At $25°C$, the ionic product of water $K_w = [H^+][OH^-] = 10^{-14}$. What is the pH of pure water at $25°C$?
A$0$
B$14$
C$10$
D$7$
Answer & Solution
Correct answer: D. $7$
In pure water, autoionisation gives equal concentrations of $H^+$ and $OH^-$. So $[H^+] = [OH^-]$, and from $[H^+][OH^-] = 10^{-14}$:
$[H^+]^2 = 10^{-14} \Rightarrow [H^+] = 10^{-7}$ M.
pH $= -\log_{10}[H^+] = -\log_{10}(10^{-7}) = 7$.
Watch the temperature qualifier. At $25°C$, neutral pH is $7$. At higher temperatures, $K_w$ grows (autoionisation is endothermic), so neutral pH drops below $7$ even though the water is still neutral.
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