What is the pH of a neutral solution at 37°C, where Kw equals 2.5 * 10^-14? (log 2 = 0.3) — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

In a neutral solution, [H+] = [OH-] = √(Kw). Given Kw = 2.5 * 10^-14, [H+] = √(2.5 * 10^-14) = √(2.5) * 10^-7 M. Taking the negative logarithm: pH = -log(√(2.5) * 10^-7) = 7 - 0.5 * log(2.5) = 7 - 0.5 * log(10 / 4) = 7 - 0.5 * (1 - 2 * log 2) = 7 - 0.5 * (1 - 0.6) = 7 - 0.2 = 6.8.