The ionic product of water at 298 K is 1.0 × 10⁻¹⁴ . What is the pH of pure water at this temperature?
Kw = [H+][OH-] = 1.0 × 10⁻¹⁴ . In pure water, [H+] = [OH-] = sqrt1.0 × 10⁻¹⁴ = 1.0 × 10⁻⁷ , so pH = -log(1.0 × 10⁻⁷) = 7 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases