Acids, Bases, and Salts CHM 1010 PGCC Barbara A. Gage Characteristics of Acids and Bases Acid Base Litmus color Phenolphthalein color pH range Reaction with active metal (like Mg) Taste Formula component CHM 1010 PGCC Barbara.

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Transcript Acids, Bases, and Salts CHM 1010 PGCC Barbara A. Gage Characteristics of Acids and Bases Acid Base Litmus color Phenolphthalein color pH range Reaction with active metal (like Mg) Taste Formula component CHM 1010 PGCC Barbara.

Acids, Bases, and Salts
CHM 1010
PGCC
Barbara A. Gage
Characteristics of Acids and Bases
Acid
Base
Litmus color
Phenolphthalein color
pH range
Reaction with active
metal (like Mg)
Taste
Formula component
CHM 1010
PGCC
Barbara A. Gage
Acids
Strong
ionizes completely in water
hydrochloric acid, HCl
hydrobromic acid, HBr
hydroiodic acid, HI
nitric acid, HNO3
sulfuric acid, H2SO4
perchloric acid, HClO4
CHM 1010
Weak
ionizes partially in water
hydrofluoric acid, HF
phosphoric acid, H3PO4
acetic acid, CH3COOH
(or HC2H3O2)
carbonic acid, H2CO3
PGCC
Barbara A. Gage
Figure 18.2
The extent of dissociation for strong acids.
Strong acid: HA(g or l) + H2O(l)
H3O+(aq) + A-(aq)
H+ and H2O  H3O+ (hydronium ion)
CHM 1010
PGCC
Barbara A. Gage
Figure 18.3
The extent of dissociation for weak acids.
Weak acid: HA(aq) + H2O(l)
CHM 1010
PGCC
H3O+(aq) + A-(aq)
Barbara A. Gage
Bases (or alkalis)
Strong
Moderate
Dissociates completely
Dissociates completely but is
not very soluble
sodium hydroxide, NaOH
magnesium hydroxide, Mg(OH)2
potassium hydroxide, KOH
aluminum hydroxide, Al(OH)3
calcium hydroxide, Ca(OH)2
strontium hydroxide, Sr(OH)2
barium hydroxide, Ba(OH)2
Weak
Dissociates partially
ammonia, NH3 (NH4OH)
carbonates, CO32bicarbonates, HCO31CHM 1010
PGCC
Barbara A. Gage
Figure 4.8
An aqueous strong acid-strong base reaction on the atomic scale.
MX is a “salt” – an
CHM 1010
PGCC
electrolyte that
is A.
not
an acid
Barbara
Gage
or base
Acid and Base Definitions
• Arrhenius
• Acid = compound that forms hydrogen
(H+) ions in water
• Base = compound that forms hydroxide
(OH-) ions in water
CHM 1010
PGCC
Barbara A. Gage
Acid and Base Definitions
• Bronsted-Lowry
• Acid = proton donor (H+ is a proton)
• Base = proton acceptor
An acid-base reaction can now be viewed from the standpoint
of the reactants AND the products.
An acid reactant will produce a base product and the two will
constitute an acid-base conjugate pair.
CHM 1010
PGCC
Barbara A. Gage
Proton transfer as the essential feature of a BrønstedLowry acid-base reaction.
Figure 18.8
Lone pair
binds H+
+
+
HCl
H2O
(acid, H+ donor)
Cl-
H3O+
(base, H+ acceptor)
Lone pair
binds H+
+
+
NH3
(base, H+ acceptor)
H2O
NH4+
(acid, H+ donor)
CHM 1010
PGCC
Barbara A. Gage
OH-
Table 18.2 The Conjugate Pairs in Some Acid-Base Reactions
Conjugate Pair
Acid
+
Base
Base
+
Acid
Conjugate Pair
Reaction 1
HF
+
H2O
F-
+
H3O+
Reaction 2
HCOOH +
CN-
HCOO-
+
HCN
Reaction 3
NH4+
+
CO32-
NH3
+
HCO3-
Reaction 4
H2PO4-
+
OH-
HPO42-
+
H2O
Reaction 5
H2SO4
+
N2H5+
HSO4-
+
N2H62+
Reaction 6
HPO42-
+
SO32-
PO43-
+
HSO3-
CHM 1010
PGCC
Barbara A. Gage
SAMPLE PROBLEM 18.4
PROBLEM:
Identifying Conjugate Acid-Base Pairs
The following reactions are important environmental
processes. Identify the conjugate acid-base pairs.
(a) H2PO4-(aq) + CO32-(aq)
(b) H2O(l) + SO32-(aq)
PLAN:
HPO42-(aq) + HCO3-(aq)
OH-(aq) + HSO3-(aq)
Identify proton donors (acids) and proton acceptors (bases).
conjugate pair2
conjugate pair
1
SOLUTION: (a) H2PO4-(aq) + CO32-(aq)
proton
donor
proton
acceptor
HPO42-(aq) + HCO3-(aq)
proton
acceptor
conjugate pair2
conjugate pair1
(b) H2O(l) + SO32-(aq)
proton proton
donor acceptor
CHM 1010
proton
donor
PGCC
OH-(aq) + HSO3-(aq)
proton
acceptor
proton
donor
Barbara A. Gage
Molecules as Lewis Acids
An acid is an electron-pair acceptor.
A base is an electron-pair donor.
F
B
F
F
acid
F
H
+
H
N
B
HH
F
base
F
N
HH
adduct
M(H2O)42+(aq)
M2+
H2O(l)
CHM 1010
PGCC
Barbara A. Gage
adduct
SAMPLE PROBLEM 18.12
Identifying Lewis Acids and Bases
PROBLEM: Identify the Lewis acids and Lewis bases in the following reactions:
(a) H+ + OH-
H2O
(b) Cl- + BCl3
BCl4-
(c) K+ + 6H2O
PLAN:
K(H2O)6+
Look for electron pair acceptors (acids) and donors (bases).
SOLUTION:
acceptor
(a) H+ + OHdonor
donor
(b) Cl- + BCl3
acceptor
acceptor
(c) K+ + 6H2O
donor
H2O
BCl4-
K(H2O)6+
CHM 1010
PGCC
Barbara A. Gage
Acid Anhydrides
• Non-metal oxides react with water to
form acidic solutions
CO2 (g) + H2O (l)  H2CO3 (aq)
N2O5 (s) + H2O (l)  2 HNO3 (aq)
SO3 (g) + H2O (l)  H2SO4 (aq)
Dissolved non-metal oxides cause acid
rain.
CHM 1010
PGCC
Barbara A. Gage
Basic Anhydrides
• Metal oxides react with water to form
alkaline solutions
Na2O (s) + H2O (l)  2 NaOH (aq)
CaO (s) + H2O (l)  Ca(OH)2 (aq)
Al2O3 (s) + 3 H2O (l)  2 Al(OH)3 (aq)
Lime (CaO) is used on lawns and is converted
to Ca(OH)2 when it rains. CaO is less
hazardous to handle.
CHM 1010
PGCC
Barbara A. Gage
An acid-base titration.
Figure 4.7
Start of titration
Excess of acid
Point of
neutralization
CHM 1010
PGCC
Slight excess of
base
Barbara A. Gage
Sample Problem 4.5
PROBLEM:
PLAN:
Finding the Concentration of Acid from an
Acid-Base Titration
You perform an acid-base titration to standardize an HCl
solution by placing 50.00 mL of HCl in a flask with a few drops
of indicator solution. You put 0.1524 M NaOH into the buret,
and the initial reading is 0.55 mL. At the end point, the buret
reading is 33.87 mL. What is the concentration of the HCl
solution?
volume(L) of base
multiply by M of base
mol of base
(33.87-0.55) mL x
NaCl(aq) + H2O(l)
1L
103 mL
molar ratio
divide by L of acid
= 0.03332 L
0.03332 L X 0.1524 M = 5.078x10-3 mol
NaOH
Molar ratio is 1:1
mol of acid
M of acid
SOLUTION:
NaOH(aq) + HCl(aq)
5.078x10-3 mol HCl
0.05000 L
CHM 1010
PGCC
= 0.1016 M HCl
Barbara A. Gage
H2O(l) + H2O(l)
Kc =
H3O+(aq) + OH-(aq)
[H3O+][OH-]
[H2O]2
The Ion-Product Constant for Water
Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0 x 10-14 at 250C
A change in [H3O+] causes an inverse change in [OH-].
In an acidic solution, [H3O+] > [OH-]
In a basic solution, [H3O+] < [OH-]
In a neutral solution, [H3O+] = [OH-]
CHM 1010
PGCC
Barbara A. Gage
Figure 18.4
The relationship between [H3O+] and [OH-] and the
relative acidity of solutions.
[H3O+]
Divide into Kw
[OH-]
[H3O+] > [OH-]
[H3O+] = [OH-]
[H3O+] < [OH-]
ACIDIC
SOLUTION
CHM 1010
NEUTRAL
SOLUTION
PGCC
BASIC
SOLUTION
Barbara A. Gage
SAMPLE PROBLEM 18.2
Calculating [H3O+] and [OH-] in an Aqueous
Solution
PROBLEM:
PLAN:
A research chemist adds a measured amount of HCl gas to
pure water at 250C and obtains a solution with [H3O+] =
3.0x10-4M. Calculate [OH-]. Is the solution neutral, acidic, or
basic?
Use the Kw at 250C and the [H3O+] to find the corresponding [OH-].
SOLUTION: K = 1.0x10-14 = [H O+] [OH-] so
w
3
[OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 =3.3x10-11M
[H3O+] is > [OH-] and the solution is acidic.
CHM 1010
PGCC
Barbara A. Gage
Figure 18.5
The pH values of
some familiar aqueous
solutions.
pH = -log [H3O+]
pOH = -log [OH-]
pH + pOH = 14
CHM 1010
PGCC
Barbara A. Gage
Figure 18.6
The relations among [H3O+], pH, [OH-], and pOH.
CHM 1010
PGCC
Barbara A. Gage
SAMPLE PROBLEM 18.3
PROBLEM:
PLAN:
Calculating [H3O+], pH, [OH-], and pOH
In an art restoration project, a conservator prepares
copper-plate etching solutions by diluting concentrated
HNO3 to 2.0M, 0.30M, and 0.0063M HNO3. Calculate
[H3O+], pH, [OH-], and pOH of the three solutions at 250C.
HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the
[OH-] and then convert to pH and pOH.
SOLUTION: For 2.0M HNO3, [H3O+] = 2.0M and -log [H3O+] = -0.30 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/2.0 = 5.0x10-15M; pOH = 14.30
For 0.3M HNO3, [H3O+] = 0.30M and -log [H3O+] = 0.52 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14M; pOH = 13.48
For 0.0063M HNO3, [H3O+] = 0.0063M and -log [H3O+] = 2.20 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 = 1.6x10-12M; pOH = 11.80
CHM 1010
PGCC
Barbara A. Gage
Buffers
• Solutions that resist change in pH
• Can maintain any pH value between 0
and 14 (not just neutral pH 7)
• Composed of a weak acid and a salt
made from the weak acid or weak base
and salt made from the weak base
• Examples: HC2H3O2 and NaC2H3O2
NH4OH and NH4Cl
CHM 1010
PGCC
Barbara A. Gage
Buffers
Reaction with acid:
HC2H3O2 + C2H3O2- + H+ 
HC2H3O2 + HC2H3O2
Reaction with base:
HC2H3O2 + C2H3O2- + OH- 
C2H3O2- + C2H3O2- + HOH
A buffer regenerates it’s own components. The pH it
maintains depends on the ratio of salt to acid (or
base) and the nature of the acid (or base).
CHM 1010
PGCC
Barbara A. Gage