Important note: all constants refered to: Kc, Kw, Ka, and Kb are temperature-dependent. All discussions are assumed to be at 25 °C, i.e. standard temperature. (See result #5 for an example of the temperature being changed.)
The following equation describes the reaction of water with itself (called autoprotolysis):
H2O + H2O ⇌ H3O+ + OH¯
The equilibrium constant for this reaction is written as follows:
[H3O+] [OH¯] Kc = ––––––––––– [H2O] [H2O]
However, in pure liquid water, [H2O] is itself 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:
Kc [H2O] [H2O] = [H3O+] [OH¯]
Since the term Kc [H2O] [H2O] is a constant, let it be symbolized by Kw, giving:
Kw = [H3O+] [OH¯]
This constant, Kw, 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, for every H3O+ produced, one OH¯ is also produced. Another way to say is that the molar ratio of H3O+ to OH¯ in the water is 1 to 1. This means that the same amount of each ion is present in pure water. In other words, in pure water, [H3O+] = [OH¯].
Therefore the values of [H3O+] and [OH¯] can be determined by taking the square root of Kw. Hence, both [H3O+] 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 [H3O+]
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 Kw equation as follows:
−log Kw = −log [H3O+] + −log [OH¯]
−log 1.00 x 10¯14 = −log [H3O+] + −log [OH¯]
Note the use of the add sign on the right side of the equation. The result is ususally written as:
pKw = pH + pOH = 14
This is an extremely important equation. Learn it well.
Result #3: If the [H3O+] or the [OH¯] is known, the other can be found.
Simply divide Kw by the known value to get the other.
Suppose [H3O+] is known, then:
[OH¯] = Kw / [H3O+]
Suppose [OH¯] is known, then:
[H3O+] = Kw / [OH¯]
Result #4: In a solution, suppose some solute is added. If one of our variables ( [H3O+] or [OH¯] ) changes in value (either up or down) due to the addition of solute, the other variable will change in the opposite direction.
The change in values will preserve this fundamental equality:
Kw = [H3O+] [OH¯]
Suppose [H3O+] became larger, therefore the [OH¯] becomes smaller.
Suppose [OH¯] became larger, therefore the [H3O+] becomes smaller.
This change happens automatically in the solution and cannot be stopped.
Result #5: Consider the behavior of pure water when the temperature is changed. Remember that Kw is a temperature-dependent constant. As such, the value of Kw will change based on the temperature. The following problem involves a temperature different from the standard temperature of 25 °C. Determine the [OH¯] and pH of pure water at 50 °C, given that Kw at 50 °C equals 5.476 x 10¯14.
Solution:
1) Write the autoprotolysis reaction for water. Then, write the ionization constant expression for water:
2H2O(ℓ) ⇌ H3O+ + OH¯Kw = [H3O+] [OH¯] = 5.476 x 10¯14
2) Based on the 1:1 stoichiometry of the products, we can assert this:
[H3O+] = [OH¯]
3) Therefore, we can assign variables to the hydronium ion concentration as well as the hydroxide concentration:
[H3O+] = [OH¯] = x <--- both are 'x' because the two concentrations are equal.
4) Substitute into the Kw expression and solve:
5.476 x 10¯14 = (x) (x)x = 2.3400855 x 10¯7 M
Rounded off to 4 sig figs, the [OH¯] equals 2.340 x 10¯7 M
5) Determine the pH:
pH = −log [H3O+]pH = −log 2.3400855 x 10¯7
pH = 6.6308