A solution has a #[OH^-]# of #1*10^-2#. What is the #pOH# of this solution?

1 Answer
May 19, 2016

Answer:

#pOH# #=# #2#.

Explanation:

#pOH# #=# #-log_(10)[HO^-]# (i.e. just as #pH# #=# #-log_(10)[H_3O^+]#.

Since know that when #log_ab# #=c#, then, by definition, #a^c# #=# #b#, thus #-log_10(1xx10^-2)# #=# #-(-2)# #=# #2#.

Given that in water at #298*K#, #K_w# #=# #10^-14#, what is #pH# of this solution.