What is the #pOH# of a #2.1*10^-6# #M# solution?

1 Answer
anor277
Jul 2, 2016

We assume that you mean a #2.1xx10^-6*mol*L^-1# solution of #KOH#.

Explanation:

#pOH=-log_10[HO^-]#

#=# #-log_10(2.1xx10^-6)# #=# #-(-5.68)# #=# #5.68#.

What is #pH# of this solution? You should be able to answer this easily.

Related questions
See all questions in pH calculations
Impact of this question
5544 views around the world
You can reuse this answer
Creative Commons License