ]K

[ ]N

OHNH

H

51.8 10

= =

+

b

In this equilibriumNH4+(aq) is called the

conjugateacid(it acts like a weak acid).


37/74

Acids & Bases 37

Some weak basesare shown in the table below:

Base Formula Structure Kb

methylamine NH2CH3 N CH

H

H 3

4.4 x 10-4

carbonate ion CO32-

C

OO

O 2-

1.8 x 10-4

ammonia NH3 N H

H

H

1.8 x 10-5

hydrosulfide ion HS- SH 1.8 x 10-7

nicotine C10H14N2

N

N

CH3

7 x 10-7

1.4 x 10-11

hydroxylamine NH2OH N OHH

H

1.1 x 10-8

pyridine C5H5NN

1.9 x 10-9

Note that all bases have atoms with lone pairs of

electrons that can interact with a H+. Rememberthat a H+doesn't have any electrons and has a

positive charge. It will be attracted to atoms with

negative charges and/or lone pairs of electrons.


38/74

Acids & Bases 38

Example: What is thepHof a 0.1 M solution of

ammonia? Kb1 105

Init: 0.1M 0M 0M

:NH3(aq) + H2O NH4+(aq) + OH(aq)

@ Eq: 0.1 -x x x

Substitute ourxvalues into the equilibrium

expression and solve forx:

4

3

OH[ ] [ ]K

[N ]

NH

H

+5

5

1 10

( )( )1 10

(0.1 )

x x

x

= =

=

b

But, this will be a quadratic expressionand the

quadratic formulais needed to exactly solve forx

(ugh!).

But lets use the very good approximationfrom

weak acid equilibria problem solving that will

dramatically simplify the algebra.


39/74

Acids & Bases 39

Because Kbis pretty small (105) we know that

ammonia is a weak base and that we are only going

to make small quantitiesof OH. That means that

xwill be a pretty small number. This in turnmeans that 0.1 -xwill be 0.1.

Our assumption, therefore, will be thatxwill be

much lessthan 0.1M(initial concentration of acid).

This now really simplifies our algebra:

}

5

5

2 6

( )( ) 1 10(0.1 )

( )( )1 10

(0.1)

1 10

x xx

x x

x

=

=

=

assume that x


40/74

Acids & Bases 40

Problem: What is thepHof a 1Msolution of

sodium carbonate? Kb 1 104

CO32(aq) + H2O HCO3(aq) + OH(aq)

Problem: What is thepHof a 0.01Msolution of

NH2OH? Kb 1 108


41/74

Acids & Bases 41

LewisAcids &Bases

Acid electron acceptor

Base electron donorAlthough the Lewis definition includes the

traditional Arrhenius and Brnsted-Lowry acids

& bases mentioned so far, it also encompasses

molecules that dont. Foremost are metal atoms

that form bonds to other molecules using emptyorbitals on the metal (Lewis acid) and filled lone

pairs on the donor atoms (Lewis bases).

Transition metal atoms typically form the

strongest bonds, followed by actinide and

lanthanides. Some examples are shown below:

Os

Cl

Cl H

H

PR3

PR3

WH3C

Ph

O

O

OH2N

NH2

Cl

Pd

Cl

PR3R3P

Problem: Identify the Lewis Base and Acid parts of

each compound shown above. Extra:What is theoxidation state of the metal in each compound?


42/74

Acids & Bases 42

Another class of important Lewis acids are

trivalent compounds of boron and aluminum that

have an empty orbital present. These are typically

very reactive to donor molecules including water.BF3, for example, is considered a superacid.

F

BF F

Cl

AlCl Cl

BCl3will form a moderately strong bond to :NH3:Cl

B

Cl Cl

+ N

H

HH

N

H

HHCl

B

ClCl

Problem: What is the hybridization of the Bin BCl3?What about the Bin Cl3B:NH3?


43/74

Acids & Bases 43

Relationship Between Kaand Kb

Consider our ammonia equilibrium:

:NH3(aq) + H2O NH4+(aq) + OH

(aq)

+4

+

3

3

4

NH

N

[ ] [ ]K

[ ]

K

[ ] [ ]

[ ]

OH

OH

HO ]

KK

H

NH

H

NH

[ [H] ]

but remember that :

substituting this in for [ we now have :

b

b=

=

=

+

+

w

w

This expression now corresponds to the followingequilibrium multiplied times Kw:

:NH3(aq) + H+(aq) NH4+(aq)The reverse of this reaction, however, is the "acid"

equilibrium:NH4

+(aq) :NH3(aq) + H+(aq)Writing the reaction this way means that we can

now set-up a Kaequilibrium expression:


44/74

Acids & Bases 44

4

3+

+

H[ ][

N

NHK

[ H ]=a

]

But note that the reciprocalof this expression isalready in the Kbrelationship:

4

3+

+

H[ ][

N

NHK

[ H ]=a

]

4

3

+[ ] K

NH

NK

[ ] [

H

H=

+w

b]

Therefore, Kaand Kbare related via Kw.

Here are the very important Ka/ Kbrelationships

you need to know/understand:

K K

K -or- KK K

K K K

= =

=

b b

b

a

w

w

a

w

a


45/74

Acids & Bases 45

Knowing the mathematical relationship between

Kaand Kbis very important. Many references

only list Kavalues for bases andnotthe Kbvalues

that you would expect.When a Kavalue is given for a baseit is really the

value for the conjugate acidof that base. For

example:

Base =NH3 Kb = 1.8 105

Conjugate Acid=[NH4]+ Ka = 5.6

10

10(Kb)(Ka) = (1.8 10

5)(5.6 1010) = 1 1014

Kw

If a reference gives a Kavalue for the baseyou are

looking up, you'll have to convert it to a Kbvalue

in order to calculate the [OH

], and then thepH.You must, therefore, pay close attention to what

data the problem is giving you and what you need

to use. READ CAREFULLY!!

DANGER!!VERY Common mistake!!


46/74

Acids & Bases 46

Problem: What is thepHof a 0.1Msolution of the

weak basetriethyl amine (NEt3)? Ka= 1 1011

Problem: What is thepHof a 0.01Msolution of

hydroxyamine (NH2OH)? Ka= 1 106


47/74

Acids & Bases 47

Salts of WeakAcids and Bases

When a weakacid(e.g., acetic acid) reacts with a

strong base(e.g., an alkali hydroxide like NaOH)water and thesalt of the weak acidis formed:

Note that the acetate anion formed in the reaction

is itself a weakbase. Thus it can react with waterto produce a small amount of [OH].

Therefore, dissolving sodium acetate (the saltof

acetic acid) in water will make abasicsolution.


48/74

Acids & Bases 48

Similarly, the reaction of a weakbasewith a strong

acidwill produce the saltof aweakbase, which will

act as a weakacid. Shown below we have the

reaction of ammoniawith HClto produceammonium chloride(the saltof a weakbase):

Dissolving ammonium chloride in water, therefore,

produces an acidicsolution (NH4+is a weak acid).


49/74

Acids & Bases 49

The big question is How the heck do I tell whether

a salt will produce anacidic,basic, orneutral

solution??

The KEY is to remember the strong acidsandbases:

Strong Acids:

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

HBr(aq) H(aq) + Br(aq)

HI(aq) H

(aq) + I

(aq)HNO

3(aq) H(aq) + NO

3(aq)

H2SO

4(aq) 2H(aq) + SO

4(aq) *

HClO4(aq) H(aq) + ClO

4(aq)

These are

extremely

weak

conjugate

bases--

Neutral

Anions!

Strong Bases:

LiOH(aq) Li

(aq) + OH

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

KOH(aq) K(aq) + OH(aq)

RbOH(aq) Rb(aq) + OH(aq)

CsOH(aq) Cs(aq) + OH(aq)

Putting these Neutral(neutral here refers to their

acid-base properties, NOT their charges!!) anions

and cationstogether generates Neutral(notacidic

orbasic)salts (Ca2 , Sr2 , Ba2 salts not shown):

These areextremely

weak

conjugate

acids--

Neutral

Cations!


50/74

Acids & Bases 50

Neutral

Cation

Neutral

Anion Neutral Salts

Li Cl LiCl, LiBr, LiI, LiNO3, Li2SO4,

Na

Br

NaCl, NaBr, NaI, NaNO3, Na2SO4,K I KCl, KBr, KI, KNO3, K2SO4,

Rb NO3 RbCl, RbBr, RbI, RbNO3, Rb2SO4,

Cs SO42 CsCl, CsBr, CsI, CsNO3, Cs2SO4,

ClO4 Perchlorate salts are explosive !!!

Solutions of theseNeutral Saltsare neither acidicnor basic, but rather have a pH = 7 (Neutral!).

A simple set of guidelines, therefore, are:

CATIONS other than Li, Na, K, Rb, Cs (&

Ca2, Sr2, Ba2) will generate ACIDICsolutions

(that is, the cation is a good conjugate acid)

ANIONS other than Cl, Br, I, NO3, SO4

2will

generate a BASICsolution (that is, the anion is a

good conjugate base)


51/74

Acids & Bases 51

SALTS

Cation(+)Anion(-)Cation comes from the base Anion comes from the acid

Identify the anionIs it from a ?strong acid

Salt Solutionwill be

(Salt of astrong

& )

Neutral

acid base

Identify the CationIs the cation from a ( or , , )strong base Group 1A Ca Sr Ba2+ 2+ 2+

Anion is froma(anion is a

moderate togood

)

weak acid

conjugatebase

Cation is from a(cation is a )

weak baseconjugate acid

Identify the anionIs the anion from a ?strong acid

Salt solutionwill be(salt of a

)

acidic

weak base

Salt solutionwill be

(salt of a

)

basic

weak acid

YES

YES

YES

NO

NO

NO

Salt of aweak acid &

a weak base!Complicated!Dont worry


52/74

Acids & Bases 52

Problem: Will the following salts make an acidic,basicor neutralsolution when dissolved in water?

a) NaF Ka(HF) = 3.5 10-4

b) NaCl Ka(HCl) = 1 108

c)N H+ Cl-

Ka(pyridineH+) = 5 10-6

d) C6H5COONaKa(C6H5COOH) = 6.5 10-5

e) RbBr Ka(HBr) = 1 1010

f) CH3COOCs Ka(CH3COOH) = 1.7 10-5

g) Cs2S Ka(H2S) = 9.1 10-8

h) NH4NO3 Ka(NH4+) = 5.6 10-10

i) KCN Ka(HCN) = 4.9 10-10

j) KNO3


53/74

Acids & Bases 53

Example: What is thepHof 1MNaF?Ka(HF) = 1 10

4.

Initial: 1M 0 0

NeutralCation

BasicAnion

F(aq) + H2O HF(aq) + OH(aq)

@ Eq: 1 -x x x

This is a basicequilibrium, so we need to convert

Kainto Kb:K

KK

14

4101 10 1 10

1 10

= = =

w

a

b

Our equilibrium expression, therefore, is:

DANGER!!

VERY

Common

mistake!!

F OH[ ][ ]K[ ]F

b

assume that x


54/74

Acids & Bases 54

So the [OH] = 1 105M. ThepOHis:

pOH= -log(1 105) = 5 DANGER!!VERY

Commonmistake!!

ThepH, then, is given by:pH = 14 -pOH = 14 - 5 = 9


55/74

Acids & Bases 55

Problem: What is thepHof a 0.1 M solution ofNH4Cl. Ka(NH4

+) = 1 x 109.


56/74

Acids & Bases 56

Reactions of Acids & Bases

Strong Acidsand Strong Bases:

The reaction of a strong acidand strong baseis thesimplest type of acid-base reaction.

Consider the rxn of NaOHand HCl:

Na+(aq) + OH(aq) + H+(aq) + Cl(aq)H2O+ Na

+(aq) + Cl(aq)

The rxn above showsallthe aqueous species. Note thatNaCl(aq) is one of the products. When acidsreact withbases, waterand the saltof the acid/base are formed (the

baseprovides the cationand acidthe anion).

Thenet ionic equation is shown below:

OH(aq) + H+(aq) H2O

A key point to remember is that when we react

acids and bases we are usually reactingsolutionsof

acids and bases. When one mixes two solutions

together the overall volume increasesand the

concentrations of all species willdecrease!

Because of thechange in volume due to the mixingtogether of two solutionsone has to take this into

account.

We do this by dealing directly with MOLES and

notMolarity. So you have to:

DANGER!! VERY Common mistake!!


57/74

Acids & Bases 57

convertmolarityintomoles,

do your calculations, then

convert back tomolarityusing theNEW total

(combined) solution volumeExample: What is thepHif we mix 100 mL of 0.1M HClwith 50 mL of 0.1MNaOH?

First, convert concentrations into moles:

(100 mL)(0.1M) = 10 mmol HCl

(50 mL)(0.1M) = 5 mmol NaOH

Now for the reaction: 5 mmol NaOHwill reactwith only 5 mmol HCl, leaving behind 5 mmol HCl.But now we have 150 mL of solution, so [H+]is:

5 mmol150 mL

= 0.033 M H+ H} p = 1.5DANGER!! VERY Common mistake!!


58/74

Acids & Bases 58

Problem: What will be thepHwhen 100 mL of0.1MHNO3is mixed with 300 mL of 0.2MKOH?

Problem: What will be thepHwhen 100 mL of0.1MHNO3is mixed with 200 mL of 0.05M

KOH?


59/74

Acids & Bases 59

Strong Acidsand Weak Bases:

Strong Basesand Weak Acids:

Things get more complicated (well, not too bad)

when you react a strong acidwith a weak baseor astrong basewith aweak acid.

For this case, however, we will simplify things a

little by only using equivalent amountsof the acid

and base.

The key thing to remember when you react a weakacidorbasewith a strong base or acid is that one

ends up with the SALTof that weakacidorbase.

Remember that the SALTof a weakacidis actually

a weakbaseand gives a basicsolution.

Similarly, the SALTof a weakbasegenerates anacidicsolution.

DANGER!!

VERY Common mistake!!


60/74

Acids & Bases 60

Example: What is the pHif 150 mL of 0.5Macetic acidis mixed with 150 mL of 0.5MNaOH?

Ka(HOAc) 1 105.

HOAc(aq) + OH

(aq) H2O + OAc

(aq)The OAcproduced is a weak baseand willproduce a weakly basic solution:

H2O + OAc(aq) HOAc(aq) + OH(aq)

Not realizing this - DANGER


Since this is a base equilibrium, we need to use Kb:

KK

K

[ ][OAc OH

OAc

]K

[ ]

H

14

59

9

1 101 10

1 10

1 10

= = =

= =

a

wb

b

Not realizing this - DANGER


Calculate the concentration of OAc:

(150 mL)(0.5M) = 75 mmol HOAc

(150 mL)(0.5M) = 75 mmol OH


61/74

Acids & Bases 61

Thus, we will produce 75 mmol of OAc. Theconcentration of OAc, therefore, will be:

75 mmol

0.25M OAc

300 mL

Now we can set-up our initial and @eq conditions:

Initial: 0.25 M 0 0

H2O + OAc(aq) HOAc(aq) + OH(aq)

@ Eq: 0.25 -x x x

OAc OH

OA

[ ][ ]K

[ ]

H

]

c

OH[

9

9

9

102

5

1 10

( )( )

1 10(0.25 )

( )( )1 10

(0.25)

2.5 101.6 10

x x

x

x x

xx

=

=

=

=

=

=

}

b

assume that x


62/74

Acids & Bases 62

Problem: What is the pHof the reaction of500 mL of 2 M NH3and 500 mL of 2 M HCl?

Ka(NH4+) 1

1010.


63/74

Acids & Bases 63

Titrations

A titrationis the careful measured addition of a

known concentration of one substance that willreact with another unknown material in order to

determine the concentration of the unknown

material.

Determining the concentrations of unknown materials is a

routine procedure in chemistry. Titrations are commonly

used to determine the concentrations of acids and bases insolution, as well as many other chemicals. Titrations are also

often the simplest and least expensive way of determining

concentrations of unknown materials.

If we are titrating an unknown acid with a known

amount of base with a known concentration, we

can use the (M1)(V1) = (M2)(V2)relationship tofind the unknown's concentration at the

equivalence point(the point at which we've added

just enough base to react with all the acid).

In order to determine when we have reached the

equivalence point in an acid base reaction, wegenerally use an indicator.


64/74

Acids & Bases 64

Indicators

An indicatoris a weak organic acid or base that

has sharply different colors in its associated anddissociated forms:

HIn(aq) H+(aq) + In(aq)

red acidic blue basic

Indicators usually have very intense colorsso one

only has to use a very small amount (a few drops)so it will not affect the titration of the solution.

Remember that the indicator is anacidor baseso if

you add a lot it will affect the titration!!O

O

O OH

H

H O

O O

H

+ H+

Phenophtalein

acidic form - colorless anionic basic form - red

N N

O

OH

N N N

O

O

N + H+

Methyl Red

acidic form - red anionic basic form - yellow


65/74

Acids & Bases 65

Some Common Acid-Base Indicators

NamepH color change

region Acid color Basecolor

Methyl violet 0 - 2 yellow violet

methyl yellow 1.2 - 2.3 red yellow

methyl orange 2.9 - 4.0 red yellow

methyl red 4.2 - 6.3 red yellow

bromthymol blue 6.0 - 7.6 yellow blue

thymol blue 8.0 - 9.6 yellow blue

phenolphthalein 8.3 - 10 colorless pink

Alizarin yellow 10.1 - 12 yellow red

HIn(aq) H+(aq) + In(aq)

I[ ][ ]

K [

H

I ]

n

nH

=

+

a

At the color change [H+] = [In] = [HIn], so:


66/74

Acids & Bases 66

Titrating an unknown strong acidwith a known

amount of strong base:


67/74

Acids & Bases 67

Titrating an unknown strong basewith a known

amount of strong acid:


68/74

Acids & Bases 68

Titrating an unknown weak basewith a known

amount ofstrong acid:


69/74

Acids & Bases 69

Titrating an unknown weak acidwith a known

amount of strong base:


70/74

Acids & Bases 70

Buffer Solutions

Consider an acetic acid solution:

HOAc(aq) H+(aq) + OAc(aq)If we add enough NaOAc(the salt of acetic acid) to

increase the OAcconcentration roughly equal to

HOAc, we now form the following mixture:

HOAc (aq) H+

(aq) +OAc(aq)The added OAc, which is a weak base, will consume

some of the free H+causing the pHto rise (become less

acidic).

Note that if we add H+to this solution, it will react with

the large pool of weak base OAc

to form HOAc. TheH+concentration, therefore, will stay about the same.

Similarly, if we add some OHit will react with the H+

present. Since we have an equilibrium, however, some

HOAcwill, in turn, dissociate to replace the missing H+,

keeping it about the same.

A solution of a weak acidand the salt(conjugate base)

of a weak acid or a weak baseand the salt

(conjugate acid) of a weak baseis called a BUFFER!


71/74

Acids & Bases 71

If the concentration of the two components is high

enough and you dont add too much acid or base, the

buffer solutionwill be very resistant to changes in the pH

due to added acid or base.

Buffer solutions are critically important to biological

systems (i.e., keeping us alive!).

For a solution of a weak acidorbaseand their salt, one

can write the following equations for calculating the

[H+] and [OH] of the buffer solution generated:

acidic saltBaseOH

basic sal

[ ][ ] K[ ]

[ ][ ] K

Acid

[H

t]

b

=

=

a+

These are called the Henderson-Hasselbalchequations.

The log form of these equations:

= +

= +

basic salt

O

H

acidacidic salt

[ ]K log

[ ][ ]

K lHbase

og[ ]

a

b

p p

p p


72/74

Acids & Bases 72

If the acid/baseconcentration is thesameas the

saltconcentration, then we can write:

[OH] = Kb [H+] = Ka

or pOH= pKb pH= pKa

By varying the weak acidorbaseand the salt being

used to make the buffer solution, as well as their

concentration ratio, one can set the pHof the

buffer to almost anything you want.The more concentratedthe buffer components, the

more effective the buffer solution will be at

resisting pHchanges. But remember that you can

always overloada buffer by adding too much acid

or base to it.One of thetrickiestthings for you to determine is

just what saltswill work to make a buffer. The

salts of strong acidsand bases(i.e., NaCl, KBr,

Na2SO4, CsNO3) do NOT usually make buffers,

nor do mixtures of strong acidsand baseswith

their salts!


73/74

Acids & Bases 73

Note, however, that reacting equivalent of a

strong basewith a weak acid, generates a buffer

solution!

This is the region where we

have a buffer solution present


74/74

Acids & Bases 74

Problem: Which of the following solutions is atraditional buffer prepared from a weak acid/base

and a conjugate salt?

a) 0.1MNaOH + 0.1MNaOAcb) 0.01MHCl + 0.01MNaCl

c) 0.5MNH4Cl + 0.5MNH3

d) 2MNaOAc + 2MHOAc

e) 0.01Mcitric acid + 0.01Msodium citrate

f) 0.3MH3PO4 + 0.3MNaH2PO4

g) 2MHNO3 + 2MNaNO3

h) 0.05MH2CO3 + 0.05MKHCO3

i) 0.5M HI + 0.5MCsI

j) 0.2Mbenzoic acid + 0.2Mcesium benzoatek) 1M NaOAc + 1MKBr

l) 0.001M HCl + 0.001MKOH

m) 0.1M KOAc + 0.1MNH3

n) 0.2M HOAc + 0.2MKHCO3

1422 chapt 18 acids bases

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