Calculate the [OH-] and the pH of a solution with an [H+=.00083 M at 25 degrees

1.2 * 10^-11 M

Explanation:

We are given;

Concentration of Hydrogen ions [H⁺] as 0.00083 M

We are required to calculate the concentration of [OH⁻]

We know that;

pH = - log [H⁺]

POH = - log[OH⁻]

Also, pH + pOH = 14

With the concentration of H⁺ we can calculate the pH

pH = - log 0.00083 M

= 3.08

But; pH + pOH = 14

Therefore; pOH = 14 - pH

= 14 - 3.08

= 10.92

But, pOH = - log[OH⁻]

Therefore; [OH] = - Antilog (pOH)

Hence; [OH⁻] = Antilog - 10.92

= 1.2 * 10^-11 M

Therefore, [OH⁻] is 1.2 * 10^-11 M