I would do it this way.
pOH = -log(OH^-) = 8
pH + pOH = pKw = 14
pH = 14- pOH = 14-8 = 6
So what's wrong with 6? Your problem is that you failed to realize that since 7 is neutral where (H^+) = (OH^-) = 10^-7M each, so when OH^- is 1E-8 (NOTE THIS IS LESS THAN 10^-7) SO IT'S A WEAKER BASE WHICH MAKES IT A STRONGER ACID and that's what pH 6 is; i.e., an acid solution.
OR another way.
(H^+)(OH^-) = Kw = 1E-14
when x*y = constant
if y is down then x must be up
Find the pH of 1.0 x10-8 M KOH solution
I tried to find it but i got 6 and that's acidic not basic.
1 answer
1 answer