Explanation:
pH = -log [H3O+]
[H3O+] = 1x10-3.71 = 1.95x10-4 M
You should be able to find [OH-] recalling that Kw = [H3O+][OH-] = 1x10-14
Answer:
[H+] = 2.04 *10^-4 M
[OH-] = 4.90*10^-11 M
Explanation:
Step 1: Data given
pH = 3.69
Step 2: Calculate [H3O+]
pH = 3.69 = -log [H+] = -log [H3O+]
[H+] = [H3O+] = 10^-3.69 = 2.04 *10^-4 M
Step 3: Calculate pOH
pH + pOH = 14
pOH = 14 - 3.69 = 10.31
Step 4: Calculate [OH-]
pOH = 10.31 = -log [OH-]
[OH-] = 10^-10.31 = 4.90*10^-11 M
