Transcript Chapter 15 Acids and Bases

Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 15
Acids and
Bases
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
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
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
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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
:B
base
HCHO2 +
acid
H2O 
base
H2O +
acid
NH3 
base
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:A– +
conjugate
base
CHO2–
conjugate
base
H–B+
conjugate
acid
+
H3O+
conjugate
acid
HO– +
conjugate
base
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
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H2O 
base
HSO4–
conjugate
base
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+
H3O+
conjugate
acid
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
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H2O
acid

H2CO3
conjugate
acid
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HO–
conjugate
base
+
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 Acids
• The stronger the acid, the
more willing it is to donate H
– use water as the standard
base
• strong acids donate
practically all their H’s
– 100% ionized in water
– strong electrolyte
• [H3O+] = [strong acid]
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HCl  H+ + ClHCl + H2O H3O+ + Cl-
Weak Acids
• weak acids donate a small
fraction of their H’s
– 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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HF  H+ + FHF + H2O  H3O+ + F-
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-132
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
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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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
10-3
+
H
OH-
10-5
10-7
+
H
OH-
10-9
10-11
+
H
H+
OH OH
[OH-]
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10-13 10-14
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H+
OH
Complete the Table
[H+] vs. [OH-]
Acid
[H+] 100 10-1
10-3
10-5
Base
10-7
+
H
+
H
OH-
OH-
[OH-]10-14 10-13 10-11 10-9
10-7
10-9
10-11
+
H
10-13 10-14
H+
OH OH OH
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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H+
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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
Approach
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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)
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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
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14
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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:
[OH] = 1.3 x 10-2 M
pH
Concept Plan:
[OH]
Relationships:
Solution:
K w  [ H 3O ][OH - ]
K w  [ H 3O  ][ OH- ]
14
1
.
0

10
[ H 3O  ] 
1.3  102
Check:
[H3O+]
pH
pH  - log[H3O ]
[H 3O ]  7.7 1013 M

pH  - log 7.7 1013
pH  12.11

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
10-3
+
H
OH-
10-5
10-7
+
H
OH-
[OH-]10-14 10-13 10-11 10-9
10-7
10-9
10-11
+
H
OH
10-5
10-3
pOH
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10-13 10-14
H+
OH
10-1 100
H+
OH
pH and pOH
Complete the Table
pH
0 1
[H+] 100 10-1
3
5
7
9
11
13
10-3
10-5
10-7
10-9
10-11
10-13 10-14
+
H
OH-
+
H
OH-
[OH-]10-14 10-13 10-11 10-9
pOH 14
13
11
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+
H
OH
10-7
10-5
7
5
10-3
3
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14
H+
OH
10-1 100
1
0
H+
OH
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
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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
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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
55
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
56
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
57
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
58
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
x
equilibrium 0.200
0.200
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

Tro, Chemistry: A Molecular
Approach
x
2
2.00101
59
x
4.6 10 2.0010 
4
x  9.6 103
1
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
60
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
61
M
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

  2.02
3
  log 9.6 10
Tro, Chemistry: A Molecular
Approach
[HNO2] [NO2-] [H3O+]
0.200
0
≈0
-x
+x
+x
62
0.0096
0.0096
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
Ka

NO H O 


2
HNO2 

9.6  10 

3 2
0.190
Tro, Chemistry: A Molecular
Approach
63
3
 4.9  104
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
64
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
65
0.012
[A-]
0
[H3O+]
≈0
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
equilibrium
0.012
x
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
66
Practice - What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2? Ka = 1.4 x 10-5 @ 25°C
determine the value of
Ka
HC6H4NO2 + H2O  C6H4NO2 + H3O+
[HA]
initial
change
equilibrium
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

Tro, Chemistry: A Molecular
Approach
x
2
1.2 102
67
0.012
-x
0.012
x
0.012

[A2-] [H3O+]
0
≈0
+x
+x
x
x

x  1.4 105 1.2 102
x  4.1104

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
68
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
69
4
M
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
70
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
2
Tro, Chemistry: A Molecular
Approach
71
5
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
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
Tro, Chemistry: A Molecular
Approach
initial
change
equilibrium
72
[HClO2] [ClO2-] [H3O+]
0.100
0
≈0
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
73
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
x
1
2
2
1.110
Tro, Chemistry: A Molecular
Approach

x
x
1.110 1.0010 
2
x  3.3 102
2
1.00101
74
x
1
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 = 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
75
x
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
76
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
x = 0.028
0.028
x
HClO2   0.100 x  0.100 0.028  0.072M

ClO2
Tro, Chemistry: A Molecular
Approach
 H O  x  0.028M

3
77
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


  log0.028  1.55
Tro, Chemistry: A Molecular
Approach
78
0.028
0.028
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
the answer matches
Tro, Chemistry: A Molecular
Approach
initial
change
equilibrium
Ka
[HClO2] [ClO2-] [H3O+]
0.100
0
≈0
-x
+x
+x
0.072
0.028

ClO H O 


2

3
HClO2 
2

0.028

 1.1  102
0.072
79
0.028
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
80
[HA]
0.100
[A-]
0
[H3O+]
≈0
5.6E-05
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]
initial
change
equilibrium
−5.6E-05
0.100 
0.100
5.6E-05
+5.6E-05
5.6E-05
[H3O+]
0
+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
0.100
[A-]
0
81
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 
initial molarity of acid
 100%
• since [ionized acid]equil = [H3O+]equil

Percent Ionization 
Tro, Chemistry: A Molecular
Approach
82
[H 3O ]equil
[HA] init
100%
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
83
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
since Ka is very small,
approximate the
[HNO2]eq = [HNO2]init
and solve for x
Ka
[HNO2] [NO2-] [H3O+]
2.5
0
≈0
-x
+x
+x
initial
change
equilibrium 2.5-x ≈2.5
4.6 104

NO H O  x x 


2

3
HNO2 
2.5
x
x2

2.5
4.6 10 2.5
4
x  3.4 102
Tro, Chemistry: A Molecular
Approach
84
x
x
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
x
0.034
HNO2   2.5  x  2.5  0.034  2.5 M
  H O  x  0.034M
NO2
Tro, Chemistry: A Molecular
Approach

3
85
x
0.034
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
3.4  102

100%  1.4%
2.5
86
0.034
 100%
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%
ionization
[HA] init
• this means that the increase in H3O+
concentration is slower than the
increase in acid concentration
Tro, Chemistry: A Molecular
Approach
87
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
88
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
89
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
HF + H2O  F + H3O+
Write the reactions for
the acids with water
and determine their Kas
HClO + H2O  ClO + H3O+ Ka = 2.9 x 10-8
H2O + H2O  OH + H3O+
If the Kas are
sufficiently different,
use the strongest acid to
-]
[HF]
[F
construct an ICE table
for the reaction
0.150
initial
0
Enter the initial
change
concentrations –
equilibrium
assuming the [H3O+]
from water is ≈ 0
Tro, Chemistry: A Molecular
Approach
Ka = 3.5 x 10-4
90
Kw = 1.0 x 10-14
[H3O+]
≈0
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 
-
91

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]
initial
change
equilibrium
since Ka is very small,
approximate the
[HF]eq = [HF]init and
solve for x

F H O  xxxx 


-x
0.150
x
0.150

-
Ka
0.150
3
HF
4
3.5 10
0.150 x 

Tro, Chemistry: A Molecular
Approach
x
x
[F-]
0
+x
x
[H3O+]
≈0
+x
x
3.5 10 1.5010 
4
x  7.2 103
2
1.50101
92
1
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
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
93
[H3O+]
≈0
+x
x
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
3
Tro, Chemistry: A Molecular
Approach
94
M
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]
95
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
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
Tro, Chemistry: A Molecular
Approach
96
 3.6  104
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
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
97
NaOH  Na+ + OH-
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:
Concept Plan:
Relationships:
[Sr(OH)2] = 1.5 x 10-3 M
pH
[OH]
[Sr(OH)2]
[OH]=2[Sr(OH)2]
K w  [ H 3O ][ OH- ]

14
1
.
0

10
[ H 3O  ] 
3.0  103
Check:
pH
K w  [ H 3O ][OH - ] pH  - log[H3O ]
Solution:
[OH]
= 2(0.0015)
= 0.0030 M
[H3O+]
[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
100
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
101
Weak Bases
• in weak bases, only a small
fraction of molecules accept H’s
– 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
102
NH3 + H2O  NH4+ + OH-
Tro, Chemistry: A Molecular
Approach
103