Transcript Chapter 15 Acids and Bases

RNA uses amino-acids to
build proteins/enzymes
Digestive Acids help to break down food
into reusable molecular fragments
Acids and
Bases
It is the acids in citrus fruits
that give them the sour taste
and allows the fruit to stay in a
state of preservation till
germination
Properties of Acids
 sour taste
 react with active metals (Al, Zn, Fe), but not Cu, Ag, or Au
2 Al(s) + 6 HCl(aq)  2 AlCl3(aq) + 3 H2(g)
 Corrosive
 react with carbonates, producing CO2
 marble, baking soda, chalk, limestone
CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)
 change color of vegetable dyes
 blue litmus turns red
 react with bases to form ionic salts
HCl(aq) + NaOH(aq)NaCl(aq) + H2O(l)
2
Common Acids
Chemical Name
Formula
Uses
Strength
Nitric Acid
HNO3
explosive, fertilizer, dye, glue
Strong
Sulfuric Acid
H2SO4
Hydrochloric Acid
HCl
Phosphoric Acid
H3PO4
Acetic Acid
HC2H3O2
explosive, fertilizer, dye, glue,
batteries
metal cleaning, food prep, ore
refining, stomach acid
fertilizer, plastics & rubber,
food preservation
plastics & rubber, food
preservation, Vinegar
Hydrofluoric Acid
HF
metal cleaning, glass etching
Weak
Carbonic Acid
H2CO3
soda water
Weak
Boric Acid
H3BO3
eye wash
Weak
Strong
Strong
Moderate
Weak
3
Structure of Acids
 binary acids have acid hydrogens attached to a
nonmetal atom
 HCl, HF
4
Structure of Acids
 oxy acids have acid hydrogens attached to an oxygen
atom
 H2SO4, HNO3
5
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
6
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
HCl(aq) + NaOH(aq)NaCl(aq) + H2O(l)
7
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
Strong
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
8
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
Amino acids have a base at one end and an acid at
the other, neighboring amino acids can neutralize to
form a polypeptide
9
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
Anthocyanins give these pansies their dark purple pigmentation and are
the pigment in red cabbage that is so sensitive to acidity
10
acids and bases: 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
HCl(aq) → H+(aq) + Cl–(aq)
HC2H3O2(aq)
H+(aq) + C2H3O2–(aq)
11
Arrhenius Theory
HCl ionizes in water
producing H+ and Cl– ions
NaOH dissociates in water
producing Na+ and OH– ions
12
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+ ≅ H+(aq)
 there are also minor amounts of H+ with multiple water molecules,
H(H2O)n+
13
Arrhenius Acid-Base Reactions
 the H+ from the acid combines with the OH- from 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)
H+(aq)+Cl-(aq)+Na+(aq)+OH-(aq)Na+(aq)+Cl-(aq)+H2O(l)
H+(aq) + OH-(aq) H2O(l)
 All acid base reactions have this same net ionic equation in the
Arrhenius idea of the acid and base
14
Limitations of the 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
15
Acids and bases: Brønsted-Lowry






in a Brønsted-Lowry Acid-Base reaction, an H+ is transferred
does not have to take place in aqueous solution
broader definition than Arrhenius
An 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
17
Brønsted-Lowry Bases
 Brønsted-Lowry bases are H+ acceptors
 any material that has atoms with lone pairs can potentially be a BrønstedLowry 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
Tro, Chemistry: A Molecular
Approach
18
Amphoteric Substances
 amphoteric substances can act as either an acid or a base
 have both transferable H and atom with lone pair
Example
 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
NH3(aq) + H2O(l)
Base
Acid
NH4+(aq)
Conjugate Acid
+
OH–(aq)
Conjugate Base
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Brønsted-Lowry: Acid-Base Reactions
H–A
acid
+
HCHO2
acid
H 2O +
acid
+
:B
base
:A–
+
H–B+
conjugate
conjugate
base
acid
H2O
base
CHO2–
conjugate
base
NH3
base
HO–
+
conjugate
base
+
H 3 O+
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
23
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
HSO4–
+
H3O+
conjugate
conjugate
base
acid
24
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
H2CO3
acid
conjugate
acid
+
HO–
conjugate
base
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Practice – Write the formula for the conjugate
acid of the following
H2O
NH3
CO32−
H2PO4−
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Practice – Write the formula for the conjugate
acid of the following
H2O
NH3
H3O+
NH4+
CO32−
HCO3−
H2PO41−
H3PO4
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Practice – Write the formula for the conjugate
base of the following
H2O
NH3
CO32−
H2PO4−
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Practice – Write the formula for the conjugate
base of the following
H2O
NH3
HO−
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
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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
HCl(aq)  H+(aq) + Cl-(aq)
HCl(aq) + H2O(l)  H3O+(aq)+ Cl-(aq)
 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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Weak Acids

weak acids donate a small fraction of their
H’s
HF(aq)
H+(aq) + F-(aq)
HF(aq) + H2O(l)
H3O+(aq) + F-(aq)
 most of the weak acid molecules do
not donate H to water
 much less than 1% ionized in water

[H3O+] << [weak acid]
33
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
34
Polyprotic Acids
 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-
35
HClO4
Conjugate
Bases
ClO4-1
H2SO4
HI
HBr
HCl
HNO3
H3O+1
HSO4-1
H2SO3
H3PO4
HNO2
HF
HC2H3O2
H2CO3
H 2S
NH4+1
HCN
HCO3-1
HS-1
H 2O
CH3-C(O)-CH3
NH3
CH4
OH-1
HSO4-1
I-1
Br-1
Cl-1
NO3-1
H 2O
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
Increasing Acidity
Acids
36
Strengths : Acids and Bases
 commonly, Acid or Base strength is measured by determining the
equilibrium constant of a substance’s reaction with water
HA + H2O
B: + H2O
A-1 + H3O+1
HB+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
37
General Trends: 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
38
Acid Ionization Constant, Ka
 acid strength measured by the size of the equilibrium
constant when react with H2O
HA + H2O
A-1 + H3O+1
 the equilibrium constant is called the acid ionization
constant, Ka
 larger Ka = stronger acid
[A-1 ] ´ [H 3O+1 ]
Ka =
[HA]
39
40
Name
Benzoic
Formula
C6H5COOH
Ka1
6.14 x 10-5
Ka2
Ka3
Propanoic
Formic
CH3CH2COOH
HCOOH
1.34 x 10-5
1.77 x 10-5
Acetic
Chloroacetic
CH3COOH
ClCH2COOH
1.75 x 10-5
1.36 x 10-5
Trichloroacetic
Oxalic
Cl3C-COOH
HOOC-COOH
1.29 x 10-4
5.90 x 10-2
Nitric
Nitrous
HNO3
HNO2
strong
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
Arsenious
H3AsO4
H3AsO3
6.0 x 10-3
6.0 x 10-10
1.05 x 10-7
3.0 x 10-14
3.0 x 10-12
very small
Perchloric
Chloric
HClO4
HClO3
> 108
5 x 102
Chlorous
Hypochlorous
HClO2
HClO
1.1 x 10-2
3.0 x 10-8
Boric
Carbonic
H3BO3
H2CO3
5.83 x 10-10
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
H2O + H2O
H+ + OH–
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
42
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
43
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
44
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:
Concept Plan:
Relationships:
Solution:
[H3O+] = 1.5 x 10-9 M
[OH-] if [OH-] > [H3O+] basic if [H3O+]>[OH-] acid
[H3O+]
[OH-]
K w = [ H 3O ][OH ]
+
K w = [ H 3O + ][OH - ]
-14
1
.
0
´
10
K
-6
w
[OH
]
=
=
6
.
7
´
10
M
[OH ] =
+
-9
[ H 3O ]
1.5 ´10
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-]
46
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
47
pH
 the acidity/basicity of a solution is often expressed as pH
pH = -log[H3O+], [H3O+] = 10-pH
Since in water [H3O+] = 10-7M
pHwater = -log[10-7] = 7
 pH < 7 is acidic; pH > 7 is basic, pH = 7 is neutral
48
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)
49
Sig. Figs. and 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
50
Tro, Chemistry: A
51
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:
[OH-] = 1.3 x 10-2 M
pH
[OH-]
Relationships:
Solution:
[H3O+]
+
pH
=
log
[
H
O
]
K w = [ H 3O ][OH ]
3
+
K w = [ H 3O + ][OH - ]
-14
1.0 ´ 10
[ H 3O ] =
1.3 ´ 10-2
+
Check:
pH
[H3O+ ] = 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
Tro, Chemistry: A Molecular
Approach
53
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
54
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
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
55
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
-14
[H 3O ][OH ] = K w = 1.0 ´ 10
+
-14
- log [H O ][OH ] = - log 1.0 ´ 10
+
))
(
)
( (
( - log ([H O ])) + ( - log ([OH ])) = 14.00
3
+
3
pH + pOH = 14.00
56
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
57
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
 for a monoprotic strong acid [H3O+] = [HA]
 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00
58
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
Tro, Chemistry: A Molecular
Approach
59
Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Write the reaction for
the acid with water
Construct an ICE table
for the reaction
HNO2 + H2O
initial
initial
NO2- + H3O+
[HNO22]
[NO22--]
[H33O++]
0.200
0
≈0
change
change
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
equilibrium
equilibrium
since no products initially, Qc = 0, and the reaction is proceeding forward
60
Find the pH of 0.200 M HNO2(aq) solution @ 25°C
represent the change
in the concentrations in
terms of x
initial
change
sum the columns to
find the equilibrium
concentrations in terms
of x
substitute into the
equilibrium constant
expression
equilibrium
[HNO2]
[NO2-]
[H3O+]
0.200
0
0
+x
+x
x
x
-x
0.200 -x
x )( x )
[NO-2 ][H 3O+ ]
(
Ka =
=
[ HNO2 ] ( 0.200 - x )
61
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
initial
change
since Ka is very small,
approximate the
[HNO2]eq = [HNO2]init and
solve for x
equilibrium
[HNO2]
[NO2-]
[H3O+]
0.200
0
≈0
-x
+x
+x
0.200-x
0.200
x
x
éë NO--22 ùû éë H 33O++ ùû ( x()x()x()x )
K aa =
=
( 0.200 )- x )
[ HNO22 ]
4.6 ´10
-4
=
x2
2.00 ´10 -1
x=
( 4.6 ´10 )( 2.00 ´10 )
-4
x = 9.6 ´10 -3
-1
62
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
initial
[NO2-]
[H3O+]
0.200
0
≈0
-x
+x
+x
0.200
x
x
change
equilibrium
x = 9.6 x 10-3
[HNO2]
9.6 ´10
-3
2.00 ´10
-1
´100% = 4.8% < 5%
the approximation is valid
63
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
initial
change
equilibrium
[HNO2]
[NO2-]
[H3O+]
0.200
0
≈0
-x
+x
+x
x
x
0.200-x
0.190 0.0096
0.0096
x = 9.6 x 10-3
-3
HNO
=
0.200
x
=
0.200
9.6
´10
[ 2]
(
) = 0.190 M
éë NO-2 ùû = éë H 3O+ ùû = x = 9.6 ´10 -3 M
64
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
[HNO2]
[NO2-]
[H3O+]
0.200
0
≈0
-x
+x
+x
0.190
0.0096 0.0096
pH = -log ( H 3O
+
= - log ( 9.6 ´10
)
-3
) = 2.02
65
Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
check by substituting the
equilibrium concentrations
back into the equilibrium
constant expression and
comparing the calculated
Ka to the given Ka
initial
change
equilibrium
though not exact,
the answer is
reasonably close
[HNO2]
[NO2-]
[H3O+]
0.200
0
≈0
-x
+x
+x
0.190
0.0096 0.0096
éë NO-2 ùû éë H 3O + ùû
Ka =
=
[ HNO2 ]
9.6 ´ 10 )
(
=
-3 2
( 0.190 )
= 4.9 ´ 10 -4
66
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
Tro, Chemistry: A Molecular
Approach
67
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
Write the reaction for
the acid with water
Construct an ICE table
for the reaction
HC6H4NO2 + H2O
initial
C6H4NO2- + H3O+
[HA]
[A-]
[H3O+]
0.012
0
≈0
change
equilibrium
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
68
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
HC6H4NO2 + H2O
represent the change
in the concentrations in
terms of x
initial
change
sum the columns to
find the equilibrium
concentrations in terms
of x
substitute into the
equilibrium constant
expression
equilibrium
C6H4NO2- + H3O+
[HA]
[A-]
[H3O+]
0.012
0
0
+x
x
+x
-x
0.012 -x
x
x )( x )
[C6 H 4 NO-2 ][H 3O+ ]
(
Ka =
=
[ HC6H 4 NO2 ] (1.2 ´10 -2 - x )
69
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
C6H4NO2- + H3O+
HC6H4NO2 + H2O
determine the value of
Ka
since Ka is very small,
approximate the [HA]eq =
[HA]init and solve for x
initial
change
equilibrium
éë A- ùû éë H 3O+ ùû ( x()x()x()x )
Ka =
=
( 0.012 )- x )
[ HA]
x
1.4 ´10 =
0.012
-5
2
[HA]
[A2-]
[H3O+]
0.012
0
≈0
-x
+x
+x
0.012
0.012
-x
x
x
x=
(1.4 ´10 )( 0.012 )
-5
x = 4.1´10 -4
70
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given 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
initial
change
equilibrium
[HA]
[A2-]
[H3O+]
0.012
0
≈0
-x
+x
+x
0.012
x
x
x = 4.1 x 10-4
4.1´10 -4
1.2 ´10
´
100
%
=
3
.
4
%
<
5
%
-2
the approximation is valid
71
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
substitute x into the
equilibrium
concentration
definitions and solve
initial
change
equilibrium
[HA]
[A2-]
[H3O+]
0.012
0
≈0
-x
+x
+x
0.012-x
x
x
x = 4.1 x 10-4
-4
HC
H
NO
=
0.012
x
=
0.012
4.1´10
[ 6 4 2]
(
) = 0.012 M
+
éëC6 H 4 NO ùû = éë H 3O ùû = x = 4.1´10
2
-4
M
72
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
substitute [H3O+] into
the formula for pH and
solve
initial
change
equilibrium
[HA]
[A2-]
[H3O+]
0.012
0
≈0
-x
+x
+x
0.012 0.00041 0.00041
pH = -log ( H 3O
+
= - log ( 4.1´10
)
-4
) = 3.39
73
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
check by substituting the
equilibrium concentrations
back into the equilibrium
constant expression and
comparing the calculated
Ka to the given Ka
initial
[HA]
[A2-]
[H3O+]
0.012
0
≈0
-x
+x
+x
change
equilibrium
0.012 0.00041 0.00041
[C6 H 4 NO-2 ][H 3O + ]
Ka =
[ HC6 H 4 NO2 ]
4.1 ´ 10 )
(
=
(1.2 ´ 10 )
-4 2
-2
the values match
= 1.4 ´ 10
-5
74