Determine the [H+] , [OH−] , and pOH of a solution with a pH of 10.10 at 25 °C.

Answer 1

Here are the relationships you need to know:

pH + pOH = 14

[H+][OH-] = 1x10^-14

pH = -log [H+]

pOH = -log [OH-]

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pH = 10.10

[H⁺] = 1x10^-10.10 = 7.94x10^-11

[OH-] = 1x10^-14 / 7.94x10^-11 = 1.26x10^-4

pOH = -log 1.26x10^-4 = 3.90

or

pOH = 14 - pH = 14 - 10.10 = 3.9