$K_e$ depends on the temperature, it decreased while the temperature of the water increases. At $10^oC$, $K_e$ $ =$ $ 3.0\cdot10^{-15}$. Let's try to calculate the pH of pure water at $10^oC$ ! First: Which is the correct relation between $[H^+]$ , $[OH^-]$ and $[H_2O]$ at $10^oC$ ?
a) $[H^+]$ $=$ $[H_2O]$ b) $[H^+]$ $=$ $[OH^-]$ c) $[H^+]$ $=$ $[H_2O]$ $-$ $[OH^-]$ d) at that temperature, the relation cannot be determined !
The correct answer is: b) $[H^+]$ $=$ $[OH^-]$ because when one molecule of water dissociates, one ion $H^+$ and one ion $OH^-$ are produced.
Second: Calculate the $pH$ at $10^oC$ !
$[H^+][OH^-]$ $=$ $3.0\cdot 10^{-15}$ $[H^+]^2$ $=$ $3.0\cdot 10^{-15}$ $pH$ $=$ $-log([H^+])$ $=$ $-\frac{1}{2}log(3.0\cdot 10^{-15})$ $=$ $7.26$