Transcript Chapter

Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 15
Acids and
Bases
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall
Stomach Acid & Heartburn
• the cells that line your stomach produce
hydrochloric acid
 to kill unwanted bacteria
 to help break down food
 to activate enzymes that break down food
• if the stomach acid backs up into your esophagus, it
irritates those tissues, resulting in heartburn
 acid reflux
 GERD = gastroesophageal reflux disease = chronic
leaking of stomach acid into the esophagus
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Curing Heartburn
• mild cases of heartburn can be cured by
neutralizing the acid in the esophagus
swallowing saliva which contains bicarbonate ion
taking antacids that contain hydroxide ions and/or
carbonate ions
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Properties of Acids
• sour taste
• react with “active” metals
 i.e., Al, Zn, Fe, but not Cu, Ag, or Au
2 Al + 6 HCl AlCl3 + 3 H2
 corrosive
• react with carbonates, producing CO2
 marble, baking soda, chalk, limestone
CaCO3 + 2 HCl CaCl2 + CO2 + H2O
• change color of vegetable dyes
 blue litmus turns red
• react with bases to form ionic salts
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Common Acids
Chemical Name
Formula
Uses
Strength
Nitric Acid
HNO3
explosive, fertilizer, dye, glue
Strong
explosive, fertilizer, dye, glue,
batteries
metal cleaning, food prep, ore
refining, stomach acid
fertilizer, plastics & rubber,
food preservation
plastics & rubber, food
preservation, Vinegar
Sulfuric Acid
H2SO4
Hydrochloric Acid
HCl
Phosphoric Acid
H3PO4
Acetic Acid
HC2H3O2
Hydrofluoric Acid
HF
metal cleaning, glass etching
Weak
Carbonic Acid
H2CO3
soda water
Weak
Boric Acid
H3BO3
eye wash
Weak
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Strong
Strong
Moderate
Weak
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Structures of Acids
• binary acids have acid hydrogens attached to
a nonmetal atom
HCl, HF
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Structure of Acids
• oxy acids have acid hydrogens attached to
an oxygen atom
H2SO4, HNO3
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Structure of Acids
• carboxylic acids have
COOH group
 HC2H3O2, H3C6H5O7
• only the first H in the
formula is acidic
 the H is on the COOH
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Properties of Bases
• also known as alkalis
• taste bitter
 alkaloids = plant product that is alkaline
 often poisonous
• solutions feel slippery
• change color of vegetable dyes
 different color than acid
 red litmus turns blue
• react with acids to form ionic salts
 neutralization
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Common Bases
Chemical
Name
sodium
hydroxide
potassium
hydroxide
calcium
hydroxide
sodium
bicarbonate
magnesium
hydroxide
ammonium
hydroxide
Formula
NaOH
Common
Name
lye,
caustic soda
Uses
soap, plastic,
petrol refining
soap, cotton,
electroplating
Strength
Strong
KOH
caustic potash
Ca(OH)2
slaked lime
cement
Strong
NaHCO3
baking soda
cooking, antacid
Weak
Mg(OH)2
milk of
magnesia
antacid
Weak
NH4OH,
{NH3(aq)}
ammonia
water
detergent,
fertilizer,
explosives, fibers
Weak
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Strong
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Structure of Bases
• most ionic bases contain OH ions
NaOH, Ca(OH)2
• some contain CO32- ions
CaCO3 NaHCO3
• molecular bases contain structures
that react with H+
mostly amine groups
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Indicators
• chemicals which change color depending on
the acidity/basicity
• many vegetable dyes are indicators
anthocyanins
• litmus
from Spanish moss
red in acid, blue in base
• phenolphthalein
found in laxatives
red in base, colorless in acid
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Arrhenius Theory
• bases dissociate in water to produce OH- ions and
cations
ionic substances dissociate in water
NaOH(aq) → Na+(aq) + OH–(aq)
• acids ionize in water to produce H+ ions and anions
because molecular acids are not made of ions, they cannot
dissociate
they must be pulled apart, or ionized, by the water
HCl(aq) → H+(aq) + Cl–(aq)
in formula, ionizable H written in front
HC2H3O2(aq) → H+(aq) + C2H3O2–(aq)
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Arrhenius Theory
HCl ionizes in water,
producing H+ and Cl– ions
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NaOH dissociates in water,
producing Na+ and OH– ions
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Hydronium Ion
• the H+ ions produced by the acid are so reactive they
cannot exist in water
 H+ ions are protons!!
• instead, they react with a water molecule(s) to produce
complex ions, mainly hydronium ion, H3O+
H+ + H2O  H3O+
 there are also minor amounts of H+ with multiple water
molecules, H(H2O)n+
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Arrhenius Acid-Base Reactions
• the H+ from the acid combines with the OHfrom the base to make a molecule of H2O
it is often helpful to think of H2O as H-OH
• the cation from the base combines with the
anion from the acid to make a salt
acid + base → salt + water
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
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Problems with Arrhenius Theory
• does not explain why molecular substances, like
•
•
•
NH3, dissolve in water to form basic solutions –
even though they do not contain OH– ions
does not explain how some ionic compounds, like
Na2CO3 or Na2O, dissolve in water to form basic
solutions – even though they do not contain OH–
ions
does not explain why molecular substances, like
CO2, dissolve in water to form acidic solutions –
even though they do not contain H+ ions
does not explain acid-base reactions that take place
outside aqueous solution
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Brønsted-Lowry Theory
• in a Brønsted-Lowry Acid-Base reaction, an
H+ is transferred
 does not have to take place in aqueous solution
 broader definition than Arrhenius
• acid is H donor, base is H acceptor
 base structure must contain an atom with an
unshared pair of electrons
• in an acid-base reaction, the acid molecule
gives an H+ to the base molecule
H–A + :B  :A– + H–B+
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Brønsted-Lowry Acids
• Brønsted-Lowry acids are H+ donors
any material that has H can potentially be a
Brønsted-Lowry acid
because of the molecular structure, often one H in
the molecule is easier to transfer than others
• HCl(aq) is acidic because HCl transfers an H+ to
H2O, forming H3O+ ions
water acts as base, accepting H+
HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
acid
base
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Brønsted-Lowry Bases
• Brønsted-Lowry bases are H+ acceptors
any material that has atoms with lone pairs can
potentially be a Brønsted-Lowry base
because of the molecular structure, often one atom
in the molecule is more willing to accept H+ transfer
than others
• NH3(aq) is basic because NH3 accepts an H+
from H2O, forming OH–(aq)
water acts as acid, donating H+
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
base
acid
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Amphoteric Substances
• amphoteric substances can act as either an
acid or a base
have both transferable H and atom with lone pair
• water acts as base, accepting H+ from HCl
HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
• water acts as acid, donating H+ to NH3
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
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Brønsted-Lowry
Acid-Base Reactions
• one of the advantages of Brønsted-Lowry theory
is that it allows reactions to be reversible
H–A + :B  :A– + H–B+
• the original base has an extra H+ after the
reaction – so it will act as an acid in the reverse
process
• and the original acid has a lone pair of electrons
after the reaction – so it will act as a base in the
reverse process
:A– + H–B+  H–A + :B
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Conjugate Pairs
• In a Brønsted-Lowry Acid-Base reaction, the
original base becomes an acid in the reverse
reaction, and the original acid becomes a base in
the reverse process
• each reactant and the product it becomes is
called a conjugate pair
• the original base becomes the conjugate acid;
and the original acid becomes the conjugate base
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Brønsted-Lowry
Acid-Base Reactions
H–A
acid
+
HCHO2
acid
H2O +
acid
:B
base

+ H2O
base

NH3
base

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:A–
conjugate
base
CHO2–
conjugate
base
+
HO–
conjugate
base
+
H–B+
conjugate
acid
+
H3O+
conjugate
acid
NH4+
conjugate
acid
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Conjugate Pairs
In the reaction H2O + NH3  HO– +
NH4+
–
H2O and HO constitute an
Acid/Conjugate Base pair
NH3 and NH4+ constitute a
Base/Conjugate Acid pair
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Ex 15.1a – Identify the Brønsted-Lowry Acids and
Bases and Their Conjugates in the Reaction
H2SO4
+
H2O

HSO4–
+
H3O+
When the H2SO4 becomes HSO4, it lost an H+  so
H2SO4 must be the acid and HSO4 its conjugate base
When the H2O becomes H3O+, it accepted an H+  so
H2O must be the base and H3O+ its conjugate acid
H2SO4
acid
+
H2O
base
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
HSO4–
conjugate
base
+
H3O+
conjugate
acid
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Ex 15.1b – Identify the Brønsted-Lowry Acids and
Bases and Their Conjugates in the Reaction
HCO3– +
H2O

H2CO3
+
HO–
When the HCO3 becomes H2CO3, it accepted an H+ 
so HCO3 must be the base and H2CO3 its conjugate acid
When the H2O becomes OH, it donated an H+  so
H2O must be the acid and OH its conjugate base
HCO3– +
base
H2O
acid
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
H2CO3
conjugate
acid
+
HO–
conjugate
base
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Practice – Write the formula for the
conjugate acid of the following
H2O
NH3
CO32−
H2PO41−
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Practice – Write the formula for the
conjugate acid of the following
H2O
H3O+
NH3
NH4+
CO32−
HCO3−
H2PO41−
H3PO4
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Practice – Write the formula for the
conjugate base of the following
H2O
NH3
CO32−
H2PO41−
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Practice – Write the formula for the
conjugate base of the following
H2O
HO−
NH3
NH2−
CO32−
since CO32− does not have an H, it
cannot be an acid
H2PO41−
HPO42−
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Arrow Conventions
• chemists commonly use two kinds
•
•
of arrows in reactions to indicate
the degree of completion of the
reactions
a single arrow indicates all the
reactant molecules are converted to
product molecules at the end
a double arrow indicates the
reaction stops when only some of
the reactant molecules have been
converted into products
  in these notes
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Strong or Weak
• a strong acid is a strong electrolyte
practically all the acid molecules ionize, →
• a strong base is a strong electrolyte
practically all the base molecules form OH– ions,
either through dissociation or reaction with water, →
• a weak acid is a weak electrolyte
only a small percentage of the molecules ionize, 
• a weak base is a weak electrolyte
only a small percentage of the base molecules form
OH– ions, either through dissociation or reaction with
water, 
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Strong Acids
• The stronger the acid, the
more willing it is to donate H
 use water as the standard base
HCl  H+ + ClHCl + H2O H3O+ + Cl-
• strong acids donate
practically all their H’s
 100% ionized in water
 strong electrolyte
• [H3O+] = [strong acid]
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Weak Acids
• weak acids donate a small
fraction of their H’s
HF  H+ + FHF + H2O  H3O+ + F-
most of the weak acid
molecules do not donate H
to water
much less than 1% ionized
in water
• [H3O+] << [weak acid]
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Polyprotic Acids
• often acid molecules have more than one ionizable H –
these are called polyprotic acids
 the ionizable H’s may have different acid strengths or be equal
 1 H = monoprotic, 2 H = diprotic, 3 H = triprotic
 HCl = monoprotic, H2SO4 = diprotic, H3PO4 = triprotic
• polyprotic acids ionize in steps
 each ionizable H removed sequentially
• removing of the first H automatically makes removal of
the second H harder
 H2SO4 is a stronger acid than HSO4
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Increasing Acidity
Conjugate Bases
ClO4-1
H2SO4
HI
HBr
HCl
HNO3
H3O+1
HSO4-1
H2SO3
H3PO4
HNO2
HF
HC2H3O2
H2CO3
H2S
NH4+1
HCN
HCO3-1
HS-1
H2O
CH3-C(O)-CH3
NH3
CH4
OH-1
HSO4-1
I-1
Br-1
Cl-1
NO3-1
H2O
SO4-2
HSO3-1
H2PO4-1
NO2-1
F-1
C2H3O2-1
HCO3-1
HS-1
NH3
CN-1
CO3-2
S-2
OH-1
CH3-C(O)-CH2-1
NH2-1
CH3-1
O-2
Increasing Basicity
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Acids
HClO4
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Strengths of Acids & Bases
• commonly, acid or base strength is measured by
•
•
determining the equilibrium constant of a substance’s
reaction with water
HAcid + H2O  Acid-1 + H3O+1
Base: + H2O  HBase+1 + OH-1
the farther the equilibrium position lies to the products,
the stronger the acid or base
the position of equilibrium depends on the strength of
attraction between the base form and the H+
 stronger attraction means stronger base or weaker acid
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General Trends in Acidity
• the stronger an acid is at donating H, the
weaker the conjugate base is at accepting H
• higher oxidation number = stronger oxyacid
H2SO4 > H2SO3; HNO3 > HNO2
• cation stronger acid than neutral molecule;
neutral stronger acid than anion
H3O+1 > H2O > OH-1; NH4+1 > NH3 > NH2-1
base trend opposite
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Acid Ionization Constant, Ka
• acid strength measured by the size of the
equilibrium constant when react with H2O
HAcid + H2O  Acid-1 + H3O+1
• the equilibrium constant is called the acid
ionization constant, Ka
larger Ka = stronger acid
1
1
[Acid ]  [H 3O ]
Ka 
[HAcid]
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41
Name
Benzoic
Propanoic
Formula
C6H5COOH
CH3CH2COOH
Ka1
6.14 x 10-5
1.34 x 10-5
Ka2
Ka3
Formic
HCOOH
1.77 x 10-5
Acetic
CH3COOH
1.75 x 10-5
Chloroacetic
Trichloroacetic
ClCH2COOH
Cl3C-COOH
1.36 x 10-5
1.29 x 10-4
Oxalic
HOOC-COOH
5.90 x 10-2
Nitric
HNO3
strong
Nitrous
HNO2
4.6 x 10-4
Phosphoric
Phosphorous
H3PO4
H3PO3
7.52 x 10-3
1.00 x 10-2
6.23 x 10-8
2.6 x 10-7
2.2 x 10-13
Arsenic
H3AsO4
6.0 x 10-3
1.05 x 10-7
3.0 x 10-12
Arsenious
H3AsO3
6.0 x 10-10
3.0 x 10-14
very small
Perchloric
Chloric
HClO4
HClO3
> 108
5 x 102
Chlorous
HClO2
1.1 x 10-2
Hypochlorous
HClO
3.0 x 10-8
Boric
H3BO3
5.83 x 10-10
Carbonic
H2CO3
4.45 x 10-7
6.40 x 10-5
4.7 x 10-11
Autoionization of Water
• Water is actually an extremely weak electrolyte
therefore there must be a few ions present
• about 1 out of every 10 million water molecules
form ions through a process called
autoionization
H2O  H+ + OH–
H2O + H2O  H3O+ + OH–
• all aqueous solutions contain both H3O+ and OH–
the concentration of H3O+ and OH– are equal in water
[H3O+] = [OH–] = 10-7M @ 25°C
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Ion Product of Water
• the product of the H3O+ and OH–
concentrations is always the same number
• the number is called the ion product of
water and has the symbol Kw
• [H3O+] x [OH–] = Kw = 1 x 10-14 @ 25°C
if you measure one of the concentrations, you
can calculate the other
• as [H3O+] increases the [OH–] must
decrease so the product stays constant
inversely proportional
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Acidic and Basic Solutions
• all aqueous solutions contain both H3O+ and
OH– ions
• neutral solutions have equal [H3O+] and [OH–]
[H3O+] = [OH–] = 1 x 10-7
• acidic solutions have a larger [H3O+] than [OH–]
[H3O+] > 1 x 10-7; [OH–] < 1 x 10-7
• basic solutions have a larger [OH–] than [H3O+]
[H3O+] < 1 x 10-7; [OH–] > 1 x 10-7
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Example 15.2b – Calculate the [OH] at 25°C when the
[H3O+] = 1.5 x 10-9 M, and determine if the solution is
acidic, basic, or neutral
Given:
Find:
[H3O+] = 1.5 x 10-9 M
[OH]
Concept Plan:
[H3O+]
Relationships:
Solution:
[OH]
K w  [ H 3O ][OH - ]
K w  [ H 3O  ][ OH- ]
1.0 1014
-] 
6
[OH

6
.
7

10
M
K
w

9
[OH ] 
1.5 10
[ H 3O  ]
Check: The units are correct. The fact that the
[H3O+] < [OH] means the solution is basic
Complete the Table
[H+] vs. [OH-]
[H+] 100 10-1
+
H
OH-
10-3
10-5
+
H
OH-
10-7
+
H
10-9
10-11
H+
OH OH
10-13 10-14
H+
OH
[OH-]
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Complete the Table
+
[H ] vs. [OH ]
[H+] 100 10-1
+
H
OH-
Acid
10-3
10-5
+
H
OH-
[OH-]10-14 10-13 10-11
10-9
10-7
10-9
Base
10-11
H+
+
H
10-13 10-14
H+
OH OH OH
10-7
10-5
10-3
10-1 100
even though it may look like it, neither H+ nor OH- will ever be 0
the sizes of the H+ and OH- are not to scale
because the divisions are powers of 10 rather than units
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pH
• the acidity/basicity of a solution is often
expressed as pH
• pH = -log[H3O+], [H3O+] = 10-pH
exponent on 10 with a positive sign
pHwater = -log[10-7] = 7
need to know the [H+] concentration to find pH
• pH < 7 is acidic; pH > 7 is basic, pH = 7 is
neutral
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Sig. Figs. & Logs
• when you take the log of a number written in scientific
•
notation, the digit(s) before the decimal point come from
the exponent on 10, and the digits after the decimal
point come from the decimal part of the number
log(2.0 x 106) = log(106) + log(2.0)
= 6 + 0.30303… = 6.30303...
since the part of the scientific notation number that
determines the significant figures is the decimal part,
the sig figs are the digits after the decimal point in
the log
log(2.0 x 106) = 6.30
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pH
• the lower the pH, the more acidic the solution; the
higher the pH, the more basic the solution
1 pH unit corresponds to a factor of 10 difference
in acidity
• normal range 0 to 14
pH 0 is [H+] = 1 M, pH 14 is [OH–] = 1 M
pH can be negative (very acidic) or larger than 14
(very alkaline)
51
pH of Common Substances
Substance
pH
1.0 M HCl
0.0
0.1 M HCl
1.0
stomach acid
1.0 to 3.0
lemons
2.2 to 2.4
soft drinks
2.0 to 4.0
plums
2.8 to 3.0
apples
2.9 to 3.3
cherries
3.2 to 4.0
unpolluted rainwater
5.6
human blood
7.3 to 7.4
egg whites
7.6 to 8.0
milk of magnesia (sat’d Mg(OH)2)
10.5
household ammonia
10.5 to 11.5
1.0 M NaOH
14
52
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Example 15.3b – Calculate the pH at 25°C when the
[OH] = 1.3 x 10-2 M, and determine if the solution is
acidic, basic, or neutral
Given:
Find:
Concept Plan:
Relationships:
Solution:
[OH] = 1.3 x 10-2 M
pH
[OH]
[H3O+]
K w  [ H 3O ][OH - ]
K w  [ H 3O  ][ OH- ]
14
1
.
0

10
[ H 3O  ] 
1.3  102
pH
pH  - log[H3O ]
[H 3O ]  7.7 1013 M

pH  - log 7.7 1013
pH  12.11

Check: pH is unitless. The fact that the pH > 7 means the
solution is basic
pOH
• another way of expressing the acidity/basicity of
a solution is pOH
• pOH = -log[OH], [OH] = 10-pOH
pOHwater = -log[10-7] = 7
need to know the [OH] concentration to find pOH
• pOH < 7 is basic; pOH > 7 is acidic, pOH = 7 is
neutral
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pH and pOH
Complete the Table
pH
[H+] 100 10-1
+
H
OH-
10-3
10-5
+
H
OH-
[OH-]10-14 10-13 10-11
10-9
10-7
+
H
10-9
10-11
H+
OH
OH
10-7
10-5
10-3
10-13 10-14
H+
OH
10-1 100
pOH
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pH and pOH
Complete the Table
pH
0
1
[H+] 100 10-1
+
H
OH-
3
5
7
9
10-3
10-5
10-7
10-9
+
H
OH-
[OH-]10-14 10-13 10-11
pOH 14
13
11
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10-9
9
+
H
11
13
10-11
10-13 10-14
H+
H+
OH
OH
10-7
10-5
7
5
14
OH
10-3
3
10-1 100
1
0
57
Relationship between pH and pOH
• the sum of the pH and pOH of a solution = 14.00
at 25°C
can use pOH to find pH of a solution
[ H 3O  ][OH - ]  K w  1.0  1014
 log [H O ][OH - ]   log 1.0  1014 
3
 log [H 3O ]   log [OH ]  14.00

pH  pOH  14.00
Tro, Chemistry: A Molecular Approach
58
pK
• a way of expressing the strength of an acid or
base is pK
• pKa = -log(Ka), Ka = 10-pKa
• pKb = -log(Kb), Kb = 10-pKb
• the stronger the acid, the smaller the pKa
larger Ka = smaller pKa
because it is the –log
Tro, Chemistry: A Molecular Approach
59
Finding the pH of a Strong Acid
• there are two sources of H3O+ in an aqueous
solution of a strong acid – the acid and the water
• for the strong acid, the contribution of the water
to the total [H3O+] is negligible
shifts the Kw equilibrium to the left so far that
[H3O+]water is too small to be significant
except in very dilute solutions, generally < 1 x 10-4 M
• for a monoprotic strong acid [H3O+] = [HAcid]
for polyprotic acids, the other ionizations can
generally be ignored
• 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00
Tro, Chemistry: A Molecular Approach
60
Finding the pH of a Weak Acid
• there are also two sources of H3O+ in and
aqueous solution of a weak acid – the acid and
the water
• however, finding the [H3O+] is complicated by
the fact that the acid only undergoes partial
ionization
• calculating the [H3O+] requires solving an
equilibrium problem for the reaction that defines
the acidity of the acid
HAcid + H2O  Acid + H3O+
Tro, Chemistry: A Molecular Approach
61
Ex 15.6 Find the pH of 0.200 M HNO2(aq)
solution @ 25°C
Write the reaction for
the acid with water
HNO2 + H2O  NO2 + H3O+
Construct an ICE
table for the reaction
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
initial
change
equilibrium
[HNO2] [NO2-] [H3O+]
0.200
0
≈0
since no products initially, Qc = 0, and the reaction is proceeding forward
Tro, Chemistry: A Molecular Approach
62
Ex 15.6 Find the pH of 0.200 M HNO2(aq)
solution @ 25°C
represent the change
in the concentrations
in terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
substitute into the
equilibrium constant
expression
Tro, Chemistry: A Molecular Approach
[HNO2] [NO2-] [H3O+]
0.200
initial
0
0
x
+x
+x
change
x
x
equilibrium 0.200 x

[NO-2 ][H3O ]
x x 
Ka 

HNO2 
2.00101  x
63
Ex 15.6 Find the pH of 0.200 M HNO2(aq)
solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
determine the value of
Ka from Table 15.5
[HNO2] [NO2-] [H3O+]
0.200
initial
0
≈0
change
-x
+x
+x
equilibrium 0.200
0.200x
x
x
since Ka is very small,
approximate the
[HNO2]eq = [HNO2]init
and solve for x
Ka


NO H O 
xxxx 


HNO 
2.0010  x
2

3
1
2
4
4.6 10

x
2
x
4.6 10 2.0010 
4
1
x  9.6 103
2.00101
Tro, Chemistry: A Molecular Approach
64
Ex 15.6 Find the pH of 0.200 M HNO2(aq)
solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
check if the
approximation is
valid by seeing if x
< 5% of [HNO2]init
[HNO2] [NO2-] [H3O+]
0.200
initial
0
≈0
change
-x
+x
+x
equilibrium 0.200
x
x
x = 9.6 x 10-3
3
9.6 10
1
2.0010
100%  4.8%  5%
the approximation is valid
Tro, Chemistry: A Molecular Approach
65
Ex 15.6 Find the pH of 0.200 M HNO2(aq)
solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
substitute x into the
equilibrium
concentration
definitions and solve
[HNO2] [NO2-] [H3O+]
0.200
initial
0
≈0
change
-x
+x
+x
equilibrium 0.200-x
0.190
0.0096
0.0096
x
x
x = 9.6 x 10-3
HNO2   0.200 x  0.200 9.6 103   0.190M
  H O  x  9.6 10
NO2
Tro, Chemistry: A Molecular Approach

3
3
M
66
Ex 15.6 Find the pH of 0.200 M HNO2(aq)
solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
substitute [H3O+]
into the formula for
pH and solve
initial
change
equilibrium

pH  -log H 3O


0.190
0.0096
0.0096

  2.02
3
  log 9.6 10
Tro, Chemistry: A Molecular Approach
[HNO2] [NO2-] [H3O+]
0.200
0
≈0
-x
+x
+x
67
Ex 15.6 Find the pH of 0.200 M HNO2(aq)
solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
check by substituting
[HNO2] [NO2-] [H3O+]
the equilibrium
0.200
initial
0
≈0
concentrations back into
-x
+x
+x
the equilibrium constant change
expression and
equilibrium 0.190 0.0096 0.0096
comparing the calculated
Ka to the given Ka

though not exact,
the answer is
reasonably close
Tro, Chemistry: A Molecular Approach
Ka

NO H O 


2
3
HNO2 

9.6  10 

3 2
0.190
 4.9  104
68
Practice - What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2?
(Ka = 1.4 x 10-5 @ 25°C)
Tro, Chemistry: A Molecular Approach
69
Practice - What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2?
Write the reaction for HC H NO + H O  C H NO  + H O+
6 4
2
2
6 4
2
3
the acid with water
Construct an ICE
table for the reaction
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
Tro, Chemistry: A Molecular Approach
[HA]
initial
change
equilibrium
0.012
[A-]
0
[H3O+]
≈0
70
Practice - What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2?
HC6H4NO2 + H2O  C6H4NO2 + H3O+
represent the change
in the concentrations
in terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
substitute into the
equilibrium constant
expression
Tro, Chemistry: A Molecular Approach
[HA]
initial
change
0.012
x
equilibrium 0.012 x
[A-]
0
+x
x
[H3O+]
0
+x
x

[C6 H 4 NO-2 ][H3O ]
x x 
Ka 

HC6H4 NO2 
1.2 102  x
71
Practice - What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2? Ka = 1.4 x 10-5 @ 25°C
HC6H4NO2 + H2O  C6H4NO2 + H3O+
determine the value of
Ka
[HA]
0.012
initial
change
-x
equilibrium 0.012
0.012x
since Ka is very small,
approximate the
[HA]eq = [HA]init and
solve for x


A H O 
xxxx 


HA
1.2 10  x 

-
Ka
3
5
1.4 10
2

x
2

[A2-] [H3O+]
0
≈0
+x
+x
x
x

x  1.4 105 1.2 102
x  4.1104
1.2 102
Tro, Chemistry: A Molecular Approach
72

Practice - What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2? Ka = 1.4 x 10-5 @ 25°C
Ka for HC6H4NO2 = 1.4 x 10-5
check if the
approximation is
valid by seeing if
x < 5% of
[HC6H4NO2]init
[HA]
initial
change
equilibrium
0.012
-x
0.012
[A2-] [H3O+]
0
≈0
+x
+x
x
x
x = 4.1 x 10-4
4
4.110
2
1.2 10
100%  3.4%  5%
the approximation is valid
Tro, Chemistry: A Molecular Approach
73
Practice - What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2? Ka = 1.4 x 10-5 @ 25°C
substitute x into the
equilibrium
concentration
definitions and solve
[HA]
initial
change
equilibrium
0.012
-x
0.012-x
[A2-] [H3O+]
0
≈0
+x
+x
x
x
x = 4.1 x 10-4
HC6H4 NO2   0.012 x  0.012 4.1104   0.012M

C6H4 NO2
Tro, Chemistry: A Molecular Approach
 H O  x  4.110

3
4
M
74
Practice - What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2? Ka = 1.4 x 10-5 @ 25°C
substitute [H3O+]
into the formula for
pH and solve
[HA]
[H3O+]
≈0
+x
0.012
initial
change
-x
equilibrium 0.012 0.00041 0.00041

pH  -log H 3O



  3.39
4
  log 4.110
Tro, Chemistry: A Molecular Approach
[A2-]
0
+x
75
Practice - What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2? Ka = 1.4 x 10-5 @ 25°C
check by substituting
the equilibrium
initial
concentrations back into
the equilibrium constant change
expression and
equilibrium
comparing the calculated
Ka to the given Ka
the values match
-x
[A2-]
0
+x
[H3O+]
≈0
+x
0.012
0.00041
0.00041
[HA]
0.012
[C6 H 4 NO-2 ][H 3O  ]
Ka 
HC6 H 4 NO2 

4.1  10 

 1.4  10
1.2  10 
4 2
5
2
Tro, Chemistry: A Molecular Approach
76
Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution @ 25°C
Write the reaction for
the acid with water
HClO2 + H2O  ClO2 + H3O+
Construct an ICE
table for the reaction
[HClO2] [ClO2-] [H3O+]
0.100
0
≈0
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
Tro, Chemistry: A Molecular Approach
initial
change
equilibrium
77
Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution @ 25°C
represent the change
in the concentrations
in terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
substitute into the
equilibrium constant
expression
Tro, Chemistry: A Molecular Approach
initial
change
equilibrium
[HClO2] [ClO2-] [H3O+]
0.100
0
≈0
-x
+x
+x
0.100-x
x
x

[ClO-2 ][H3O ]
x x 
Ka 

HClO2 
1.00101  x
78
Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution @ 25°C
Ka for HClO2 = 1.1 x 10-2
determine the value of
Ka from Table 15.5
since Ka is very small,
approximate the
[HClO2]eq = [HClO2]init
and solve for x
Ka
initial
change
equilibrium


ClO H O 
x x 


HClO  1.00 10 
[HClO2] [ClO2-] [H3O+]
0.100
0
≈0
-x
+x
+x
0.100-x

2
3
1
2
2
1.110

x
2
x
x
x
1.110 1.0010 
2
1
x  3.3 102
1.00101
Tro, Chemistry: A Molecular Approach
79
Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution @ 25°C
Ka for HClO2 = 1.1 x 10-2
check if the
approximation is
valid by seeing if x
< 5% of [HNO2]init
initial
change
equilibrium
[HClO2] [ClO2-] [H3O+]
0.100
0
≈0
-x
+x
+x
0.100-x
x
x
x = 3.3 x 10-2
2
3.3 10
1
1.0010
100%  33%  5%
the approximation is invalid
Tro, Chemistry: A Molecular Approach
80
Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution @ 25°C
Ka for HClO2 = 1.1 x 10-2
if the approximation
is invalid, solve for x
using the quadratic
formula
Ka

ClO H O 



2
3
HClO2 
2
1.110

x x 
1.00 10
1
 x
x2
1.0010
1
x

0  x 2  0.011x  0.0011
x
Tro, Chemistry: A Molecular Approach
 0.011
0.0112  4(1)(0.0011)
2(1)
x  0.028or - 0.039
81
Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution @ 25°C
Ka for HClO2 = 1.1 x 10-2
substitute x into the
equilibrium
concentration
definitions and solve
initial
change
equilibrium
[HClO2] [ClO2-] [H3O+]
0.100
0
≈0
-x
+x
+x
0.100-x
0.072
0.028
x
0.028
x
x = 0.028
HClO2   0.100 x  0.100 0.028  0.072M

ClO2
 H O  x  0.028M
Tro, Chemistry: A Molecular Approach

3
82
Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution @ 25°C
Ka for HClO2 = 1.1 x 10-2
substitute [H3O+]
into the formula for
pH and solve
initial
change
equilibrium

[HClO2] [ClO2-] [H3O+]
0.100
0
≈0
-x
+x
+x
pH  -log H3O
0.072

0.028
0.028

  log0.028  1.55
Tro, Chemistry: A Molecular Approach
83
Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution @ 25°C
Ka for HClO2 = 1.1 x 10-2
check by substituting
the equilibrium
concentrations back into
the equilibrium constant
expression and
comparing the calculated
Ka to the given Ka
initial
change
equilibrium
Ka
the answer matches
Tro, Chemistry: A Molecular Approach
[HClO2] [ClO2-] [H3O+]
0.100
0
≈0
-x
+x
+x
0.072
0.028
0.028

ClO H O 


2

3
HClO2 
2

0.028

 1.1  102
0.072
84
Ex 15.8 - What is the Ka of a weak acid if a
0.100 M solution has a pH of 4.25?
Use the pH to find the
equilibrium [H3O+]
Write the reaction for
the acid with water
[H3O ]  10-pH  104.25  5.6 105 M
HA + H2O  A + H3O+
Construct an ICE table
for the reaction
initial
Enter the initial
change
concentrations and
equilibrium
[H3O+]equil
Tro, Chemistry: A Molecular Approach
[HA]
0.100
[A-]
0
[H3O+]
≈0
5.6E-05
85
Ex 15.8 - What is the Ka of a weak acid if a
0.100 M solution has a pH of 4.25?
HA + H2O  A + H3O+
fill in the rest of the
table using the
[H3O+] as a guide
if the difference is
insignificant,
[HA]equil = [HA]initial
substitute into the Ka
expression and
compute Ka
[HA]
[A-]
0
[H3O+]
0
initial
change
−5.6E-05 +5.6E-05 +5.6E-05
equilibrium
0.100 
0.100
5.6E-05
0.100
5.6E-05 5.6E-05
[A- ][H 3O ] 5.6  105 5.6  105 
Ka 

HA
0.100
Ka  3.1  108
Tro, Chemistry: A Molecular Approach
86
Percent Ionization
• another way to measure the strength of an acid is
to determine the percentage of acid molecules that
ionize when dissolved in water – this is called the
percent ionization
the higher the percent ionization, the stronger the acid
molarity of ionized acid
Percent Ionization 
 100%
initial molarity of acid
• since [ionized acid]equil = [H3O+]equil

Percent Ionization 
Tro, Chemistry: A Molecular Approach
[H 3O ]equil
[HA] init
100%
87
Ex 15.9 - What is the percent ionization of a
2.5 M HNO2 solution?
Write the reaction for
the acid with water
HNO2 + H2O  NO2 + H3O+
[HNO2] [NO2-] [H3O+]
Construct an ICE table
2.5
initial
0
≈0
for the reaction
x
+x
+x
Enter the Initial
change
Concentrations
x
x
equilibrium 2.5  x
Define the Change in
Concentration in
terms of x
Sum the columns to
define the Equilibrium
Concentrations
Tro, Chemistry: A Molecular Approach
88
Ex 15.9 - What is the percent ionization of a
2.5 M HNO2 solution?
Ka for HNO2 = 4.6 x 10-4
determine the value of
Ka from Table 15.5
[HNO2] [NO2-] [H3O+]
2.5
0
≈0
-x
+x
+x
initial
change
equilibrium 2.5-x ≈2.5
since Ka is very small,
approximate the
[HNO2]eq = [HNO2]init
and solve for x
Ka

NO H O  x x 


2

3
HNO2 
2.5
4.6 104
x
x
x
x2

2.5
4.6 10 2.5
4
x  3.4 102
Tro, Chemistry: A Molecular Approach
89
Ex 15.9 - What is the percent ionization of a
2.5 M HNO2 solution?
substitute x into the
Equilibrium
Concentration
definitions and solve
x = 3.4 x 10-2
HNO2 + H2O  NO2 + H3O+
initial
change
equilibrium
[HNO2] [NO2-] [H3O+]
2.5
0
≈0
-x
+x
+x
2.52.5
x
0.034
x
0.034
x
HNO2   2.5  x  2.5  0.034  2.5 M
  H O  x  0.034M
NO2
Tro, Chemistry: A Molecular Approach

3
90
Ex 15.9 - What is the percent ionization of a
2.5 M HNO2 solution?
Apply the Definition
and Compute the
Percent Ionization
since the percent
ionization is < 5%,
the “x is small”
approximation is
valid
Tro, Chemistry: A Molecular Approach
HNO2 + H2O  NO2 + H3O+
initial
change
equilibrium
[HNO2] [NO2-] [H3O+]
2.5
0
≈0
-x
+x
+x
2.5
P ercentIonization
0.034
[H3O  ]equil
[HNO2 ]init
0.034
 100%
3.4  102

100%  1.4%
2.5
91
Relationship Between
[H3O+]equilibrium & [HA]initial
• increasing the initial concentration
•
•
of acid results in increased H3O+
concentration at equilibrium
Percent Ionization
increasing the initial concentration

[H
O
of acid results in decreased percent  3 ]equil 100%
[HA] init
ionization
this means that the increase in H3O+
concentration is slower than the
increase in acid concentration
Tro, Chemistry: A Molecular Approach
92
Why doesn’t the increase in H3O+
keep up with the increase in HA?
• the reaction for ionization of a weak acid is:
HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
• according to Le Châtelier’s Principle, if we reduce the
concentrations of all the (aq) components, the equilibrium
should shift to the right to increase the total number of
dissolved particles
 we can reduce the (aq) concentrations by using a more dilute
initial acid concentration
• the result will be a larger [H3O+] in the dilute solution
compared to the initial acid concentration
• this will result in a larger percent ionization
93
Finding the pH of Mixtures of Acids
• generally, you can ignore the contribution of the
•
•
weaker acid to the [H3O+]equil
for a mixture of a strong acid with a weak acid, the
complete ionization of the strong acid provides more
than enough [H3O+] to shift the weak acid equilibrium
to the left so far that the weak acid’s added [H3O+] is
negligible
for mixtures of weak acids, generally only need to
consider the stronger for the same reasons
 as long as one is significantly stronger than the other, and
their concentrations are similar
Tro, Chemistry: A Molecular Approach
94
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
Write the reactions for
the acids with water
and determine their Kas
If the Kas are
sufficiently different,
use the strongest acid to
construct an ICE table
for the reaction
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
Tro, Chemistry: A Molecular Approach
HF + H2O  F + H3O+
Ka = 3.5 x 10-4
HClO + H2O  ClO + H3O+ Ka = 2.9 x 10-8
H2O + H2O  OH + H3O+ Kw = 1.0 x 10-14
[HF]
initial
change
equilibrium
0.150
[F-]
0
[H3O+]
≈0
95
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
represent the change
in the concentrations
in terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
substitute into the
equilibrium constant
expression
Tro, Chemistry: A Molecular Approach
[HF]
initial
change
0.150
x
equilibrium 0.150 x
[F-]
0
+x
x
[H3O+]
0
+x
x

[F ][H 3O ]
x x 
Ka 

HF
1.50 101  x 
-

96
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
determine the value of
Ka for HF
[HF]
0.150
initial
change
-x
equilibrium 0.150
0.150x
since Ka is very small,
approximate the
[HF]eq = [HF]init and
solve for x

F H O  xxxx 



-
Ka
3
HF
4
3.5 10
0.150 x 

x
2
x
[F-]
0
+x
x
[H3O+]
≈0
+x
x
3.5 10 1.5010 
4
1
x  7.2 103
1.50101
Tro, Chemistry: A Molecular Approach
97
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
check if the
approximation is
valid by seeing if x
< 5% of [HF]init
[HF]
initial
change
equilibrium
0.150
-x
0.150
[F-]
0
+x
x
[H3O+]
≈0
+x
x
x = 7.2 x 10-3
3
7.2 10
1
1.5010
100%  4.8%  5%
the approximation is valid
Tro, Chemistry: A Molecular Approach
98
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
substitute x into the
equilibrium
concentration
definitions and solve
[HF]
initial
change
equilibrium
0.150
-x
0.150-x
0.143
[F-] [H3O+]
0
≈0
+x
+x
0.0072
0.0072
x
x
x = 7.2 x 10-3
HF  0.150 x  0.150 7.2 103   0.143M
F  H O  x  7.2 10

-
3
Tro, Chemistry: A Molecular Approach
3
M
99
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
substitute [H3O+]
into the formula for
pH and solve
initial
change
equilibrium

pH  -log H 3O
[H3O+]
≈0
+x
0.143
0.0072
0.0072


0.150

  2.14
3
  log 7.2 10
Tro, Chemistry: A Molecular Approach
-x
[F-]
0
+x
[HF]
100
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
check by substituting
the equilibrium
initial
concentrations back into
the equilibrium constant change
expression and
equilibrium
comparing the calculated
Ka to the given Ka
though not exact,
the answer is
reasonably close
Tro, Chemistry: A Molecular Approach
Ka
-x
[F-]
0
+x
[H3O+]
≈0
+x
0.143
0.0072
0.0072
[HF]
0.150

F H O 



-
3
HF

7.2  10 

3 2
0.143
 3.6  104
101
Strong Bases
• the stronger the base, the more
willing it is to accept H
use water as the standard acid
• for strong bases, practically all
NaOH  Na+ + OH-
molecules are dissociated into
OH– or accept H’s
strong electrolyte
multi-OH strong bases
completely dissociated
• [HO–] = [strong base] x (# OH)
Tro, Chemistry: A Molecular Approach
102
Example 15.11b – Calculate the pH at 25°C of a 0.0015 M
Sr(OH)2 solution and determine if the solution is acidic,
basic, or neutral
Given:
Find:
[Sr(OH)2] = 1.5 x 10-3 M
pH
Concept Plan: [Sr(OH) ]
2
[OH]
[H3O+]
pH

Relationships: [OH]=2[Sr(OH)2] K w  [ H 3O ][OH ] pH  - log[H3O ]
Solution:
[OH]
= 2(0.0015)
= 0.0030 M
Check:
K w  [ H 3O ][ OH- ]

14
1
.
0

10
[ H 3O  ] 
3.0  103
[H 3O ]  3.3 1012 M

pH  - log 3.3 1012
pH  11.48
pH is unitless. The fact that the pH > 7 means the
solution is basic

Practice - Calculate the pH of a 0.0010 M
Ba(OH)2 solution and determine if it is
acidic, basic, or neutral
Tro, Chemistry: A Molecular Approach
105
Practice - Calculate the pH of a 0.0010 M
Ba(OH)2 solution and determine if it is
acidic, basic, or neutral
Ba(OH)2 = Ba2+ + 2 OH- therefore
[OH-] = 2 x 0.0010 = 0.0020 = 2.0 x 10-3 M
Kw = [H3O+][OH]
-14
1.00
x
10
-12M
[H3O+] =
=
5.0
x
10
2.0 x 10-3
pH = -log [H3O+] = -log (5.0 x 10-12)
pH = 11.30
pH > 7 therefore basic
Tro, Chemistry: A Molecular Approach
106
Weak Bases
• in weak bases, only a small
fraction of molecules accept H’s
NH3 + H2O  NH4+ + OHweak electrolyte
most of the weak base molecules
do not take H from water
much less than 1% ionization in
water
• [HO–] << [weak base]
• finding the pH of a weak base
solution is similar to finding the
pH of a weak acid
Tro, Chemistry: A Molecular Approach
107
Tro, Chemistry: A Molecular Approach
108
Structure of Amines
Tro, Chemistry: A Molecular Approach
109
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Write the reaction for
the base with water
NH3 + H2O  NH4+ + OH
Construct an ICE
table for the reaction
Enter the initial
concentrations –
assuming the [OH]
from water is ≈ 0
initial
change
equilibrium
[NH3] [NH4+] [OH]
0.100
0
≈0
since no products initially, Qc = 0, and the reaction is proceeding forward
Tro, Chemistry: A Molecular Approach
110
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
represent the change
in the concentrations
in terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
substitute into the
equilibrium constant
expression
Tro, Chemistry: A Molecular Approach
[NH3] [NH4+] [OH]
0.100
initial
0
0
x
+x
+x
change
x
x
equilibrium 0.100 x

[NH4 ][OH ]
x x 
Kb 

NH3 
1.00 101  x
111
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Kb for NH3 = 1.76 x 10-5
determine the value of
Kb from Table 15.8
[NH3] [NH4+] [OH]
0.100
initial
0
≈0
change
-x
+x
+x
x
equilibrium 0.100
0.100
x
x
since Kb is very small,
approximate the
[NH3]eq = [NH3]init
and solve for x
Kbb


NH OH 
xxxx 


NH 
1.0010  x

44

11
33
5
1.7610

x
2


x  1.76105 1.00101
x  1.33103
1.00101
Tro, Chemistry: A Molecular Approach
112

Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Kb for NH3 = 1.76 x 10-5
check if the
approximation is
valid by seeing if x
< 5% of [NH3]init
initial
change
equilibrium
[NH3] [NH4+] [OH]
0.100
0
≈0
-x
+x
+x
0.100
x
x
x = 1.33 x 10-3
3
1.3310
1
1.0010
100%  1.33%  5%
the approximation is valid
Tro, Chemistry: A Molecular Approach
113
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Kb for NH3 = 1.76 x 10-5
substitute x into the
equilibrium
concentration
definitions and solve
[NH3] [NH4+] [OH]
0.100
initial
0
≈0
change
-x
+x
+x
0.099x 1.33E-3
1.33E-3
equilibrium 0.100
x
x
x = 1.33 x 10-3
NH3   0.100 x  0.100 1.33103   0.099M

3
[NH4 ]  [OH ]  x  1.3310 M
Tro, Chemistry: A Molecular Approach
-
114
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Kb for NH3 = 1.76 x 10-5
use the [OH-] to find
the [H3O+] using Kw
substitute [H3O+]
into the formula for
pH and solve
K w  [H3O  ][OH- ]
-14
1.00

10
[H3O  ] 
1.33 10-3
[H3O  ]  7.52 10-12
Tro, Chemistry: A Molecular Approach
initial
change
equilibrium
[NH3] [NH4+] [OH]
0.100
0
≈0
-x
+x
+x
0.099
1.33E-3 1.33E-3
pH  -logH O 
  log7.5210   11.124

3
12
115
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Kb for NH3 = 1.76 x 10-5
check by substituting
the equilibrium
concentrations back into initial
the equilibrium constant
change
expression and
comparing the calculated equilibrium
Kb to the given Kb
though not exact,
the answer is
reasonably close
Tro, Chemistry: A Molecular Approach
Kb
[NH3] [NH4+] [OH]
0.100
0
≈0
-x
+x
+x
0.099
1.33E-3 1.33E-3

NH OH 


4

NH3 

1.33  10 

3 2
0.099
 1.8  105
117
Practice – Find the pH of a 0.0015 M morphine
solution, Kb = 1.6 x 10-6
Tro, Chemistry: A Molecular Approach
118
Practice – Find the pH of a 0.0015 M morphine solution
Write the reaction for
the base with water
Construct an ICE
table for the reaction
Enter the initial
concentrations –
assuming the [OH]
from water is ≈ 0
B + H2O  BH+ + OH
[B]
initial
change
equilibrium
0.0015
[BH+] [OH]
0
≈0
since no products initially, Qc = 0, and the reaction is proceeding forward
Tro, Chemistry: A Molecular Approach
119
Practice – Find the pH of a 0.0015 M morphine solution
represent the change
in the concentrations
in terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
substitute into the
equilibrium constant
expression
Tro, Chemistry: A Molecular Approach
[BH+] [OH]
0.0015
initial
0
0
x
+x
+x
change
x
x
equilibrium 0.0015 x
[B]

[BH ][OH ]
x x 
Kb 

B
1.5 103  x
120
Practice – Find the pH of a 0.0015 M morphine solution
Kb for Morphine = 1.6 x 10-6
determine the value of
Kb
[BH+] [OH]
0.0015
initial
0
≈0
change
-x
+x
+x
x
equilibrium 0.0015
0.0015
x
x
[B]
since Kb is very small,
approximate the
[B]eq = [B]init and
solve for x


BH
xxxx 
BH OH
OH 


BB
11..5510
10  x 

K
Kbb

33
6
1.6 10

x
2


x  1.6 106 1.5 103
x  4.9 105
1.5 101
Tro, Chemistry: A Molecular Approach
121

Practice – Find the pH of a 0.0015 M morphine solution
Kb for Morphine = 1.6 x 10-6
check if the
approximation is
valid by seeing if x
< 5% of [B]init
[B]
initial
change
equilibrium
0.0015
-x
0.0015
[BH+] [OH]
0
≈0
+x
+x
x
x
x = 4.9 x 10-5
5
4.9 10
3
1.5 10
100%  3.3%  5%
the approximation is valid
Tro, Chemistry: A Molecular Approach
122
Practice – Find the pH of a 0.0015 M morphine solution
Kb for Morphine = 1.6 x 10-6
substitute x into the
equilibrium
concentration
definitions and solve
[BH+] [OH]
0.0015
initial
0
≈0
change
-x
+x
+x
0.0015x 4.9E-5
4.9E-5
equilibrium 0.0015
x
x
[B]
x = 4.9 x 10-5
Morphine  0.0015 x  0.0015 4.9 105   0.0015M

5
[BH ]  [OH ]  x  4.9 10 M
Tro, Chemistry: A Molecular Approach
-
123
Practice – Find the pH of a 0.0015 M morphine solution
Kb for Morphine = 1.6 x 10-6
use the [OH-] to find
the [H3O+] using Kw
substitute [H3O+]
into the formula for
pH and solve
K w  [H3O  ][OH- ]
-14
1.00

10
[H3O  ] 
4.9  10-5
[H3O  ]  2.0  10-10
Tro, Chemistry: A Molecular Approach
[B]
initial
change
equilibrium
0.0015
-x
0.0015
[BH+] [OH]
0
≈0
+x
+x
4.9E-5
4.9E-5
pH  -logH O 
  log2.0 10   9.69

3
10
124
Practice – Find the pH of a 0.0015 M morphine solution
Kb for Morphine = 1.6 x 10-6
check by substituting
the equilibrium
concentrations back into initial
the equilibrium constant
change
expression and
comparing the calculated equilibrium
Kb to the given Kb
the answer
matches the
given Kb
Tro, Chemistry: A Molecular Approach
Kb
[BH+] [OH]
0
≈0
+x
+x
[B]
0.0015
-x
0.0015
4.9E-5
4.9E-5

BH OH 



B

4.9  10 

5 2
0.0015
 1.6  106
126
Acid-Base Properties of Salts
• salts are water soluble ionic compounds
• salts that contain the cation of a strong base and an
anion that is the conjugate base of a weak acid are
basic
 NaHCO3 solutions are basic
 Na+ is the cation of the strong base NaOH
 HCO3− is the conjugate base of the weak acid H2CO3
• salts that contain cations that are the conjugate acid of a
weak base and an anion of a strong acid are acidic
 NH4Cl solutions are acidic
 NH4+ is the conjugate acid of the weak base NH3
 Cl− is the anion of the strong acid HCl
Tro, Chemistry: A Molecular Approach
127
Anions as Weak Bases
• every anion can be thought of as the conjugate base of an
•
acid
therefore, every anion can potentially be a base
 A−(aq) + H2O(l)  HA(aq) + OH−(aq)
• the stronger the acid is, the weaker the conjugate base is
 an anion that is the conjugate base of a strong acid is pH neutral
Cl−(aq) + H2O(l)  HCl(aq) + OH−(aq)
 since HCl is a strong acid, this equilibrium lies practically completely to the
left
 an anion that is the conjugate base of a weak acid is basic
F−(aq) + H2O(l)  HF(aq) + OH−(aq)
 since HF is a weak acid, the position of this equilibrium favors the right
Tro, Chemistry: A Molecular Approach
128
Ex 15.13 - Use the Table to Determine if
the Given Anion Is Basic or Neutral
a) NO3−
the conjugate base of a
strong acid, therefore
neutral
b) NO2−
the conjugate base of a
weak acid, therefore
basic
Tro, Chemistry: A Molecular Approach
129
Relationship between Ka of an Acid and
Kb of Its Conjugate Base
• many reference books only give tables of Ka values
because Kb values can be found from them
when you add
equations,
you multiply
the K’s


HA(aq)  H2O(l )  A (aq)  H3O (aq)

[A  ][H3O ]
Ka 
[HA]

A (aq)  H2O(l )  HA(aq)  OH (aq)
[ HA][H3O ]
Kb 
[A ]
2 H 2O(l )  H3O  (aq )  OH  (aq )
[A  ][ H 3O  ] [ HA ][ OH  ]
Ka  K b 

[ HA ]
[A  ]
Ka  K b  [ H 3O  ][ OH  ]  K w
130
Na+
is the cation of a
strong base – pH
neutral. The CHO2−
is the anion of a
weak acid – pH basic
Write the reaction for
the anion with water
Construct an ICE
table for the reaction
Ex 15.14 Find the pH of 0.100 M
NaCHO2(aq) solution
CHO2− + H2O  HCHO2 + OH
[CHO2−] [HCHO2] [OH]
initial
change
equilibrium
0.100
0
≈0
Enter the initial
concentrations –
assuming the [OH]
from water is ≈ 0
Tro, Chemistry: A Molecular Approach
131
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution
represent the change
in the concentrations
in terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
Calculate the value
of Kb from the value
of Ka of the weak
acid from Table 15.5
substitute into the
equilibrium constant
expression
[CHO2−] [HCHO2] [OH]
0.100
initial
x
change
equilibrium 0.100 x
0
≈0
+x
+x
x
x
Ka  K b  K w
1.0  1014
11
Kb 

5
.
6

10
1.8  104
x x 
[HCHO2 ][OH ]
Kb 


1.00 101  x
CHO2


132
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution
Kb for CHO2− = 5.6 x 10-11
[CHO2−] [HCHO2] [OH]
since Kb is very small,
approximate the
initial
−
−
[CHO2 ]eq = [CHO2 ]init
change
and solve for x
-x
equilibrium 0.100
0.100x
xxxx 
[HCHO
[HCHO22][
][OH
OH]]
K
Kbb 



11..00
0010
1011 x 
CHO
CHO22

0
≈0
+x
x
+x
x
0.100

5.6  10
x
11
2
x

1.00  10 1
5.6 10 1.00 10 
11
1
x  2.4  106
Tro, Chemistry: A Molecular Approach
133
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution
Kb for CHO2− = 5.6 x 10-11
[CHO2−] [HCHO2] [OH]
initial
change
equilibrium
check if the
approximation is
valid by seeing if x
< 5% of [CHO2−]init
0.100
0
≈0
-x
+x
x
+x
x
0.100
x = 2.4 x 10-6
6
2.4  10
 100%  0.0024 %  5%
1
1.00  10
the approximation is valid
Tro, Chemistry: A Molecular Approach
134
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution
Kb for CHO2− = 5.6 x 10-11
[CHO2−] [HCHO2] [OH]
substitute x into the
equilibrium
concentration
definitions and solve
0.100
initial
change
-x
0.100−x
equilibrium 0.100
0
≈0
+x
2.4E-6
x
+x
2.4E-6
x
x = 2.4 x 10-6
CHO  0.100  x  0.100  2.4 10   0.100 M

6
2
[HCHO2 ]  [OH ]  x  2.4  10
-
Tro, Chemistry: A Molecular Approach
6
M
135
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution
Kb for CHO2− = 5.6 x 10-11
[CHO2−] [HCHO2] [OH]
use the [OH-] to find
the [H3O+] using Kw
initial
change
equilibrium
substitute [H3O+]
into the formula for
pH and solve
K w  [H3O  ][OH- ]
1.00 10
[H3O ] 
2.4  10-6
[H3O  ]  4.2  10-9

-14
Tro, Chemistry: A Molecular Approach
0.100
0
≈0
-x
+x
+x
0.100
2.4E-6
2.4E-6

pH  -log H3O


  log 4.2  10

9
  8.38
136
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution
Kb for CHO2− = 5.6 x 10-11
check by substituting
−] [HCHO ] [OH]
[CHO
2
2
the equilibrium
concentrations back into initial
0.100
0
≈0
the equilibrium constant
change
-x
+x
+x
expression and
0.100
2.4E-6 2.4E-6
comparing the calculated equilibrium
Kb to the given Kb
though not exact,
the answer is
reasonably close
Tro, Chemistry: A Molecular Approach
Kb

HCHO2 OH  

CHO 


2.4  10 

6 2
0.100
2
 5.8  1011
138
Polyatomic Cations as Weak Acids
• some cations can be thought of as the conjugate acid
of a base
 others are the counterions of a strong base
• therefore, some cation can potentially be an acid
 MH+(aq) + H2O(l)  MOH(aq) + H3O+(aq)
• the stronger the base is, the weaker the conjugate
acid is
 a cation that is the counterion of a strong base is pH
neutral
 a cation that is the conjugate acid of a weak base is acidic
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
 since NH3 is a weak base, the position of this equilibrium favors
the right
Tro, Chemistry: A Molecular Approach
139
Metal Cations as Weak Acids
• cations of small, highly charged metals are weakly
acidic
 alkali metal cations and alkali earth metal cations pH neutral
 cations are hydrated
Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+ (aq) + H3O+(aq)
Tro, Chemistry: A Molecular Approach
140
Ex 15.15 - Determine if the Given
Cation Is Acidic or Neutral
a) C5N5NH2+
the conjugate acid of a weak base, therefore
acidic
b) Ca2+
the counterion of a strong base, therefore neutral
c) Cr3+
a highly charged metal ion, therefore acidic
Tro, Chemistry: A Molecular Approach
141
Classifying Salt Solutions as
Acidic, Basic, or Neutral
• if the salt cation is the counterion of a strong
base and the anion is the conjugate base of a
strong acid, it will form a neutral solution
NaCl
Ca(NO3)2
KBr
• if the salt cation is the counterion of a strong
base and the anion is the conjugate base of a
weak acid, it will form a basic solution
NaF
Ca(C2H3O2)2
Tro, Chemistry: A Molecular Approach
KNO2
142
Classifying Salt Solutions as
Acidic, Basic, or Neutral
• if the salt cation is the conjugate acid of a weak
base and the anion is the conjugate base of a
strong acid, it will form an acidic solution
NH4Cl
• if the salt cation is a highly charged metal ion
and the anion is the conjugate base of a strong
acid, it will form an acidic solution
Al(NO3)3
Tro, Chemistry: A Molecular Approach
143
Classifying Salt Solutions as
Acidic, Basic, or Neutral
• if the salt cation is the conjugate acid of a weak
base and the anion is the conjugate base of a
weak acid, the pH of the solution depends on the
relative strengths of the acid and base
NH4F since HF is a stronger acid than NH4+, Ka of
NH4+ is larger than Kb of the F−; therefore the
solution will be acidic
Tro, Chemistry: A Molecular Approach
144
Ex 15.16 - Determine whether a solution of the
following salts is acidic, basic, or neutral
a) SrCl2
Sr2+ is the counterion of a strong base, pH neutral
Cl− is the conjugate base of a strong acid, pH neutral
solution will be pH neutral
b) AlBr3
Al3+ is a small, highly charged metal ion, weak acid
Cl− is the conjugate base of a strong acid, pH neutral
solution will be acidic
c) CH3NH3NO3
CH3NH3+ is the conjugate acid of a weak base, acidic
NO3− is the conjugate base of a strong acid, pH neutral
solution will be acidic
Tro, Chemistry: A Molecular Approach
145
Ex 15.16 - Determine whether a solution of the
following salts is acidic, basic, or neutral
d) NaCHO2
e)
Na+ is the counterion of a strong base, pH neutral
CHO2− is the conjugate base of a weak acid, basic
solution will be basic
NH4F
NH4+ is the conjugate acid of a weak base, acidic
F− is the conjugate base of a weak acid, basic
Ka(NH4+) > Kb(F−); solution will be acidic
Tro, Chemistry: A Molecular Approach
146
Polyprotic Acids
• since polyprotic acids ionize in steps, each H has a
•
•
separate Ka
Ka1 > Ka2 > Ka3
generally, the difference in Ka values is great enough so
that the second ionization does not happen to a large
enough extent to affect the pH
 most pH problems just do first ionization
 except H2SO4  use [H2SO4] as the [H3O+] for the second
ionization
• [A2-] = Ka2 as long as the second ionization is negligible
Tro, Chemistry: A Molecular Approach
147
148
Ex 15.18 Find the pH of 0.0100 M H2SO4(aq)
solution @ 25°C
Write the reactions
for the acid with
water
Construct an ICE
table for the reaction
Enter the initial
concentrations –
assuming the
[HSO4−] and [H3O+]
is ≈ [H2SO4]
Tro, Chemistry: A Molecular Approach
H2SO4 + H2O  HSO4 + H3O+
HSO4 + H2O  SO42 + H3O+
initial
change
equilibrium
[HSO4 ] [SO42 ] [H3O+]
0.0100
0
0.0100
149
Ex 15.18 Find the pH of 0.0100 M H2SO4(aq)
solution @ 25°C
represent the change
in the concentrations
in terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
[HSO4 ] [SO42 ] [H3O+]
0.0100
initial
change
−x
equilibrium 0.0100 −x
substitute into the
equilibrium constant
expression
Tro, Chemistry: A Molecular Approach
0
0.0100
+x
+x
x
0.0100 −x
[SO 24- ][H3O ] x 0.0100 x 
Ka 

0.0100 x 
[ HSO 4 ]
150
Ex 15.18 Find the pH of 0.0100 M H2SO4(aq)
solution @ 25°C
Ka for HSO4− = 0.012
expand and solve for
x using the quadratic
formula
[SO 24- ][H 3O ] x 0.0100 x 
Ka 

0.0100 x 
[ HSO 4 ]

0.0100  x x
0.012 
0.0100  x 
1.2  10  4  1.2  10  2 x  1.00  10 2 x  x 2
0  x 2  0.022 x  0.00012
x
Tro, Chemistry: A Molecular Approach
 0.022 
0.022 2  4(1)(0.00012 )
2(1)
x  0.027 or 0.0045
151
Ex 15.18 Find the pH of 0.0100 M H2SO4(aq)
solution @ 25°C
Ka for HSO4− = 0.012
substitute x into the
equilibrium
initial
concentration
definitions and solve change
[HSO4 ] [SO42 ] [H3O+]
0.0100
0
0.0100
−x
+x
+x
0.0055−x 0.0045
x
0.0100
0.0145−x
equilibrium 0.0100
x = 0.0045
[HSO 4 ]  0.0100  x  0.0100  0.0045   0.0055 M

[H3O ]  0.0100  x   0.0145 M

[SO 24- ]  x  0.0045 M
Tro, Chemistry: A Molecular Approach
152
Ex 15.18 Find the pH of 0.0100 M H2SO4(aq)
solution @ 25°C
Ka for HSO4− = 0.012
substitute [H3O+]
into the formula for
pH and solve
[HSO4 ] [SO42 ] [H3O+]
initial
change
equilibrium

0.0100
0
0.0100
−x
+x
+x
0.0055
0.0045
0.0145


pH  -log H3O
  log 0.0145   1.839
Tro, Chemistry: A Molecular Approach
153
Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution @ 25°C
Ka for HSO4− = 0.012
check by substituting
the equilibrium
concentrations back
into the equilibrium
constant expression
and comparing the
calculated Ka to the
given Ka
[HSO4 ] [SO42 ] [H3O+]
initial
change
equilibrium
the answer matches
Tro, Chemistry: A Molecular Approach
Ka
0.0100
0
0.0100
−x
+x
+x
0.0055
0.0045
0.0145

SO H O 


HSO 
24

3
4

0.00450.0145

 1.2  102
0.0055
154
Strengths of Binary Acids
• the more d+ H-X d- polarized the
•
•
bond, the more acidic the bond
the stronger the H-X bond, the
weaker the acid
binary acid strength increases to the
right across a period
 H-C < H-N < H-O < H-F
• binary acid strength increases down
the column
 H-F < H-Cl < H-Br < H-I
Tro, Chemistry: A Molecular Approach
155
Strengths of Oxyacids, H-O-Y
• the more electronegative the Y atom, the
stronger the acid
helps weakens the H-O bond
• the more oxygens attached to Y, the stronger the
acid
further weakens and polarizes the H-O bond
Tro, Chemistry: A Molecular Approach
156
Lewis Acid - Base Theory
• electron sharing
• electron donor = Lewis Base = nucleophile
must have a lone pair of electrons
• electron acceptor = Lewis Acid = electrophile
electron deficient
• when Lewis Base gives electrons from lone
pair to Lewis Acid, a covalent bond forms
between the molecules
Nucleophile: + Electrophile  Nucleophile:Electrophile
• product called an adduct
• other acid-base reactions also Lewis
Tro, Chemistry: A Molecular Approach
157
Example - Complete the Following
Lewis Acid-Base Reactions
Label the Nucleophile and Electrophile
OH
H
C

H
+ OH-1 
OH
H
C

Electrophile
OH
H
••
•
-1 
+ • OH
••
Nucleophile
Tro, Chemistry: A Molecular Approach
H
C
H
OH
158
Practice - Complete the Following
Lewis Acid-Base Reactions
Label the Nucleophile and Electrophile
• BF3 + HF 
• CaO + SO3 
• KI + I2 
Tro, Chemistry: A Molecular Approach
159
Practice - Complete the Following
Lewis Acid-Base Reactions
Label the Nucleophile and Electrophile
• BF3 + HF  H+1BF4-1
Nuc
••
H F
••
F
H+1 F
+B
• CaO + SO3  Ca+2SO4-2
F
F
O
O -2
Nuc
Elec
• KI + I2  KI3
Elec
Tro, Chemistry: A Molecular Approach
+S
O
Ca+2 O
O
K+1
••
I
••
-1
••
••
Nuc
-2
••
••
Ca+2
••
O
••
F
-1
••
Ele
c
F
+ I
B
F
S
O
O
I
K+1 I
I
I -1
160
What Is Acid Rain?
• natural rain water has a pH of 5.6
naturally slightly acidic due mainly to CO2
• rain water with a pH lower than 5.6 is called
acid rain
• acid rain is linked to damage in ecosystems and
structures
Tro, Chemistry: A Molecular Approach
161
What Causes Acid Rain?
• many natural and pollutant gases dissolved in the air
are nonmetal oxides
 CO2, SO2, NO2
• nonmetal oxides are acidic
•
CO2 + H2O  H2CO3
2 SO2 + O2 + 2 H2O  2 H2SO4
processes that produce nonmetal oxide gases as waste
increase the acidity of the rain
 natural – volcanoes and some bacterial action
 man-made – combustion of fuel
• weather patterns may cause rain to be acidic in regions
other than where the nonmetal oxide is produced
Tro, Chemistry: A Molecular Approach
162
pH of Rain in Different Regions
Sources of SO2 from Utilities
Damage from Acid Rain
• acids react with metals, and materials that contain
•
•
•
•
carbonates
acid rain damages bridges, cars, and other metallic
structures
acid rain damages buildings and other structures made
of limestone or cement
acidifying lakes affecting aquatic life
dissolving and leaching more minerals from soil
 making it difficult for trees
Tro, Chemistry: A Molecular Approach
165
Acid Rain Legislation
• 1990 Clean Air Act attacks acid rain
force utilities to reduce SO2
• result is acid rain in northeast stabilized and
beginning to be reduced
Tro, Chemistry: A Molecular Approach
166
Damage from Acid Rain
167