Chemistry 12
UNIT 4
ACIDS AND BASES
CHAPTER 9 and 10
Recall the definitions from Unit 3:
•
•
•
•
•
Strong electrolyte
Strong electrolytic solution
Weak electrolyte
Weak electrolytic solution
Non - electrolyte
Go back and read these
definitions in last unit 3
notes package!
Acids and bases are both electrolytes
RECALL FROM CHEMISTRY 11:
Svante Arrhenius - Swedish Chemist 1884
ARRHENIUS THEORY OF ACIDS AND BASES:
ACID: a substance that contains Hydrogen and produces H+ in aqueous solution
BASE: a substance that contains hydroxyl groups and produces OH- in aqueous
solution
NEUTRALIZATION REACTIONS: the reaction of H+ with OH- to produce water
molecules
H+ (aq) + OH- (aq) ⇄ H2O (l)
ACID + BASE SALT + WATER
SALTS: contain a cation other than H+ and an anion other than OH- .
CHARACTERISTICS OF ACIDS AND BASES:
Acids
BASES
Both dissolve in water to give a solution that:
-tastes sour
-tastes bitter
-conducts electricity (both are electrolytes)
-causes certain indicators to change colour
-produces or donates a proton
-accepts a proton
(reacts with a base)
(reacts with an acid)
-produces H2 (g) when reacting with
-feels slippery
metals
eg. Mg(s) + 2HCl(l)  MgCl2(aq) + H2(g)
(note that some metals such as Na and K
are nonreactive with acids)
-turns litmus paper RED
-turns litmus paper BLUE
(reD in aciD)
(Blue in Base)
-loses the above properties when reacted with a base / acid
(but the resulting solution conducts a current)
All of the above characteristics can be used to identify acids and bases in a lab …
except taste!!!!
SEE ALSO TEXT LIST p. 419
There are several criteria to consider when deciding whether solutions are
acidic or basic.
The Arrhenius Theory does not consider all cases such as solutions that
prove to be acidic even when there aren’t H+ ions present.
For that reason, there are other ways to classify Acids and Bases:
BRONSTED - LOWRY THEORY OF ACIDS AND BASES:
Johannes Bronsted - Danish chemist 1923
and Thomas Lowry - English chemist
ACID: A proton (hydrogen ion) donor
BASE: A proton acceptor
Bronsted - Lowry Acids:
HCl (g) + H2O (l)  H3O+ (aq) + Cl- (aq)
STRUCTURALLY:
Acid: HCl (donates H+)
Base: H2O (accepts H+)
HYDRONIUM ION: Free protons will always bond to water when in aqueous
solution.
The Symbol H+ is often used because it is more convenient.
This ion is acidic since it can donate a proton to a base.
Bronsted - Lowry Bases:
NaOH is an example of a base that fits with the Arrhenius definition:
NaOH (s)  Na+ (aq) + OH- (aq)
dissolve in water
Example of a base that does not "fit" the Arrhenius Theory:
NH3 (g) + H2O (l)  NH4+ (aq) + OH- (aq)
Ammonia accepts a proton and is therefore a Bronsted base.
WEAK / STRONG ACIDS / BASES
RECALL THE DEFINITIONS OF WEAK AND STRONG ELECTROLYTES.
Acids and Bases are electrolytes because they conduct electric current in solution.
Therefore:
STRONG ACID: completely ionized in solution
eg. Hydrochloric acid is a very strong acid.
HCl (g) + H2O (l)  H3O+ (aq) + Cl- (aq)
Even when you dilute HCl, it completely ionizes.
Therefore even dilute HCl is a strong acid.
WEAK ACID: partially ionized in solution; equilibrium
CH3COOH (g) + H2O (l) ⇄ H3O+ (aq) + CH3COO- (aq)
STRONG BASE: completely dissociates in water
NaOH (s)  Na+ (aq) + OH- (aq)
WEAK BASE: partially dissociates in water; equilibrium
NH3 (g) + H2O (l) ⇄ NH4+ (aq) + OH- (aq)
Molecules that differ only by a proton are sometimes referred to as
CONJUGATE ACID - BASE PAIRS
In any neutralization reaction, there are always two acid-base pairs:
NH3 (g)
+ H2O (l) ⇄ NH4+ (aq)
+ OH- (aq)
BASE 1---------------------------------ACID 1
To be continued
On Wed
ACID 2---------------------------------BASE 2
HCN (g) + H2O (l)
⇄
H3O+ (aq) + CN- (aq)
BASE 2----------------ACID 2
ACID 1----------------------------------------------------BASE 1
AMPHIPROTIC SUBSTANCE: capable of acting as both an acid and a base,
SEE ABOVE TWO EXAMPLES - water is an acid in the first example and a base in the
second example.
Lewis theory:
ACID: electron pair acceptor
BASE: electron pair donor
More on this in University level Chemistry
……………………………………………………………………………………………....
pH 7:
There is approximately 6 x 1016 H+ ions in a liter of water.
Therefore:
6 x 1016 (hydrogen ions)
x
1
+23
6 x 10
particles (atoms / ions)
=
1 mole
7
10
x
1
+23
6 x 10
particles (atoms / ions)
=
1 mole
101
x
1
+23
6 x 10
particles (atoms / ions)
=
1 mole
14
10
The notation pH 7 comes from:
p: power of 10
H: hydrogen ions
7: the exponent on the power of 10
ACIDIC SOLUTIONS:
6 x 1022 (hydrogen ions)
pH 1
BASIC SOLUTIONS:
6 x 109 (hydrogen ions)
pH 14
pH
pH is the measure of how acidic or basic a solution is.
The pH scale runs from 0 to 14
0--------------------------------------7--------------------------------14
acid
base
neutral
Consider pure water:
H2O (l) + H2O (l) ⇄ H3O+ (aq) + OH- (aq)
at 25oC
10-7M
10-7M
Kw = [H3O+][OH-]
= (10-7M) (10-7M )
= 1 x 10-14
the equilibrium constant for water
pH of pure water:
pH = -log[H3O+]
= -log 10-7
=7
(recall log 10x = x)
therefore water is neutral
H2O (l) ⇄ H+ (aq) + OH- (aq)
-same equation as before, but writing proton instead of hydronium ion
pH = -log[H3O+]
[H3O+] = 10 -pH
check: [H3O+] = 10 -pH
if water is neutral, then [H3O+] = 10 -7
What is the pH if the concentration of H3O+ is 4.2 x 10-5 M?
pH = -log[H3O+]
= -log (4.2 x 10-5)
= 4.38
therefore acidic
If [H3O+][OH-] always equals 10-14 at 25oC, then calculate [OH-] in the above
problem.
[H3O+][OH-] = 10-14
[OH-] = 10-14
= 10-14
= 2.38 x 10 -10
_______
_______
[H3O+]
4.2 x 10-5
NOTE ALSO:
CONSIDER: 2H2O (l) + kJ ⇄ H3O+ (aq) + OH- (aq)
As shown above, if temp increases, then Kw increases since it is an endo reaction.
Therefore, consider if Kw = 1.00 X 10-11 at a certain temp, then calculate pH.
Kw = [H3O+][OH-]
1.00 x 10-11 = x2
x=
3.16 x 10-6 M = [H+] = [OH-]
pH = -log (3.16 x 10-6)
= 5.500
NOTE MATHEMATICAL THEORY:
[H+] = 3.16 x 10-6 M
= 10 .5 x 10 -6
= 10 -5.5
For pH, only the numbers
after the decimal
count for sig figs
pH = 5.500
3 sig figs (because of 3.16)
the 5 in front of the decimal does not count for the sig fig
pOH
pOH = -log[OH-]
[OH-] = 10 -pOH
Consider: [H3O+][OH-] = 10-14
10 -pH x 10 -pOH = 10-14
take antilogs of both sides
(-pH)+(-pOH) = -14
multiply through by -1
pH + pOH = 14
If the pH of a solution is 2.800, calculate:
a) [H3O+]
b) [OH-]
c) pOH
d) Is the solution acidic or basic?
a) [H3O+] = 10 -pH
= 10-2.8
= 1.58 x 10-3 M
b) [H3O+][OH-] = 10-14
[OH-] = 10-14
_______
[H3O+]
= 10-14
_______
1.58 x 10-3
= 6.33 x 10 -12 M
c) pOH = - log [OH-]
OR
pH + pOH = 14
= -log(6.33 x 10 -12)
pOH = 14- 2.8
pOH = 11.200
d) acidic!!!!
NOW DO WORKSHEET 4.1 FOR HOMEWORK
Wkst 4.1: pH and pOH
[H+], M
-
[OH ], M
pH
pOH
1.0 x 10-9
5.0 x 10-4
9.0 x 10-1
8.0 x 10-1
3.3 x 10-13
4.0 x 10-10
6.50
7.00
2.75
8.10
12.50
8.90
Acidic/Basic?
Wkst 4.1: pH and pOH
[H+], M
[OH-], M
pH
pOH
Acidic/Basic?
1.0 x 10-9
1.0 x 10-5
9.00
5.00
Basic
5.0 x 10-4
2.0 x 10-11
3.30
10.70
Acidic
9.0 x 10-1
1.1 x 10-14
0.05
13.96
Acidic
1.3 x 10-14
8.0 x 10-1
13.89
0.10
Basic
3.0 x 10-2
3.3 x 10-13
1.52
12.48
Acidic
2.5 x 10-5
4.0 x 10-10
4.60
9.40
Acidic
3.2 x 10-7
3.2 x 10-8
6.50
7.50
Acidic
1.0 x 10-7
1.0 x 10-7
7.00
7.00
Neutral
1.8 x 10-3
5.6 x 10-12
2.75
11.25
Acidic
1.3 x 10-6
7.9 x 10-9
5.90
8.10
Acidic
3.2 x 10-2
3.2 x 10-13
1.50
12.50
Acidic
7.9 x 10-6
1.3 x 10-9
5.10
8.90
Acidic