Answer :
Step 1
Formulas used here:
[tex]\begin{gathered} pOH\text{ = -log }\lbrack OH-\rbrack\text{ \lparen1\rparen} \\ pH\text{ + pOH = 14 \lparen2\rparen} \\ pH\text{ = -log}\lbrack H3O+\rbrack\text{ => }\lbrack H3O+\rbrack=10^{-pH}(3) \end{gathered}[/tex]-----------------
Step 2
From (1):
pOH = -log [OH-] = -log(1.0 x 10^-12 M) = 12
pOH = 12
-----------------
Step 3
From (2):
pH + pOH = 14 => pH = 14 - pOH = 14 - 12 = 2
pH = 2
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Step 4
From (3):
[H3O+] = 10^-pH = 10^-2 = 0.01 M or 1x10^-2 M
Answer: [H3O+] = 0.01 M or 1x10^-2 M