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Kw = [H+][OH-] = 2.9 × 10⁻¹⁴ . In pure water, [H+] = sqrt2.9 × 10⁻¹⁴ ≈ 1.7 × 10⁻⁷ , pH = -log(1.7 × 10⁻⁷) ≈ 6.77 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
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
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