What is the pH of a neutral solution at 37°C, where equals ? () — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

In a neutral solution, $[\text{H}^+] = [\text{OH}^-]$, so $[\text{H}^+] = \sqrt{K_w}$. Thus, the neutral pH is $\frac{1}{2}pK_w$. Here, $pK_w = -\log(2.5 \times 10^{-14}) = 14 - \log(2.5) = 14 - \log(10/4) = 14 - (1 - 2\log 2) = 13 + 2(0.3) = 13.6$. The pH is $\frac{13.6}{2} = 6.8$.