Transcript Chapter

Chemistry II
Chapter 15
Acids and Bases
Properties of Acids
• sour taste
• react with “active” metals
 i.e., Al, Zn, Fe, but not Cu, Ag, or Au
2 Al + 6 HCl → 2 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
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
Strong
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
Strong
Moderate
Weak
Structure of Acids
• binary acids have H-atoms attached to a nonmetal
atom
e.g. HCl, HF
Structure of Acids
• oxy acids have H-atoms attached to an O-atom:
Structure of Acids
• carboxylic acids have
COOH group:
e.g. Acetic acid
CH3COOH
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
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
Structure of Bases
• most ionic bases contain OH- ions
 NaOH, Ca(OH)2
• some contain CO32- ions
 CaCO3 , NaHCO3
CO32− + 2H2O → HCO3− + H2O + OH−
HCO3− + H2O + OH− → H2CO3 + 2OH−
Definitions of Acids and Bases
• What are the main characteristics of molecules and ions that
exhibit acid and base behavior?
• 3 different definitions:
 Arrhenius
 Brønsted-Lowry
 Lewis
Arrhenius Theory (1880s)
• 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)
CH3COOH(aq) → H+(aq) + CH3COO–(aq)
• bases dissociate in water to produce OH- ions and cations
 ionic substances dissociate in water
NaOH(aq) → Na+(aq) + OH–(aq)
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 + + H 2 O  H 3 O+
 Chemists use H+(aq) and H3O+(aq) interchangeably
Arrhenius Acid-Base Reactions
• the H+ from the acid combines with the OH- from the base
•
to make a molecule of H2O
H+ + OH- → H2O
the cation from the base combines with the anion from the
acid to make a salt
Na+ + Cl- → NaCl
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
acid + base → salt + water
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
Brønsted-Lowry Theory (1923)
•
in a Brønsted-Lowry Acid-Base reaction,
an H+ is transferred
Acid: proton (H+) donor
Base: proton (H+) acceptor
•
base structure must contain an atom with unshared
pair of e-
•
in an acid-base reaction, the acid molecule gives an H+
to the base molecule
H–A + :B ⇌ :A– + H–B+
Brønsted-Lowry Acids
• Brønsted-Lowry acids are H+ donors
 Anything that has H+ can potentially be Brønsted-Lowry
acid
• HCl(aq) is acidic because HCl transfers an H+ to H2O
(base or proton acceptor), forming H3O+ ions
HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
acid
base
Brønsted-Lowry Bases
• Brønsted-Lowry bases are H+ acceptors
 any material that has atoms with lone pairs can
potentially be a Brønsted-Lowry base
• NH3(aq) is basic because NH3 accepts an H+ from H2O
(acid, proton donor), forming OH–(aq)
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
base
acid
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)
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 e- after the reaction
– so it will act as a base in the reverse process
:A– + H–B+ ⇌ H–A + :B
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
Brønsted-Lowry
Acid-Base Reactions
H–A
acid
+
HCHO2
acid
H 2O +
acid
+
:B
base
⇌
:A– +
conjugate
base
H–B+
conjugate
acid
H2O
base
⇌
CHO2–
conjugate
base
+
H 3 O+
conjugate
acid
NH3
base
⇌
HO– +
conjugate
base
NH4+
conjugate
acid
Conjugate (joined) Pairs
In the reaction H2O + NH3 ⇌ OH– + NH4+
H2O and OH– constitute an
Acid/Conjugate Base pair
NH3 and NH4+ constitute a
Base/Conjugate Acid pair
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
HSO4–
conjugate
base
+
H3O+
conjugate
acid
Practice – Write the formula for the conjugate
acid of the following bases
H2O
H3O+
NH3
NH4+
CO32−
HCO3−
H2PO4−
H3PO4
Practice – Write the formula for the conjugate
base of the following acids
H2O
OH−
NH3
NH2−
CO32−
H2PO4−
No H+, it cannot be an acid
HPO42−
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, ⇌
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]
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]
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
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
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
HA + H2O ⇌ A- + H3O+
B: + H2O ⇌ HB+ + OH-
• 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
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+ > H2O > OH-;
 base trend opposite
NH4+ > NH3 > NH2-
Acid Ionization Constant, Ka
• acid strength measured by the size of the equilibrium
constant when react with H2O
HA + H2O ⇌ A- + H3O+
• acid ionization constant, Ka
 larger Ka = stronger acid
[A  ]  [H 3O  ]
Ka 
[HA]
40
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
Ion Product of Water
• the product of H3O+ and OH– concentrations is always the same n
• the number is called the ion product of water - symbol, Kw
Kw = [H3O+] [OH–] = 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
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
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-
[OH-]
10-3
10-5
+
H
OH-
10-7
+
H
10-9
10-11
H+
OH OH
10-13 10-14
H+
OH
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
+
H
Base
10-9
10-11
H+
H+
OH OH
10-7
10-5
10-13 10
OH
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
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
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
Question
Complete the following table of pH values
[H+]
pH
1 x 10-7
7.0
1
1.0 x 10-7
7.00
2
1.6 x 10-7
6.80
2
4.1 x 10-5
4.39
2
10.750
3
1.78 x 10-11
Significant Figures
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 can be negative (very acidic) or larger than 14 (very alkaline)
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
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 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
pH and pOH
Complete the Table
pH
[H+] 100 10-1
+
H
OH-
10-3
+
H
OH-
[OH-]10-14 10-13 10-11
pOH
10-5
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
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
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

14
 log [H O ][OH ]   log 1.0  10 
3
 log [H O ]   log [OH- ]  14.00

3
pH  pOH  14.00
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
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+] = [HA]
 for polyprotic acids, the other ionizations can generally be ignored
• 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00
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
HA + H2O ⇌ A + H3O+
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
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
[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
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
2.00101
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
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

3
3
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
[HNO2] [NO2-] [H3O+]
0.200
0
≈0
-x
+x
+x
initial
change
equilibrium

pH  -log H 3O


0.190

  2.02
3
  log 9.6 10
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
3
HNO2 

9.6  10 

3 2
0.190
 4.9  104
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
[H 3O  ]equil
[HA] init
100%
88
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
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

NO H O  x x 


2

3
HNO2 
2.5
4.6 104
x
x
x2

2.5
4.6 10 2.5
4
x  3.4 102
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
0.034
x
HNO2   2.5  x  2.5  0.034  2.5 M
  H O  x  0.034M
NO2

3
0.034
x
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
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
0.034
 100%
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
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
HF + H2O ⇌ F + H3O+
Ka = 3.5 x 10-4
HClO + H2O ⇌ ClO + H3O+ Ka = 2.9 x 10-8
If the Kas are
H2O + H2O ⇌ OH + H3O+ Kw = 1.0 x 10-14
sufficiently different,
use the strongest acid to
-]
+]
[HF]
[F
[H
O
3
construct an ICE table
for the reaction
0.150
initial
0
≈0
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
change
equilibrium
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
[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 
-

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
1.50101
x
[F-]
0
+x
x
[H3O+]
≈0
+x
x
3.5 10 1.5010 
4
x  7.2 103
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
99
[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
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
-x
[F-]
0
+x
[H3O+]
≈0
+x
0.143
0.0072
0.0072
[HF]
initial
change
equilibrium

0.150
pH  -log H 3O 
  log7.2 10

  2.14
3
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
 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
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:
[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
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
Weak Bases
• in weak bases, only a small fraction of
molecules accept H’s
 weak electrolyte
+
 most of the weak base molecules do notNH3 + H2O ⇌ NH4 + OH
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
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
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
[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
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
1.00101


x  1.76105 1.00101
x  1.33103

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
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
-
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
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
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
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
Practice – Find the pH of a 0.0015 M morphine
solution, Kb = 1.6 x 10-6
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
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
[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
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
1.5 101


x  1.6 106 1.5 103
x  4.9 105

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
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
-
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
[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
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
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
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
Acid-Base Properties of Salts
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
Ex 15.13 - Use the Table to Determine if the
Given Anion Is Basic or Neutral
a)
b)
NO3−
the conjugate base of a strong
acid, therefore neutral
NO2−
the conjugate base of a weak
acid, therefore basic
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
[A ][H O ]
when you


HA(aq)  H2O(l )  A (aq)  H3O (aq)
add
equations,
you multiply
the K’s

Ka 

A (aq)  H2O(l )  HA(aq)  OH (aq)
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


3
[HA]
[ HA][H3O ]
Kb 
[A ]
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
Enter the initial
concentrations –
assuming the [OH]
from water is ≈ 0
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
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


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
x  2.4  106
1
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
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
-
6
M
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
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
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
Kb

HCHO2 OH  

CHO 


2.4  10 

6 2
0.100
2
 5.8  1011
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
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)
Ex 15.15 - Determine if the Given Cation Is
Acidic or Neutral
a) C5N5NH2+
b)
c)
the conjugate acid of a weak base, therefore acidic
Ca2+
the counterion of a strong base, therefore neutral
Cr3+
a highly charged metal ion, therefore acidic
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 KNO2
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
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
• Ion with higher K value dominates




NH4F = NH4+, FNH4+ is conjugate acid of weak base (NH3) = acidic
F- is conjugate base of weak acid = basic
Ka of NH4+ (5.68 x 10-10) is larger than Kb of the F− (2.9 x 10-11);
therefore the solution will be acidic
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
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
Practice - Lewis Acid-Base Reactions
Label the Nucleophile and Electrophile
• BF3 + HF ⇌ H+BF4Elec
Nuc
F
H+1 F
+B
• CaO + SO3 ⇌ Ca+2SO4-2
F
F
O
O -2
Nuc
Elec
• KI + I2 ⇌ KI3
Elec
+S
O
Ca+2 O
O
K+1
••
I
••
-1
••
••
Nuc
-2
••
••
Ca+2
••
O
••
F
-1
••
••
H F
••
F
+ I
B
S
F
O
O
I
K+1 I
I
I -1