Identifying Acids and Bases
• Acids
• Acid (anhydrides)
• Bases
• Base (anhydrides)
• Salts
contains H+ ions as the cation, with and other element as the anion
Non-metal oxide Contains OH- as the anion Combined with a cation A metal with oxygen
Ionic compound Formed from a cation and an anion
Metal and non-metal
H2SO4
HI P2O5 Ca(OH)2
Al2O3
Mg F2
KCl LiBr
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 Definitions
•Lewis:
Acids: Any substance which can accept an electron pair.
Bases: Any substance which can donate an electron pair.
Acid / Base Chemistry
•Hydrogen ion concentration = [H+]
•Hydroxide ion concentration = [OH-]
•Pure H20 – neutral substance – – ph=7 and [H+]=[OH-]
•H2O will self-dissociate – H0 > H++ OH-
Acid / Base Chemistry
- Kw- is the ion product constant for water
- Kw= 1.0 x 10-14M2 Kw= [H+] [OH-]
FOR H2O:Kw= [1.0 x 10-7M][1.0 x 10-7M]
-
there is an indirect relationship between [H+] and [OH-]Acid / Base Chemistry
•Calculate [OH
-] if [H
+] is 1.0 x 10
-2M
• Kw= [H+] [OH-]
• [OH
-]= K
w= 1.0 x 10
-14M
2• [H
+] 1.0 x 10
-2M
• [OH
-]= 1.0 X 10
–12M
Acid / Base Chemistry
•Calculate [H+] if [OH-] is 6.5 x 10-11M
• Kw= [H+] [OH-]
• [H+]= Kw = 1.0 x 10-14M2 [OH-] 6.5 x 10-11M
• [H+]=
• [H+] will tell us if a soln is acidic or basic.
Acidic solutions
Neutral solutions
Basic solutions
pH= 1- 6
[H+]= 1.0 x 10 M – 1.0 x 10 M [H+] > [OH-]
-1 -6
pH= 7 [H+] = [OH-]
1.0 x 10 M for both-7
pH= 8 -14
[H+]= 1.0 x 10 M – 1.0 x 10 M [H+] < [OH-]
-8 -14
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-]
Acid / Base Chemistry
• If [OH-]= 2.63 x 10-4M, 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 10-3M. Is this acidic, basic, or neutral?
• Calculate the pH of a solution with [H+] = 4.2 x 10-6M. Acidic, basic, or neutral?
What is the [OH-]?
Acid / Base Formulas
• A solution with a pH of 9.1 has what [H+]?
– 2ndlog –9.1 ENTER
• What is the [OH-]?
• What is the pOH?
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
Acids, Bases, and Their Conjugates
• NH3+ H2O -- NH4++ OH-
• HCl + H2O --->
Acids, Bases, and Their Conjugates
• CO3-2+ H2O ---> HCO3- + OH-
• Ca(OH)2 + 2HNO3--> Ca(NO3)2 + 2HOH
Identifying monoprotic, diprotic, andtriprotic:
monoprotic:
HCl H++ Cl- diprotic:
H2SO4 H++ HSO4-1
HSO4-1 H++ SO4-2
triprotic:
H3PO4 H++ H2PO4-1
H2PO4-1 H++ HPO4-2
HPO4-2 H++ PO4-3
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
•Weak acids and bases slightly dissociate
ex. HClO <--> ClO-+ H+
Strengths of Acids and Bases
•Strong Acids: HCl, HI, HBr, H2SO4, HNO3, HClO4
•Strong Bases: Groups I and II with OH- except Be
Electrolytes and Non-Electrolytes
•Strong Electrolytes:Strong Electrolytes:Strong Electrolytes:Strong Electrolytes:
- Compounds that conduct electricity in aqueous solution
- Substances that completely dissociate - Strong acids and bases
- Soluble salts
Electrolytes and Non- Electrolytes
•Weak Electrolytes:Weak Electrolytes:Weak Electrolytes:Weak Electrolytes:
- Slight electron conductivity - Slight dissociation in aqueous - Weak acids and bases
- Insoluble salts if in molten state
Electrolytes and Non- Electrolytes
•NonElectrolytes:NonElectrolytes:NonElectrolytes:NonElectrolytes:
- Don’t conduct electricity - No ions to carry current - Organic, molecular
Examples of Electrolytes and Nonelectrolytes:
• H2S03
• weak acid/weak elect.
• CsOH
• strong base/strong elect.
• HCl04
• strong acid/strong elect.
• C6H1206
• nonelectrolyte- covalent, molecular, organic b/c carbon
Examples of Electrolytes and Nonelectrolytes:
• Al(OH)3
• weak base/weak electrolyte
• HF
• weak acid/weak electrolyte
• CCl4
• nonelectrolyte- molecular compound
• NaN03
• soluble salt/strong electrolyte
• S03
• acid anhydride/weak electrolyte
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.
Neutralization Reactions:
General Equation:
• Acid+ Base Salt+ Water
• HCl+ NaOH NaCl+ HOH Acid Base Salt Water
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
Titration Curve
Neutralization practice problems:
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:
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?
Neutralization practice problems:
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?
G: 45 mL 2.6 M HCl04
3.0 M Ca(OH)2
U: ? vol Ca(OH)2
Ca(OH)2+ HCl04 2 HOH + Ca(Cl04)2 0.45 L HCl04x 2.6 mol HCl04x 1 mol Ca(OH)2x 1 L Ca(OH)2
1 L Ca(OH)2 2 mol HCl04 3 mol Ca(OH)2
vol mol mol volM CR M
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
G: 220.5 mL 6.0 M H3P04
300.0 mL Ba(OH)2
U: ? M Ba(OH)2
3Ba(OH)2+ 2H3P04 Ba3(P04)2+ 6HOH 0.2205 L H3P04x 6.0 mol H3P04x 3 mol Ba(OH)2
1 L H3P04 2 mol H3P04
= 1.985 mol Ba(OH)2
vol mol molM CR