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
Question
At 100°C, the $K_w$ of water is 55 times its value at 25°C. What will be the approximate pH of a neutral solution at this temperature? (Given $\log 55 = 1.74$)
Answer: C
💡 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 = $12.26 / 2 = 6.13$.
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