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

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Answer 1 · 2017-06-16T15:18:04

#pH=12.08#

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=....??#

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