Water autoionizes to produce H3O+ and OH-, described by Kw = [H3O+][OH-] = 1.0 x 10^-14 at 25 degrees C. pH and pOH are the negative base-10 logarithms of those concentrations, and they always sum to 14 at 25 degrees C. Strong acids (HCl, HBr, HI, HClO4, HNO3, H2SO4) and strong bases (group I and II hydroxides) ionize completely, so you read [H3O+] or [OH-] directly from the starting concentration. Group II hydroxides like Ca(OH)2 release two OH- per formula unit, so [OH-] = 2 x [Ca(OH)2].
- Kw: Ion product of water, equal to [H3O+][OH-] = 1.0 x 10^-14 at 25 degrees C; temperature dependent, so neutral pH is not always 7.
- pH = -log[H3O+]: Converts hydronium concentration to the pH scale; reversible as [H3O+] = 10^(-pH).
- pH + pOH = 14: Holds at 25 degrees C; use this to convert between pH and pOH after finding one of them.
- Complete ionization: Strong acids and bases dissociate 100% in water, so no equilibrium calculation is needed to find [H3O+] or [OH-].
- Group II hydroxides: Release two OH- per formula unit; [OH-] = 2 x initial molarity of the base.
Given 0.025 M HCl, what is the pH? Given 0.010 M Ca(OH)2, what is the pOH and then the pH?
| Species | Ionization | Finding [H3O+] or [OH-] | Example |
|---|
| Strong acid | Complete | [H3O+] = initial acid concentration | 0.10 M HCl: [H3O+] = 0.10 M |
| Strong base (group I) | Complete | [OH-] = initial base concentration | 0.10 M NaOH: [OH-] = 0.10 M |
| Strong base (group II) | Complete | [OH-] = 2 x initial base concentration | 0.10 M Ca(OH)2: [OH-] = 0.20 M |