How do you find [H30+] and [OH-] of a base with a molarity of .026M

Question

How do you find [H30+] and [OH-] of a base with a molarity of .026M

Answer 1

Assuming the base is a strong base which ionizes/dissociates completely, the [OH-] = 0.026 M, since

BOH => B⁺ + OH^-

0.026 M....0.026 M..0.026 M

You should also know that [H3O⁺][OH^-] = Kw = 1x10^-14.

[H3O⁺] = 1x10^-14 / 0.026

[H3O⁺] = 3.8x10^-13 M