Calculate the pH of a solution that has [OH^-] = 3.2 × 10^-2 M. Be sure your answer has the correct number of significant figures.
Note: Reference the Fundamental constants table for additional information.
The relationship between [OH-] and pH is:
pH = 14 - pOH
Since pOH = -log[OH-], we can calculate pOH as follows:
pOH = -log(3.2 × 10^-2)
pOH = -log(3.2) + log(10^-2)
pOH = -0.5052 + 2
pOH = 1.4948
Therefore:
pH = 14 - 1.49
pH = 12.51
Therefore, the pH of a solution with [OH-] = 3.2 × 10^-2 M is 12.51.