find the pH of the following solution

Question

a. 1x10^-3 HCL

b. 6x10^-9 Potassium Hydroxide

Answer 1

pH = -log [H⁺]

pH + pOH = 14

[H⁺][OH^-] = 1x10-14

a). pH of 1x10^-3 M HCl

HCl is a strong acid so ionizes completely producing 1x10^-3 M H⁺

-log 1x10^-3 = pH = 3

b). 6x10^-9 M KOH

KOH is a strong base so dissociates completely producing 6x10^-9 M OH^-

-log 6x10^-9 = pOH = 8.22

pH = 14 - pOH

pH = 5.78