Calculate the pH and pOH of aqueous solutions having the following ion concentrations.

Question

a) [OH^-] = 3.0 ✕ 10^-6 M

pOH	=
pH	=

(b) [OH^-] = 6.1 ✕ 10^-4 M

pOH	=
pH	=

(c) [H⁺] = 3.8 ✕ 10^-9 M

pOH	=
pH	=

(d) [H⁺] = 0.027 M

pOH	=
pH	=

Answer 1

You have to learn the different relationships between and among pH, pOH, [H+], [OH-] and Kw (ionization constant for H2O).

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pH = -log [H+]

pOH = -log [OH-]

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

pH + pOH = 14

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(a) [OH^-] = 3.0 ✕ 10^-6 M

pOH = -log 3.0 ✕ 10^-6 M = 5.52

pH = 14 - pOH = 14 - 5.52 = 8.48

(b) try it yourself

(c) [H⁺] = 3.8 ✕ 10^-9 M

pOH = -log [OH-] and [OH-] = 1x10^-14 / 3.8 ✕ 10^-9 = 2.63x10^-6

pOH = -log 2.63x10^-6

pOH = 5.58

pH = 14 - pOH = 14 - 5.58

pH = 8.42

(d) try it yourslef