pH = -log [H3O+] --> [H3O+] = 10^-pHpH 1: [H3O+] = 10^-1 = 0,1 mol/L
Cola [H3O+] = 0,002 mol/L --> pH = -log 0,002 = 2,7
pOH = -log [OH-] --> [OH-] = 10^-pOH
Ammoniak [OH-] = 0,0001 mol/L --> pOH = -log 0,0001 = 3
pKw = [pH] + [pOH] = 14
Dit geeft: pH = 14-3 = 11. Dus [H3O+] = 10^-11 = 1x10^-11