Answer
$(ii) : [H^+] = [OH^-]$
Work Step by Step
Since the temperature falls, the production of $[OH^-]$ and $[H^+]$
will reduce:
$1H_2O(l) 1H^+(aq) + 1OH^-(aq)$
But, the proportion will still 1 to 1 to 1, and the concentrations of $(H^+)$ and $(OH^-)$ will still be equal.
