The autoionization of water at 25°C is described by which equilibrium, and what is the pH?

• A. 2 H2O ⇌ H3O+ + OH− ; pH = 7.00
• B. H2O ⇌ H+ + OH− ; pH = 7.0
• C. 2 H2O ⇌ H2O2 + OH− ; pH = 7.5
• D. H2O ⇌ H2O+ + OH− ; pH = 6.9

Answer: (A)

Water self-ionizes by transferring a proton from one molecule to another, giving hydronium and hydroxide: 2 H2O ⇌ H3O+ + OH−. This shows water acts as both an acid and a base (amphoteric) and defines the ion-product Kw = [H3O+][OH−], which is 1.0 × 10^−14 at 25°C. In pure water, the two ions are produced in equal amounts, so [H3O+] = [OH−] = sqrt(Kw) = 1.0 × 10^−7 M, leading to pH = −log10(1.0 × 10^−7) = 7.00. The proton in solution is actually present as H3O+, so describing the process with H3O+ and H2O is more accurate than writing H+ explicitly. The other depictions introduce incorrect species or omit the two-water involvement, so they don’t reflect the true autoionization chemistry.