Important note: all constants refered to: K_{c}, K_{w}, K_{a}, and K_{b} are temperature-dependent. All discussions are assumed to be at 25 °C, i.e. standard temperature.

The following equation describes the reaction of water with itself (called autoprotolysis):

H_{2}O + H_{2}O <==> H_{3}O^{+}+ OH¯

The equilibrium constant for this reaction is written as follows:

K_{c}= ( [H_{3}O^{+}] [OH¯] ) / ( [H_{2}O] [H_{2}O] )

However, in pure liquid water, [H_{2}O] is a constant value. To demonstrate this, consider 1000 mL of water with a density of 1.00 g/mL. This 1.00 liter (1000 mL) would weigh 1000 grams. This mass divided by the molecular weight of water (18.0152 g/mol) gives 55.5 moles. The "molarity" of this water would then be 55.5 mol / 1.00 liter or 55.5 M.

The solutions studied in introductory chemistry are so dilute that the "concentration" of water is unaffected. So 55.5 molar can be considered to be a constant if the solution is dilute enough.

Cross-multiplying the above equation gives:

K_{c}[H_{2}O] [H_{2}O] = [H_{3}O^{+}] [OH¯]

Since the term K_{c} [H_{2}O] [H_{2}O] is a constant, let it be symbolized by K_{w}, giving:

K_{w}= [H_{3}O^{+}] [OH¯]

This constant, K_{w}, is called the water autoprotolysis constant or water autoionization constant. (Sometimes the prefix auto is dropped, as was done in the title of this section.) It can be determined by experiment and has the value 1.011 x 10¯^{14} at 25 °C. Generally, a value of 1.00 x 10¯^{14} is used.

From the chemical equation just above, it can be seen that H_{3}O^{+} and OH¯ concentrations are in the molar ratio of one-to-one. This means that [H_{3}O^{+}] = [OH¯].

Therefore the values of [H_{3}O^{+}] and [OH¯] can be determined by taking the square root of K_{w}. Hence, both [H_{3}O^{+}] and [OH¯] equal 1.00 x 10¯^{7} M in pure water. This leads to several important results in the acid base world.

Result #1: The pH of pure water is 7

By definition, pH = -log [H_{3}O^{+}]

The pH of pure water then equals -log 10¯^{7}, which is 7.

Result #2: If the pH or the pOH is known, the other can be found.

Take the negative logarithm of each side of the K_{w} equation as follows:

- log K_{w} = -log [H_{3}O^{+}] + -log [OH¯]

-log 1.00 x 10¯^{14} = -log [H_{3}O^{+}] + -log [OH¯]

Note the use of the add sign on the right side of the equation. The result is ususally written as:

pK_{w}= pH + pOH = 14

This is an extremely important equation. Learn it well.

Result #3: If the [H_{3}O^{+}] or the [OH¯] is known, the other can be found.

Simply divide K_{w} by the known value to get the other.

Suppose [H_{3}O^{+}] is known, then:

[OH¯] = K_{w} / [H_{3}O^{+}]

Suppose [OH¯] is known, then:

[H_{3}O^{+}] = K_{w} / [OH¯]

Result #4: If one variable ( [H_{3}O^{+}] or [OH¯] ) changs value (either up or down), the other variable will change in the opposite direction.

The change in values will still preserve this fundamental equality:

K_{w}= [H_{3}O^{+}] [OH¯]

Suppose [H_{3}O^{+}] became larger, therefore the [OH¯] becomes smaller.

Suppose [OH¯] became larger, therefore the [H_{3}O^{+}] becomes smaller.

This happens automatically and cannot be stopped.