$K_e$ is not independent of temperature, but rather decreases when the temperature of the water increases. At $10^oC$, $K_e$ $ = $ $3.0\cdot 10^{-15}$. Let us try to calculate the $ pH $ of pure water at $10^oC$! First: What is the relationship between $[H^+]$ , $[OH^-]$ and $[H_2O]$ ?
a) $[H^+]$ $=$ $[H_2O]$ b) $[H^+]$ $=$ $[OH^-]$ c) $[H^+]$ $=$ $[H_2O]$ $-$ $[OH^-]$ d) The relationship can not be determined at this temperature.
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The correct answer ist: b) $[H^+]$ = $[OH^-]$, because every time a molecule of water dissociates, one ion $H^+$ and one ion $OH^-$ are formed.
Then: Calculate the $pH$ at $10^oC$ !
$[H^+][OH^-]$ $=$ $3,0\cdot 10^{-15}$ $[H^+]^2$ $=$ $3,0\cdot 10^{-15}$ $pH$ $=$ $-log([H^+])$ $=$ $-\frac{1}{2}log(3,0\cdot 10^{-15})$ $=$ $7,26$
