What is the [OH-] in a 50.0 mL solution of 0.050 M HClO4?

Question

Help would be really appreciated!

Answer 1

Recall that [H⁺][OH^-] = 1x10^-14

From 0.050 M HClO₄ we know the [H⁺] = 0.05 M since this is a strong acid and ionizes completely.

[0.05][OH^-] = 1x10^-14

[OH^-] = 1x10^-14 / 0.05 = 1x10^-14 / 5x10^-2

[OH^-] = 2x10^-13 M (ignoring the autoionization of H2O)