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