Calculate the [oh-] and the ph of a solution with an [h+= .00083 m at 25 degrees

1.2 × 10^-11 M

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

i believe it is 40–40–20%

explanation: