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 pHpH of a solution 1.2xx10^-2*mol*L^-11.2xx10^-2*mol*L^-1 with respect to KOHKOH?