Identifying Acids and Bases
•
Acids
contains H+ ions as the cation,
with and other element as the
anion
•
Acid (anhydrides)
Non-metal oxide
Contains OH- as the anion
•
Bases
•
Base (anhydrides)
•
Salts
H2SO4
HI
P2O5
Ca(OH)2
Combined with a cation
A metal with oxygen
Ionic compound
Formed from a cation and an
anion
Al2O3
Mg F2
KCl
LiBr
Metal and non-metal
Acid/base Definitions
•Arrhenius:
Acids: Any substance which releases
H+ ion in water solution.
Bases: Any substance which releases
OH- ion in water solution.
Acid/Base Definitions
•Bronsted - Lowry:
Acids: Any substance which
donates a proton.
Bases: Any substance which accepts
a proton.
Acid / Base Chemistry
Acid/Base Definitions
•Lewis:
Acids: Any substance which can
accept an electron pair.
Bases: Any substance which can
donate an electron pair.
•Hydrogen ion concentration = [H+]
•Hydroxide ion concentration = [OH-]
•Pure H20 – neutral substance –
– ph=7 and [H+]=[OH-]
•H2O will self-dissociate
– H20 > H+ + OH– Note: the number of H2O molecules that will self dissociate is very
small; only 2 molecules per a billion, therefore [H2O] is unchanged.
1
Acid / Base Chemistry
- Kw - is the ion product constant for
water
- Kw = 1.0 x 10-14 M2
:Kw= [1.0 x
•Calculate [OH-] if [H+] is 1.0 x 10-2 M
• Kw= [H+] [OH-]
• [OH-]= Kw
Kw= [H+] [OH-]
FOR H2O
Acid / Base Chemistry
10-7
M][1.0 x
10-7M]
- there is an indirect relationship
[H+]
•
= 1.0 x 10-14M2
1.0 x 10-2M
• [OH-]= 1.0 X 10 –12 M
between [H+] and [OH-]
Acid / Base Chemistry
•Calculate [H+] if [OH-] is 6.5 x 10-11 M
Acidic
solutions
pH= 1- 6
-1
-6
[H+]= 1.0 x 10 M – 1.0 x 10 M
[H+] > [OH-]
• Kw= [H+] [OH-]
• [H+]= Kw
[OH-]
= 1.0 x 10-14M2
6.5 x
10-11M
• [H+]=
•
[H+]
will tell us if a soln is acidic or
basic.
Neutral
solutions
Basic
solutions
pH= 7
[H+] = [OH-]
-7
1.0 x 10 M for both
pH= 8 -14
-8
-14
[H+]= 1.0 x 10 M – 1.0 x 10 M
[H+] < [OH-]
Acid / Base Chemistry
• If [H+]= 1.0 x 10-9M, will the solution be
basic or acidic?
• Basic
• Why?
• because the pH= 9 and this solution
contains less [H+] than [OH-]
2
Acid / Base Chemistry
[OH-]=
10-4M,
• If
2.63 x
will the solution
be basic or acidic or neutral?
•
• Why?
Acid / Base Formulas
•
•
•
•
•
pH= -log [H+]
pOH= -log [OH-]
pH + pOH =14
[H+] = antilog (-pH)
[OH-] = antilog (-pOH)
• How to use the calculator!!
• Calculate the pH of a solution with [H+] =
0.001 M. Is this acidic, basic, or neutral?
Acid / Base Formulas
• Calculate the pH of a solution with [OH-] = 7.9
x 10-3 M. Is this acidic, basic, or neutral?
Acid / Base Formulas
• A solution with a pH of 9.1 has what [H+]?
– 2nd log –9.1 ENTER
• What is the [OH-]?
• Calculate the pH of a solution with [H+] =
4.2 x 10-6 M. Acidic, basic, or neutral?
• What is the pOH?
What is the [OH-]?
Acids, Bases, and Their
Conjugates
•Water is amphoteric – has basic +
acidic properties.
•Conjugate Acid: Substances formed in a
reaction when a base gains a H+
•Conjugate Base: Substances formed in a
reaction when an acid donates H+
Acids, Bases, and Their Conjugates
• NH3 + H2O
--
NH4+ + OH-
• HCl + H2O
--->
3
Identifying monoprotic, diprotic, and triprotic:
Acids, Bases, and Their Conjugates
monoprotic:
H+ + Cl-
HCl
• CO3-2 + H2O
---> HCO3- + OH-
• Ca(OH)2 + 2HNO3 --> Ca(NO3)2 + 2HOH
diprotic:
H2SO4
H+ + HSO4-1
HSO4-1
H+ + SO4-2
triprotic:
H3PO4
H2PO4-1
HPO4-2
Strengths of Acids and Bases
• Strong acids and bases will completely dissociate
ex. HCl --> H+ + Cl• The six strong acids - HCl, HBr,
• HI, H2SO4, HNO3, HClO4
• All others are weak.
• Strong Bases: Groups I
• and II with OH- except Be
Strengths of Acids and Bases
•Strong Acids: HCl, HI, HBr,
H2SO4, HNO3, HClO4
•Strong Bases: Groups I and II with
OH- except Be
H+ + H2PO4-1
H+ + HPO4-2
H+ + PO4-3
Strengths of Acids and Bases
•Weak acids and bases slightly
dissociate
ex. HClO <--> ClO- + H+
Electrolytes and Non-Electrolytes
•Strong
Strong Electrolytes:
- Compounds that conduct electricity in
aqueous solution
- Substances that completely dissociate
- Strong acids and bases
- Soluble salts
4
Electrolytes and NonElectrolytes
•Weak
Weak Electrolytes:
- Slight electron conductivity
- Slight dissociation in aqueous
- Weak acids and bases
- Insoluble salts if in molten state
Electrolytes and NonElectrolytes
•NonElectrolytes:
NonElectrolytes:
- Don’t conduct electricity
- No ions to carry current
- Organic, molecular
Examples of Electrolytes and Nonelectrolytes:
Examples of Electrolytes and Nonelectrolytes:
• H2S03
• Al(OH)3
• weak acid/weak elect.
• weak base/weak electrolyte
• CsOH
• HF
• strong base/strong elect.
• weak acid/weak electrolyte
• HCl04
• CCl4
• strong acid/strong elect.
• nonelectrolyte- molecular compound
• C6H1206
• NaN03
• nonelectrolyte- covalent, molecular, organic b/c
carbon
• soluble salt/strong electrolyte
• S03
• acid anhydride/weak electrolyte
Neutralization Reactions:
Neutralization
•Acid + Base > Salt + H2O
•Neutralization is found using titration
•Titration: Analytical method used to
determine the concentration of acids,
bases in a neutralization reaction.
General Equation:
• Acid + Base
Salt + Water
• HCl + NaOH
NaCl + HOH
Acid
Base
Salt
Water
5
Neutralization Reactions:
Titration: Analytical method used to determine the
concentration of acids and bases in a neutralization
reaction.
1. Measured volume of an acid or
base with unknown
concentration. Put in Erlenmeyer.
2. Add a few drops of indicator to
the Ernlenmeyer.
3. A measured volume of titrant is
delivered into the flask through a
buret (base or acid)
4. Concentration of the titrant is
known –Standard Solution.
5. Neutralization occurs when
indicator is revealed End
Point
6. Equivalence Point
MOL ACID = MOL BASE
Neutralization practice problems:
Titration Curve
1. What volume of calcium hydroxide in a 3.0
M solution is needed to neutralize 45.0 mL
of 2.6 M perchloric acid?
Neutralization practice problems:
Neutralization practice problems:
2. Determine the [Ba(OH)2] solution if 300.0 mL
was titrated to neutrality with 220.5 mL of
6.0 M solution of phosphoric acid?
1. What volume of calcium hydroxide in a 3.0 M solution is needed to neutralize
45.0 mL of 2.6 M perchloric acid?
M
CR
M
vol
mol
mol
vol
G: 45 mL
Ca(OH)2 + HCl04
2 HOH + Ca(Cl04)2
2.6 M HCl04
0.45 L HCl04 x 2.6 mol HCl04 x 1 mol Ca(OH)2 x 1 L Ca(OH)2
3.0 M Ca(OH)2
1 L Ca(OH)2
2 mol HCl04
3 mol Ca(OH)2
U: ? vol Ca(OH)2
2. Determine the [Ba(OH)2] solution if 300.0 mL was titrated to neutrality with
220.5 mL of 6.0 M solution of phosphoric acid vol M mol CR mol
G: 220.5 mL
6.0 M H3P04
3Ba(OH)2 + 2H3P04
Ba3(P04)2 + 6HOH
0.2205 L H3P04 x 6.0 mol H3P04 x 3 mol Ba(OH)2
300.0 mL Ba(OH)2
1 L H3P04
U: ? M Ba(OH)2
2 mol H3P04
= 1.985 mol Ba(OH)2
1.985 mol Ba(OH)2
0.300L Ba(OH)2
= 6.62 M Ba(OH)2
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