Question

Describe the autoprotolysis of water. Include the relevant chemical reaction and the derivation of the autoprotolysis constant.

Answer 1

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

The equilibrium constant for this reaction is written as follows:

However, in pure liquid water, [H₂O] 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:

Since the term K_c [H₂O] is a constant, let it be symbolized by K_w, giving:

This constant, K_w, is called the water autoprotolysis constant or water autoionization constant. 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.