What is #pH# of a solution #1.2xx10^-2*mol*L^-1# with respect to #KOH#?

1 Answer
Jun 16, 2017

Answer:

#pH=12.08#

Explanation:

Well, first off we calculates #pOH#

And #pOH=log_(10)[HO^-]=-log_(10)1.2xx10^-2=1.92#.

Now it is a given that #pOH+pH=14#.

Why? Because #K_w=[HO^-][H_3O^+]=10^(-14)#, and when we take #log_10# of both sides, we come up with that relationship.

And so #pH=14-pOH=14-11.92=....??#