Acids were defined as compounds that contain hydrogen and can dissolve in water to release hydrogen ions into solution. For example, HCl dissolves in water as follows: HCl
(g)
+ H
2
O → H +
(aq)
+ Cl -
(aq)
The process that converts a molecule such as HCl into ions is called ionization.
Bases are defined as substances that dissolve in water to release hydroxide ions into solution. For example, a typical base according to the Arrhenius definition is sodium hydroxide, NaOH:
NaOH
(s)
+ H
2
O → Na +
(aq)
+ OH -
(aq)
The process of converting a solid ionic compound, such as NaOH, into ions in solution is called dissociation.

Brønsted and Lowry independently proposed a much broader and more useful concept of acids and bases.
Acid is any substance capable of donating a hydrogen ion or proton to another substance. Acids are proton donors.
Example, HCl is an acid because it donates H+ to a water molecule.
HCl
(g)
+ H
2
O
(l)
→ H
3
O +
(aq)
+ Cl -
(aq)
Base is any substance capable of accepting a hydrogen or proton from another substance. Bases are proton acceptors.
Example, NH
3 is a base because it accepts H+ from the water molecule.
NH
3(g)
+ H
2
O
(l)
NH
4
+
(aq)
+ OH -
(aq)

Every aqueous solution is either acidic, basic or neutral. There is a quantitative relationship between the concentration of hydronium and hydroxide ions in the solution.
There is a common practice of substituting H + for H
3
O + to simplify the equation.
Whenever you see the symbol H + (hydrogen ion), you should realize that we are actually referring to H
3
O + (hydronium ion).
In pure water H + and OH form in equal amounts, therefore their concentrations are equal. Concs. of H + and OH at 25°C are each = 1.0 x 10 -7 mol/L (M).
In pure water: [H + ] = 1.0 x 10 -7 M [OH ] = 1.0 x 10 -7 M
The square brackets, [ ], are used to indicate concentration in moles per liter (M).
Multiplying [H + ] and [OH ] concentrations gives a constant called ion product of water, symbol Kw.

Kw = [H + ] x [OH ] = (1.0 x 10 -7 ) x (1.0 x 10 -7 ) = 1.0 x 10 -14
Kw = [H+ ] [OH- ] applies, not only to pure water, but to any aqueous solution.
Notice that [H + ] and [OH ] are inversely proportional. This means that if [H + ] goes up, the [OH ] must come down so that the product of the two is still =
1.0 x10 -14 .
Acidic solution: [H + ] > 10 -7 M and [OH ] < 107 M
Neutral solution: [H + ] = 10 -7 M and [OH ] = 10 -7 M
Basic solution: [H + ] < 10 -7 M and [OH ] > 10 -7 M
The pH scale is a numerical scale which, for most applications extends from
0 through to 14. The numbers on the scale represent the relative acidity of solutions and can be converted into actual hydronium ion concentrations.
The pH scale is based on the self-ionization of pure water. Two water molecules will sometimes combine into hydronium and hydroxide ions.

The pH scale:
11
12
13
8
9
10
6
7
4
5 pH
1
2
3
So the pH is the -log of the [hydronium ion].
[H
3
O + ]
10 -1
10 -2
10 -3
10 -4
10 -5
10 -6
10 -7
10 -8
10 -9
10 -10
10 -11
10 -12
10 -13
[OH ]
10 -13
10 -12
10 -11
10 -10
10 -9
10 -8
10 -7
10 -6
10 -5
10 -4
10 -3
10 -2
10 -1
3
2
1
6
5
4
10
9
8
7 pOH
13
12
11

Problem 1. Calculate the concentration of OH in an aqueous solution if the H + concentration is 2.8 x 10 -3 .
Solution
We know that Kw = [H + ] x [OH ] = 1.0 x10
The [H + ] is given and we solve for [OH ].
-14
1.0 x 10-14 [OH ] = = 3.6 x 10 -12 M
Problem 2. For a 6.8 x 10 -4 M KOH solution, determine the following: a) [OH ] b) [H + ], c) the pH
Solution a) Because KOH is a strong base, it completely dissociates: KOH → K b) Substituting the [OH- ] value in Kw = [H +
+ + OH -
So the concentration of hydroxide ion of a 6.8 x 10 -4 M KOH solution is 6.8 x 10 -4 M.
[OH- ] = 6.8 x 10 -4
] x [OH ] = 1.0 x10 -14 yields:
[H + ] = 1.0 x10 -14 / 6.8 x 10 -4 = 1.5 x 10 -11 c) pH = -log [H + ] pH = -log(1.5 x 10 -11 ) ] pH = -log10 -11 – log 1.5
pH = 11 – 0.18 = 10.82

The pOH scale is the corollary of the pH scale pH + pOH = 14
You'll remember from math class that when you multiply two numbers you only add their logs. [H
3
O + ] x [OH ] = (1.0 x 10 -7 ) x (1.0 x 10 -7 ) = 1.0 x 10 -14 or -log[H
3
O + ]
+
-log[OH ] = 7 + 7 = 14
Thus a solution that has a pH = 7 must also have a pOH = 7.
Problem 3.
Calculate the pOH value of a solution of HCl which has a [H + ] = 1.0 x 10 -3 ?
Solution: pH = -log[H + ] = -log[1.0 x 10 -3 ] = 3 pH + pOH = 14
3 + pOH = 14 pOH = 14 -3 = 11