1422 chapt 18 acids bases
TRANSCRIPT
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Chapters 18 & 19
Acids&BasesH2O + HF(aq) H3O
+(aq) + F(aq)
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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
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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+.
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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
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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
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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
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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?
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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!
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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).
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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
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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)
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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.
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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
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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!!
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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).
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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
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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
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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).
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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)
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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)
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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:
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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).
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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.
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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
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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)
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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
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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
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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
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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.
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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:
HClO HClO2 HClO3 HClO4
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).
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Acids & Bases 35
Here are the ECP surface plots for the acids:
HClO HClO2 HClO3 HClO4
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
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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).
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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.
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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.
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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
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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
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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?
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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?
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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:
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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
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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!!
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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
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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.
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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).
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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!
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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)
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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
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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
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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
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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
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Acids & Bases 55
Problem: What is thepHof a 0.1 M solution ofNH4Cl. Ka(NH4
+) = 1 x 109.
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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!!
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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!!
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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?
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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!!
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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
VERY Common mistake!!
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
VERY Common mistake!!
Calculate the concentration of OAc:
(150 mL)(0.5M) = 75 mmol HOAc
(150 mL)(0.5M) = 75 mmol OH
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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
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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.
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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.
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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
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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:
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Acids & Bases 66
Titrating an unknown strong acidwith a known
amount of strong base:
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Acids & Bases 67
Titrating an unknown strong basewith a known
amount of strong acid:
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Acids & Bases 68
Titrating an unknown weak basewith a known
amount ofstrong acid:
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Acids & Bases 69
Titrating an unknown weak acidwith a known
amount of strong base:
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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!
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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
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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!
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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
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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