What is the equilibrium constant of pure water at 25°C?

1 Answer
May 29, 2018

Answer:

#K_w = 1.0xx10^(-14)# at #25ºC#.

Explanation:

Sometimes, in pure water, one proton (#H^+# ion) of one water molecule will be attracted to one of the lone pairs in another water molecule.
This causes the #H^+# ion to leave its current water molecule (leaving behind #OH^-#) to attach to the lone pair of another water molecule (forming #H_3O^+#, the hydronium ion).

This process is called the auto-ionization or self-ionization of water, since #H_2O# is ionizing into #OH^-# and #H^+#.

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So, this reaction will sometimes occur:

#H_2O + H_2O rightleftharpoons H_3O^+ + OH^-#

However, it's important to note that this reaction only happens very rarely. As such, the equilibrium constant, which is a measure of the concentration of products to the concentration of reactants, will be very low.

In fact, at #25ºC#, this equilibrium constant will be #1.0xx10^(-14)#.