Data:
[tex]pH = ?[/tex]
[tex][ H_{3} O] = 1.0*10^{-12}[/tex]
The logarithm of number 1 on any basis will always be equal to 0, therefore:
Log10 (1) = 0
Solving:
[tex]pH = - log[ H_{3} O^+][/tex]
[tex]pH = -log1.0*10^{-12}[/tex]
[tex]pH = 12 - log1.0[/tex]
[tex]pH = 12 - 0[/tex]
[tex]\boxed{\boxed{pH = 12}}\end{array}}\qquad\quad\checkmark[/tex]
Note: For pH greater than 7 indicates that the substance is basic (pH > 7)
