What are the concentrations of H3O + and OH-? in oranges that have a pH of 3.69?

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

[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