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.

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