8
General, Organic, and
Biochemistry, 7e
Bettelheim,
Brown, and March
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8-1
8 Chapter 8
Acids and Bases
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8-2
8 Arrhenius Acids and Bases
• In 1884, Svante Arrhenius proposed these
definitions
• acid: a substance that produces H3O+ ions aqueous
solution
• base: a substance that produces OH- ions in aqueous
solution
• this definition of an acid is a slight modification of the
original Arrhenius definition, which was that an acid
produces H+ in aqueous solution
• today we know that H+ reacts immediately with a water
molecule to give a hydronium ion
+
H ( aq) + H2 O( l)
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+
H3 O ( aq)
Hydronium ion
8-3
8 Arrhenius Acids and Bases
• when HCl, for example, dissolves in water, its reacts
with water to give hydronium ion and chloride ion
H2 O(l) + HCl(aq)
H3 O+ (aq) + Cl- (aq)
• we use curved arrows to show the change in position
of electron pairs during this reaction
: :
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:
: :
:
H O: + H Cl :
H
+
H O H + :Cl:
H
8-4
8 Arrhenius Acids and Bases
• With bases, the situation is slightly different
• many bases are metal hydroxides such as KOH, NaOH,
Mg(OH)2, and Ca(OH)2
• these compounds are ionic solids and when they
dissolve in water, their ions merely separate
NaOH(s)
H2 O
+
-
Na (aq) + OH (aq)
• other bases are not hydroxides; these bases produce
OH- by reacting with water molecules
NH3 ( aq) + H2 O(l)
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NH4 + ( aq) + OH- (aq)
8-5
8 Arrhenius Acids and Bases
• we use curved arrows to show the transfer of a proton
from water to ammonia
H
+
H N H + :O H
H
: :
+ HO H
:
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:
H
H N:
H
8-6
8 Acid and Base Strength
• Strong acid: one that reacts completely or almost
completely with water to form H3O+ ions
• Strong base: one that reacts completely or almost
completely with water to form OH- ions
• here are the six most common strong acids and the
four most common strong bases
Formula
HCl
HBr
HI
HNO3
H2 SO4
HClO4
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N ame
Hydrochloric acid
Hydrob romic acid
Hydroiodic acid
N itric acid
Su lfu ric acid
Perch loric acid
Formula
LiOH
NaOH
KOH
Ba( OH) 2
N ame
Lith iu m h yd roxide
Sodiu m hydroxide
Potass iu m h yd roxide
Barium hydroxide
8-7
8 Acid and Base Strength
• Weak acid: a substance that dissociates only
partially in water to produce H3O+ ions
• acetic acid, for example, is a weak acid; in water, only 4
out every 1000 molecules are converted to acetate ions
CH3 COOH(aq) + H2 O( l)
Acetic acid
-
CH3 COO ( aq) + H3O+ ( aq)
Acetate ion
• Weak base: a substance that dissociates only
partially in water to produce OH- ions
• ammonia, for example, is a weak base
NH3 (aq) + H2 O( l)
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NH4 + (aq) + OH-( aq)
8-8
8 Brønsted-Lowry Acids & Bases
•
•
•
•
Acid: a proton donor
Base: a proton acceptor
Acid-base reaction: a proton transfer reaction
Conjugate acid-base pair: any pair of molecules or ions
that can be interconverted by transfer of a proton
conju gate acid -base p air
conju gate acid -base p air
HCl(aq)
Hyd rogen
chloride
(acid)
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+ H 2 O(l)
Water
(bas e)
-
Cl (aq) +
Ch loride
ion
(conju gate
b ase of HCl)
+
H3 O ( aq)
Hyd ronium
ion
(con jugate
acid of water)
8-9
8 Brønsted-Lowry Acids & Bases
• Brønsted-Lowry definitions do not require water
as a reactant
con jugate acid-bas e pair
con jugate acid-bas e pair
CH3 COOH
Acetic acid
(acid)
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+ NH3
Ammonia
(base)
-
+
CH3 COO +
NH4
Acetate
Ammonium
ion
ion
(con jugate base (conjugate acid
acetic acid)
of ammonia)
8-10
8 Brønsted-Lowry Acids & Bases
• we can use curved arrows to show the transfer of a
proton from acetic acid to ammonia
: O:
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: O:
CH3 -C-O: +
: :
: :
H
CH3 -C-O H +
:N H
H
Acetic acid
Ammon ia
(p roton donor) (proton acceptor)
A cetate ion
H
+
H N H
H
Ammon ium
ion
8-11
8
Acid
Strong HI
Acid s HCl
H2 SO4
HNO3
H3 O+
N ame of acid
Hydroiodic acid
Hydrochloric acid
Su lfu ric acid
N itric acid
Hydron iu m ion
HSO4 H3 PO4
CH3 COOH
H2 CO3
H2 S
-
H2 PO4
+
NH4
HCN
C6 H5 OH
HCO3
2-
HPO4
Weak H2 O
A cids C2 H5 OH
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Conjugate
Base
I
ClHSO4
NO3 H2 O
N ame of ion
Iod ide
Chloride
Hydrogen su lfate
N itrate
Water
Weak
Bases
D ihydrogen p hosph ate
Ammon ium ion
Hydrocyanic acid
Ph enol
SO4 2 Su lfate
H2 PO4
D ihydrogen p hosph ate
CH3 COO Acetate
HCO3
Bicarbonate
HS
Hydrogen su lfid e
2HPO4
Hydrogen ph os phate
NH3
Ammon ia
CN
Cyan ide
C6 H5 O
Ph enoxide
Bicarbonate ion
CO3
Hydrogen su lfate ion
Ph os phoric acid
Acetic acid
Carb on ic acid
Hydrogen su lfid e
23-
Hydrogen ph os phate ion PO4
OHWater
C2 H5 O
Eth anol
Carb on ate
Ph os phate
Hydroxide
Eth oxide
S trong
Bas es
8-12
8 Brønsted-Lowry Acids & Bases
• Note the following about the conjugate acid-base
pairs in the table
1. an acid can be positively charged, neutral, or
negatively charged; examples of each type are
H3O+, H2CO3, and H2PO42. a base can be negatively charged or neutral;
examples are OH-, Cl-, and NH3
3. acids are classified a monoprotic, diprotic, or
triprotic depending on the number of protons
each may give up; examples are HCl, H2CO3, and
H3PO4
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8-13
8 Brønsted-Lowry Acids & Bases
• carbonic acid, for example can give up one proton to
become bicarbonate ion, and then the second proton to
become carbonate ion
+
H2 CO3 + H2 O
HCO3 + H3 O
Carbonic
Bicarbon ate
acid
ion
-
HCO3 + H2 O
Bicarbonate
ion
2-
+
CO3
+ H3 O
Carbonate
ion
4. several molecules and ions appear in both the
acid and conjugate base columns; that is, each
can function as either an acid or a base
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8-14
8 Brønsted-Lowry Acids & Bases
• the HCO3- ion, for example, can give up a proton to
become CO32-, or it can accept a proton to become
H2CO3
• a substance that can act as either an acid or a base is
said to be amphiprotic
• the most important amphiprotic substance in Table 8.2
is H2O; it can accept a proton to become H3O+, or lose a
proton to become OH-
5. a substance cannot be a Brønsted-Lowry acid
unless it contains a hydrogen atom, but not all
hydrogen atoms in most compounds can be
given up
• acetic acid, for example, gives up only one proton
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8-15
8 Brønsted-Lowry Acids & Bases
6. there is an inverse relationship between the
strength of an acid and the strength of its
conjugate base
• the stronger the acid, the weaker its conjugate base
• HI, for example, is the strongest acid in Table 8.2, and
its conjugate base, I-, is the weakest base in the table
• CH3COOH (acetic acid) is a stronger acid that H2CO3
(carbonic acid); conversely, CH3COO- (acetate ion) is a
weaker base that HCO3- (bicarbonate ion)
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8-16
8 Acid-Base Equilibria
• we know that HCl is a strong acid, which means that the position
of this equilibrium lies very far to the right
Cl- + H3 O+
HCl + H2O
• in contrast, acetic acid is a weak acid, and the position of its
equilibrium lies very far to the left
CH3 COO- + H3 O+
Acetate ion
CH3 COOH + H2 O
Acetic acid
• but what if the base is not water? How can we determine which are
the major species present?
CH3 COOH
Acetic acid
(acid)
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+
NH3
Ammonia
(b ase)
?
-
CH3 COO
+
+
NH4
Acetate ion
Ammon ium ion
(conju gate b ase (con jugate acid
of N H 3
of CH 3COOH
8-17
8 Acid-Base Equilibria
• To predict the position of an acid-base
equilibrium such as this, we do the following
• identify the two acids in the equilibrium; one on the left
and one on the right
• using the information in Table 8.2, determine which is
the stronger acid and which is the weaker acid
• also determine which is the stronger base and which is
the weaker base; remember that the stronger acid
gives the weaker conjugate base, and the weaker acid
gives the stronger conjugate base
• the stronger acid reacts with the stronger base to give
the weaker acid and weaker base; equilibrium lies on
the side of the weaker acid and weaker base
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8-18
8 Acid-Base Equilibria
• identify the two acids and bases, and their relative
strengths
CH3 COOH
+
NH3
A cetic acid
Ammonia
(stron ger acid) (s tronger base)
?
-
CH3 COO +
NH4
+
A cetate ion Ammonium ion
(w eak er b ase) (w eaker acid)
• the position of this equilibrium lies to the right
CH3 COOH
+
NH3
Acetic acid
Ammonia
(s tronger acid) (stronger bas e)
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-
CH3 COO +
NH4
+
Acetate ion Ammonium ion
(weaker base) (weaker acid)
8-19
8 Acid-Base Equilibria
• Example: predict the position of equilibrium in
this acid-base reaction
-
H2 CO3 + OH
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?
-
HCO3 + H2 O
8-20
8 Acid-Base Equilibria
• Example: predict the position of equilibrium in
this acid-base reaction
?
-
H2 CO3 + OH
-
HCO3 + H2 O
• Solution: the position of this equilibrium lies to
the right
H2 CO3 +
OH
-
S tronger Stron ger
acid
bas e
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HCO3
-
Weaker
base
+
H2 O
Weak er
acid
8-21
8 Acid Ionization Constants
• when a weak acid, HA, dissolves in water
HA + H2 O
-
+
A + H3 O
• the equilibrium constant, Keq, for this ionization is
K eq =
[A-] [H3 O+ ]
[HA][ H2 O]
• because water is the solvent and its concentration
changes very little when we add HA to it, we treat [H2O]
as a constant equal to 1000 g/L or 55.5 mol/L
• we combine the two constants to give a new constant,
which we call an acid ionization constant, Ka
Ka = Ke q [H 2 O] =
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[ A- ][ H3 O+ ]
[HA]
8-22
8 Acid Ionization Constants
• Ka for acetic acid, for example is 1.8 x 10-5
• because the acid ionization constants for weak acids
are numbers with negative exponents, we commonly
express acid strengths as pKa where
pKa = -log Ka
• the value of pKa for acetic acid is 4.75
• values of Ka and pKa for some weak acids are given in
Table 8.3
• as you study the entries in this table, note the inverse
relationship between values of Ka and pKa
• the weaker the acid, the smaller its Ka, but the larger its
pK
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8-23
8
Acid
Name
Ka
pKa
H3 PO4
Phosph oric acid
7.5 x 10-3
2.12
HCOOH
Formic acid
1.8 x 10-4
3.75
CH3 CH( OH)COOH Lactic acid
8.4 x 10-4
3.08
CH3 COOH
Acetic acid
1.8 x 10-5
4.75
H2 CO3
Carbonic acid
4.3 x 10-7
6.37
Dih yd rogen ph osp hate ion 6.2 x 10-8
7.21
H2 PO4
-
Boric acid
7.3 x 10-1 0
9.14
Ammonium ion
5.6 x 10-1 0
9.25
HCN
Hydrocyanic acid
4.9 x 10-1 0
9.31
C6 H5 OH
Phenol
1.3 x 10-1 0
9.89
H3 BO3
NH4
+
HCO3
-
Bicarb on ate ion
5.6 x 10-1 1
10.25
HPO
2-
Hydrogen phosp hate ion
2.2 x 10-1 3
12.66
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8-24
8 Properties of Acids & Bases
• Neutralization
• acids and bases react with each other in a process
called neutralization; these reactions are discussed in
Section 8.10
• Reaction with metals
• strong acids react with certain metals (called active
metals) to produce a salt and hydrogen gas, H2
+
Mg(s)
2HCl(aq)
Magnes ium Hydrochloric
acid
MgCl2 (aq) + H2 (g)
Magnesium Hydrogen
chloride
• reaction of a strong acid with a metal is a redox
reaction; the metal is oxidized to a metal ion and H+ is
reduced
to H2
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8-25
8 Properties of Acids & Bases
• Reaction with metal hydroxides
• reaction of an acid with a metal hydroxide gives a salt
plus water
HCl(aq) + KOH( aq)
KCl(aq) + H2 O
Potass ium
Hydrochloric
Potassiu m
Water
chloride
acid
h yd roxide
• the reaction is more accurately written as
+
+
H3 O + Cl + K + OH
+
2 H2 O + Cl + K
• omitting spectator ions gives this net ionic equation
+
H3 O + OH
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2 H2 O
8-26
8 Properties of Acids & Bases
• Reaction with metal oxides
• strong acids react with metal oxides to give water plus
a salt
+
2 H3 O (aq) + CaO( s)
2+
3 H2 O(l) + Ca ( aq)
Calciu m
oxide
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8-27
8 Properties of Acids & Bases
• Reaction with carbonates and bicarbonates
• strong acids react with carbonates to give carbonic
acid, which rapidly decomposes to CO2 and H2O
+
22 H3 O (aq) + CO3 (aq)
H2 CO3 (aq)
2 H3 O+ (aq) + CO3 2 -(aq)
H2 CO3 (aq) + 2 H2 O(l)
CO2 (g) + H2 O( l)
CO2 (g) + 3 H2 O(l)
• strong acids react similarly with bicarbonates
+
H3 O ( aq) + HCO3 ( aq)
H2 CO3 (aq)
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+
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H3 O ( aq) + HCO3 ( aq)
H2 CO3 (aq) + H2 O( l)
CO2 (g) + H2 O( l)
CO2 (g) + 2 H2 O(l)
8-28
8 Properties of Acids & Bases
• Reaction with ammonia and amines
• any acid stronger than NH4+ is strong enough to react
with NH3 to give a salt
• in the following reaction, the salt formed is ammonium
chloride, which is shown as it would be ionized in
aqueous solution
HCl( aq) + NH3 (aq)
+
-
NH4 ( aq) + Cl (aq)
• in Ch 16 we study amines, compounds in which one or
more hydrogens of NH3 are replaced by carbon groups
HCl(aq) + CH3 NH2 (aq)
Methylamin e
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+
CH3 NH3 (aq) + Cl ( aq)
Methylammon ium
ion
8-29
8 Self-Ionization of Water
• pure water contains a very small number of H3O+ ions
and OH- ions formed by proton transfer from one water
molecule to another
H2 O + H2 O
A cid
Base
-
OH
+
+
H3 O
Conjugate Conjugate
base of H2 O acid of H 2 O
• the equilibrium expression for this reaction is
K eq =
[H3 O+ ] [HO-]
[H2 O] 2
• we can treat [H2O] as a constant = 55.5 mol/L
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8-30
8 Self-Ionization of Water
• combining these constants gives a new constant called
the ion product of water, Kw
• in pure water, the value of Kw is 1.0 x 10-14
+
-
Kw = Ke q [H 2O]2 = [ H3 O ][OH ]
Kw = 1.0 x 10-14
• in pure water, H3O+ and OH- are formed in equal
amounts (remember the balanced equation for the selfionization of water)
• this means that in pure water
+
[ H3 O ] = 1.0 x 10-7 mol/L
-
-7
[ OH ] = 1.0 x 10 mol/L
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in pure w ater
8-31
8 Self-Ionization of Water
• the equation for the ionization of water applies not only
to pure water but also to any aqueous solution
• the product of [H3O+] and [OH-] in any aqueous solution
is equal to 1.0 x 10-14
• for example, if we add 0.010 mole of HCl to 1 liter of
pure water, it reacts completely with water to give 0.010
mole of H3O+
• in this solution, [H3O+] is 0.010 or 1.0 x 10-2
• this means that the concentration of hydroxide ion is
-
[OH ] =
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1.0 x 10-14
-1 2
=
1.0
x
10
1.0 x 10-2
8-32
8 pH and pOH
• because hydronium ion concentrations for most
solutions are numbers with negative exponents, we
commonly express these concentrations as pH, where
pH = -log [H3O+]
• we can now state the definitions of acidic and basic
solutions in terms of pH
• acidic solution: one whose pH is less than 7.0
• basic solution: one whose pH is greater than 7.0
• neutral solution: one whose pH is equal to 7.0
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8-33
8 pH and pOH
• just as pH is a convenient way to designate the
concentration of H3O+, pOH is a convenient way to
designate the concentration of OHpOH = -log[OH-]
• the ion product of water, Kw, is 1.0 x 10-14
+
-
-14
Kw = [H3 O ][ OH ] = 1.0 x 10
• taking the logarithm of this equation gives
pH + pOH = 14
• thus, if we know the pH of an aqueous solution, we can
easily calculate its pOH
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8-34
8 pH and pOH
• pH of some common materials
Material
pH
Material
pH
Battery acid
0.5
Gastric juice
1.0-3.0
Lemon juice
2.2-2.4
Vin egar
2.4-3.4
Tomato juice
4.0-4.4
Carb on ated beverages 4.0-5.0
Black coffee
5.0-5.1
Saliva
Pu re w ater
Blood
Bile
Pan creatic flu id
Seaw ater
Soap
Urine
Rain (unp ollu ted)
5.5-7.5
6.2
Milk of magnesia
Househ old ammon ia
10.5
11.7
Milk
6.3-6.6
Lye (1.0 M N aOH)
14.0
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6.5-7.5
7.0
7.35-7.45
6.8-7.0
7.8-8.0
8.0-9.0
8.0-10.0
8-35
8 pH of Salt Solutions
• When some salts dissolve in pure water, there is
no change in pH from that of pure water
• Many salts, however, are acidic or basic and
cause a change the pH when they dissolve
• We are concerned in this section with basic salts
and acidic salts
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8-36
8 pH of Salt Solutions
• Basic salt: the salt of a strong base and a weak
acid; when dissolved in water, it raises the pH
• as an example of a basic salt is sodium acetate
• when this salt dissolves in water, it ionizes; Na+ ions do
not react with water, but CH3COO- ions do
+
CH3 COOH2 O
A cetate ion
Water
(w eak er b ase (w eaker base)
CH3 COOH +
OHAcetic acid
Hyd roxide ion
(s tronger acid) (stronger bas e)
• the position of equilibrium lies to the left
• nevertheless, there are enough OH- ions present in 0.10
M sodium acetate to raise the pH to 8.88
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8-37
8 pH of Salt Solutions
• Acidic salt: the salt of a strong acid and a weak
base; when dissolved in water, it lowers the pH
• an example of an acidic salt is ammonium chloride
• chloride ion does not react with water, but the
ammonium ion does
+
+
+
+
H2 O
NH3
H3 O
NH4
Water
A mmonia Hydron ium ion
A mmonium ion
(w eak er acid) (w eaker bas e) (s tronger b ase) (s tronger acid
• although the position of this equilibrium lies to the left,
there are enough H3O+ ions present to make the
solution acidic
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8-38
8 Acid-Base Titrations
• Titration: an analytical procedure in which a
solute in a solution of known concentration
reacts with a known stoichiometry with a
substance whose concentration is to be
determined
• in this chapter, we are concerned with titrations in
which we use an acid (or base) of known concentration
to determine the concentration of a base (or acid) in
another solution
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8-39
8 Acid-Base Titrations
• An acid-base titration must meet these
requirement
1. we must know the equation for the reaction so that we
can determine the stoichiometric ratio of reactants to
use in our calculations
2. the reaction must be rapid and complete
3. there must be a clear-cut change in a measurable
property at the end point (when the reagents have
combined exactly)
4. we must have accurate measurements of the amount
of each reactant
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8-40
8 Acid-Base Titrations
• As an example, let us use 0.108 M H2SO4 to
determine the concentration of a NaOH solution
• requirement 1: we know the balanced equation
H2 SO4 (aq) + 2NaOH(aq)
(concentration
(concentration
known)
not known)
Na2 SO4 (aq) + 2H2 O( l)
• requirement 2: the reaction between H3O+ and OH- is
rapid and complete
• requirement 3: we can use either an acid-base indicator
or a pH meter to observe the sudden change in pH that
occurs at the end point of the titration
• requirement 4: we use volumetric glassware
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8-41
8 Acid-Base Titrations
• experimental measurements
V olu me of
0.108 M H 2 SO 4
Volume
of N aOH
Trial I
Trial II
25.0 mL
25.0 mL
33.48 mL
33.46 mL
Trial III
25.0 mL
33.50 mL
average = 33.48 mL
• doing the calculations
2 mol NaOH
mol NaOH
0 .1 0 8 mol H 2 SO4
0 .0 2 5 0 L H2 SO4
=
x
x
L N aOH
1 L H2 SO4
0 .0 3 3 4 8 L NaOH 1 mol H2 SO4
=
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0 .1 6 1 mol NaOH
= 0 .1 6 1 M
L NaOH
8-42
8 pH Buffers
• pH buffer: a solution that resists change in pH
when limited amounts of acid or base are added
to it
• a pH buffer as an acid or base “shock absorber”
• a pH buffer is common called simply a buffer
• the most common buffers consist of approximately
equal molar amounts of a weak acid and a salt of the
weak acid; that is, approximately equal molar amounts
of a weak acid and a salt of its conjugate base
• for example, if we dissolve 1.0 mole of acetic acid and
1.0 mole of its conjugate base (in the form of sodium
acetate) in water, we have an acetate buffer
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8-43
8 pH Buffers
Ad ded as
CH 3COO -Na+
Ad ded as
CH 3COOH
CH3 COOH + H2 O
-
+
CH3 COO + H3 O
• How an acetate buffer resists changes in pH
• if we add a strong acid, such as HCl, added H3O+ ions
react with acetate ions and are removed from solution
+
CH3 COO + H3 O
CH3 COOH + H2 O
• if we add a strong base, such as NaOH, added OH- ions
react with acetic acid and are removed from solution
CH3 COOH + OH
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CH3 COO + H2 O
8-44
8 pH Buffers
• The effect of a buffer can be quite dramatic
• consider a phosphate buffer prepared by dissolving
0.10 mole of NaH2PO4 (a weak acid) and 0.10 mole of
Na2HPO4 (the salt of its conjugate base) in enough
water to make 1 liter of solution
pH
w ater 7.0
0.10 M ph os phate bu ffer 7.21
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pH after
addition of
0.010 mole HCl
pH after
addition of
0.010 mole N aOH
2.0
7.12
12.0
7.30
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8 pH Buffers
• Buffer pH
• if we mix equal molar amounts of a weak acid and a
salt of its conjugate base, the pH of the solution will be
equal to the pKa of the weak acid
• if we want a buffer of pH 9.14, for example, we can mix
equal molar amounts of boric acid (H3BO3), pKa 9.14,
and sodium dihydrogen borate (NaH2BO3), the salt of
its conjugate base
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8-46
8 pH Buffers
• Buffer capacity depends both its pH and its
concentration
pH
Con centration
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The clos er the p H of the b uffer is to the pKa
of th e w eak acid, th e greater th e buffer capacity
The greater the con centration of the w eak
acid and its con jugate b ase, the greater the
bu ffer capacity
8-47
8 Blood Buffers
• The average pH of human blood is 7.4
• any change larger than 0.10 pH unit in either direction
can cause illness
• To maintain this pH, the body uses three buffer
systems
• carbonate buffer: H2CO3 and its conjugate base, HCO3• phosphate buffer: H2PO4- and its conjugate base,
HPO42• proteins: discussed in Chapter 21
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8-48
8 Henderson-Hasselbalch Eg.
• Henderson-Hasselbalch equation: a mathematical
relationship between
• pH,
• pKa of the weak acid, HA
• concentrations HA, and its conjugate base, A• It is derived in the following way
+
A + H3 O
HA + H2 O
Ka =
[ A- ][ H3 O+ ]
[HA]
• taking the logarithm of this equation gives
[A
]
+
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log K a = log [ H3 O ] + log
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[ HA]
8-49
8
Henderson-Hasselbalch Eg.
• multiplying through by -1 gives
-
[A ]
+
-log Ka = -log [H3 O ] - log
[HA]
• -log Ka is by definition pKa, and -log [H3O+] is by
definition pH; making these substitutions gives
[A ]
pKa = pH - log
[HA]
• rearranging terms gives
-
pH = pKa + log
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[A ]
[HA]
Henderson-Hasselbalch Equation
8-50
8
Henderson-Hasselbalch Eg.
• Example: what is the pH of a phosphate buffer
solution containing 1.0 mole of NaH2PO4 and 0.50
mole of Na2HPO4 dissolved in enough water to
make 1.0 liter of solution
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8-51
8 Henderson-Hasselbalch Eg.
• Example: what is the pH of a phosphate buffer
solution containing 1.0 mole of NaH2PO4 and 0.50
mole of Na2HPO4 in enough water to make one
liter of solution
• Solution
• the equilibrium we are dealing with and its pKa are
H2 PO4 - + H2 O
1.0 mol/L
HPO4 2 - + H3 O+
0.50 mol/L
pKa = 7.21
• substituting these values in the H-H equation gives
pH = 7.21 + log 0.50
1.0
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= 7.21 - 0.30 = 6.91
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8 Chapter 8 Acids and Bases
End
Chapter 8
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8-53