What does pOH mean?

1 Answer
May 26, 2016

Answer:

#pOH# #=# #-log_10[HO^-]#

Explanation:

So #pOH# is the basic counterpart of #pH# #(=-log_10[H_3O^+])#.

This derives from the autoprotolysis reaction in water:

#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#

This equilibrium has been carefully measured, and at #298K# we write,

#K_w=[HO^-][H_3O^+]# #=# #10^-14#.

We can take #log_10# of both sides to give:

#log_10K_w# #=# #log_10[HO^-] + log_10[H_3O^+]# #=# #log_10{10^-14}#

#log_10K_w# #=# #log_10[HO^-] + log_10[H_3O^+]# #=# #-14# (because by definition, #log_a(a^b)=b#. On rearrangement:

#14=-log_10[HO^-] - log_10[H_3O^+]#

And, again by definition,

#pH + pOH =14#

The autoprotolysis of water is a bond breaking reaction. How would you predict #K_w# to evolve at elevated temperature? What would this mean with respect to #pH# and #pOH#?