At , the ionic product of water () is 55 times its value at . What will be the exact pH of a neutral — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

At $25^\circ C$, $K_w = 10^{-14}$. At $100^\circ C$, $K_w = 55 \times 10^{-14}$. A neutral solution has $[H^+] = [OH^-] = \sqrt{K_w}$. Thus, $[H^+] = \sqrt{55 \times 10^{-14}}$. Taking the negative logarithm, $pH = -\frac{1}{2}\log(55 \times 10^{-14}) = -\frac{1}{2}(\log 55 - 14) = -\frac{1}{2}(1.74 - 14) = -\frac{1}{2}(-12.26) = 6.13$.