At 100°C, the of water is 55 times its value at 25°C. What will be the approximate pH of a neutral s — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

At 100°C, $K_w = 55 \times 10^{-14}$. Neutral pH = $1/2 \times pK_w$. $pK_w = -\log(55 \times 10^{-14}) = 14 - 1.74 = 12.26$. Neutral pH = $12.26 / 2 = 6.13$.