Chapter 13Acids and Bases
What are acids and Bases?
A. Properties of Acids
-tastes sour
-conducts electricity
-turn blue litmus paper red
-reacts with bases to form water and salts
-have pH less than 7
Naming Acids
Two Types of Acids
a. Binary Acid
-contains only H and one other element
Naming Binary Acids
1. Name begins with prefix hydro-
2. Root of second element follows hydro and ends with the suffix -ic
Examples of Binary Acids
Hydrofluoric acid HF
Hydrochloric acid HCl
Hydrobromic acid HBr
Hydrosulfuric acid H2S
Acids cont.
b. Oxyacid
- acid that contains H, O, and a third element,usually a nonmetal
Naming Oxyacids
1. Root of the anion plus the suffix –ic followed by the word acid
Examples of Oxyacids
Acetic acid acetate
Carbonic acid carbonate
Nitric acid nitrate
Phosphoric acid phosphate
Sulfuric acid sulfate
Properties of Bases
- tastes bitter- Feels slippery- Turns red litmus paper blue- Reacts with acids to form water and
salts- Have a pH greater than 7
Bases cont.
Naming Bases
- Element name followed by hydroxide
- There are a few exceptions:
Ammonia NH3
Aniline C6H5NH2
Acid-Base Theories
Arrhenius Acids and Bases
1. Arrhenius Acid
acids in water produce H+ ions
HNO3(aq) H+(aq) + NO3-
2. Arrhenius Base
bases in water produce OH-
KOH(s) K+(aq) + OH-(aq)
Cont.
Bronsted-Lowry Acids and Bases-Acids are proton(H+) donors
H2CO3(aq)+H2O(l)H3O+(aq)+HCO3-(aq)
(acid) (base) (C.A.) (C.B.)-Bases are proton acceptor
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
base acid C.A. C.B.Conjugate acid- ion formed when a base has accepted a
H+ Conjugate base- ion formed when an acid has donated a
H+
How are weak acids and bases compared?
Acid-dissociation constant(Ka)- ratio of concentrations of the products to the
reactants for a weak acid- Example
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
Ka= [CH3COO-] [H3O+] [CH3COOH]
Cont.
Base-dissociation constant(Kb)
-ratio of concentrations of the products to the reactants for a weak base
NH3(aq) +H2O(l) NH4+(aq) + OH-(aq)
Kb= [NH4+] [OH-]
[NH3]
The pH scale
Is the negative of the common logarithm of the hydronium ion concentration
Ranges from 0-14
We use the following equation:
pH= -log[H3O+]
Self-Ionization Constant of Water
2H2O(l) H3O+(aq) + OH-(aq)
Kw= [H3O+] [OH-]
Kw = 1.00 x 10-14 at 25° C
Determining [H3O+] and [OH-] with Kw
Example
What is [OH-] in a 3.00x 10-5 M solution of HCl?
HCl(g) + H2O(l) H3O+(aq) +Cl-(aq)
2H20(l) H3O+(aq) + OH-(aq)
What do we know?
Kw= 1.00 x 10-14
[H3O+]= 3.00 x 10-5
Using Logarithms in pH calculations
1. Calculating pH from [H3O+]
pH= -log [H3O+]
2. Calculating [H3O+] from pH
[H3O+] = 10-pH
Examples of pH
What is the pH of a .00010 M solution of HNO3, a strong acid?
[HNO3]= .00010 M or 1.0 x 10-4 M
pH= -log[1.0 x10-4]
pH= 4.00
Examples cont.
What is the pH of a .0136 M solution of KOH, a strong base?
[OH-]= .0136 M
Kw= 1.00 x 10-14
Kw= [H3O+][OH-]
pOH= -log[OH-]
pH + pOH = 14
Calculating [H3O+] and [OH-] from pH
What are the concentrations of the hydronium and hydroxide ions in a sample of rain that has a pH of 5.05?
[H3O+] = 10-5.05
= 8.9 x 10-6 MSince pH + pOH = 14 pOH= 8.95 so
[OH-] = 10-8.95
= 1.1 x 10-9 M
Calculating Ka/Kb for a weak acid or base
A vinegar sample is found to have .837 M CH3COOH. Its hydronium concentration is found to be 3.86 x 10-3 mol/L. Calculate Ka for acetic acid.
What do we know?
[CH3COOH]= .837 M
[H3O+]= 3.86 x 10-3 M
CH3COOH(aq)+ H2O(l)H3O+(aq)+CH3COO-(aq)
Neutralization and Titrations
Neutralization Reaction- a reaction between an acid and a
hydroxide base that results in a salt and water
HCl + NaOH NaCl + H2O
Titrations
Titrations- a procedure in which a solution of known
concentration is used to determine the concentration of a second unknown solution
Equivalence Point
- the point at which a neutralization reaction is complete
Titrations cont.
How do we know when the equivalence point has been reached?You use an indicatorA substance that changes its color as it reacts with either an acid or baseThe instant the indicator changes color is called the End Point
Common indicators are red cabbage juice, litmus, bromthymol blue, and phenolphthalein
Titration Calculations
Suppose that in the titration of 40. mL of vinegar, 20.mL of .50 M NaOH were needed to reach the equivalence point. What is the molarity of the acetic acid in the vinegar?
NaOH(aq)+CH3COOH(aq)NaCH3COO(aq) + H2O(l)
What do we know?Volume of NaOH= 20. Ml[NaOH]= .50 MVolume of vinegar = 40. ml