At $25°C$, the ionic product of water $K_w = [H^+][OH^-] = 10^{-14}$. What is the pH of pure water at $25°C$?

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.