Acids, Bases, and Salts CHM 1010 PGCC Barbara A. Gage
Dec 22, 2015
Acid Base
Litmus color
Phenolphthalein color
pH range
Reaction with active metal (like
Mg)
Taste
Formula component
Characteristics of Acids and Bases
CHM 1010 PGCC Barbara A. Gage
AcidsStrong
hydrochloric acid, HClhydrobromic acid, HBrhydroiodic acid, HInitric acid, HNO3
sulfuric acid, H2SO4
perchloric acid, HClO4
Weak
hydrofluoric acid, HFphosphoric acid, H3PO4acetic acid, CH3COOH (or HC2H3O2)
ionizes completely in water ionizes partially in water
carbonic acid, H2CO3
CHM 1010 PGCC Barbara A. Gage
Strong acid: HA(g or l) + H2O(l) H3O+(aq) + A-
(aq)
The extent of dissociation for strong acids.
H+ and H2O H3O+ (hydronium ion)
CHM 1010 PGCC Barbara A. Gage
The extent of dissociation for weak acids.
Weak acid: HA(aq) + H2O(l) H3O+(aq) + A-
(aq)
CHM 1010 PGCC Barbara A. Gage
Bases (or alkalis)Strong
Weak
sodium hydroxide, NaOH
calcium hydroxide, Ca(OH)2
potassium hydroxide, KOH
strontium hydroxide, Sr(OH)2
barium hydroxide, Ba(OH)2
ammonia, NH3 (NH4OH)
Moderate
Dissociates completely Dissociates completely but is not very soluble
aluminum hydroxide, Al(OH)3
magnesium hydroxide, Mg(OH)2
Dissociates partially
carbonates, CO32-
bicarbonates, HCO31-
CHM 1010 PGCC Barbara A. Gage
An aqueous strong acid-strong base reaction on the atomic scale.
MX is a “salt” – an electrolyte that is not an acid or baseCHM 1010 PGCC
Barbara A. Gage
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 Brønsted-Lowry acid-base reaction.
(acid, H+ donor) (base, H+ acceptor)
HCl H2O
+
Cl- H3O+
+
Lone pair binds H+
(base, H+ acceptor) (acid, H+ donor)
NH3 H2O
+
NH4+ OH-
+
Lone pair binds H+
CHM 1010 PGCC Barbara A. Gage
The Conjugate Pairs in Some Acid-Base Reactions
Base Acid+Acid Base+
Conjugate Pair
Conjugate Pair
Reaction 4
H2PO4- OH-+
Reaction 5
H2SO4 N2H5++
Reaction 6
HPO42- SO3
2-+
Reaction 1
HF H2O+ F- H3O++
Reaction 3
NH4+ CO3
2-+
Reaction 2
HCOOH CN-+ HCOO- HCN+
NH3 HCO3-+
HPO42- H2O+
HSO4- N2H6
2++
PO43- HSO3
-+
CHM 1010 PGCC Barbara A. Gage
SAMPLE PROBLEM Identifying Conjugate Acid-Base Pairs
PROBLEM: The following reactions are important environmental processes. Identify the conjugate acid-base pairs.
(a) H2PO4-(aq) + CO3
2-(aq) HPO42-(aq) + HCO3
-(aq)
(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3
-(aq)
SOLUTION:
Identify proton donors (acids) and proton acceptors (bases).
(a) H2PO4-(aq) + CO3
2-(aq) HPO42-(aq) + HCO3
-(aq)
proton
donor
proton accepto
r
proton accepto
r
proton
donor
conjugate pair1conjugate pair2
(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3
-(aq)
conjugate pair2conjugate pair1
proton
donor
proton accepto
r
proton accepto
r
proton
donorCHM 1010 PGCC Barbara A. Gage
Molecules as Lewis Acids
F
B
F F
H
N
H H
+
F
B
F F
H
N
H H
acid base adduct
An acid is an electron-pair acceptor.A base is an electron-pair donor.
M2+
H2O(l)
M(H2O)42+(aq)
adductCHM 1010 PGCC
Barbara A. Gage
SAMPLE PROBLEM 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 K(H2O)6+
SOLUTION:
PLAN: Look for electron pair acceptors (acids) and donors (bases).
(a) H+ + OH- H2Oacceptor
donor
(b) Cl- + BCl3 BCl4-
donor
acceptor
(c) K+ + 6H2O K(H2O)6+
acceptor
donorCHM 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.
Start of titrationExcess of acid
Point of neutralization
Slight excess of base
CHM 1010 PGCC Barbara A. Gage
Sample Problem Finding the Concentration of Acid from an Acid-Base Titration
PROBLEM: 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?
SOLUTION:
NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)
(33.87-0.55) mL x
1L
103
mL
= 0.03332 L
0.03332 LX 0.1524 M = 5.078x10-3 molNaOH
Molar ratio is 1:15.078x10-3 mol HCl
0.05000 L= 0.1016 M HCl
CHM 1010 PGCC Barbara A. Gage
Kc = [H3O+][OH-]
[H2O]2
Kc[H2O]2 = [H3O+][OH-]
The Ion-Product Constant for Water
Kw =
A change in [H3O+] causes an inverse change in [OH-].
= 1.0 x 10-14 at 250C
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
In an acidic solution, [H3O+] > [OH-]
In a basic solution, [H3O+] < [OH-]
In a neutral solution, [H3O+] = [OH-]
CHM 1010 PGCC Barbara A. Gage
The relationship between [H3O+] and [OH-] and the
relative acidity of solutions.
[H3O+] [OH-]Divide into Kw
ACIDIC SOLUTION
BASIC SOLUTION
[H3O+] > [OH-]
[H3O+] = [OH-]
[H3O+] < [OH-]
NEUTRAL SOLUTION
CHM 1010 PGCC Barbara A. Gage
SAMPLE PROBLEM Calculating [H3O+] and [OH-] in an Aqueous Solution
PROBLEM: 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?
SOLUTION:
Use the Kw at 250C and the [H3O+] to find the corresponding [OH-].
Kw = 1.0x10-14 = [H3O+] [OH-] so
[OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 =
[H3O+] is > [OH-] and the solution is acidic.
3.3x10-
11M
CHM 1010 PGCC Barbara A. Gage
The pH values of some familiar
aqueous solutions.
pH = -log [H3O+]
CHM 1010 PGCC Barbara A. Gage
pOH = -log [OH-]pH + pOH = 14
SAMPLE PROBLEM Calculating [H3O+], pH, [OH-], and pOH
PROBLEM: 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.
SOLUTION:
PLAN: HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-] and then convert to pH and pOH.
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
[OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14M; pOH = 13.48
For 0.3M HNO3, [H3O+] = 0.30M and -log [H3O+] = 0.52 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 = 1.6x10-12M; pOH = 11.80
For 0.0063M HNO3, [H3O+] = 0.0063M and -log [H3O+] = 2.20 = pH
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