# Ionic product of water at 310 K is 2.7 x 10^-14. What is the pH of neutral water at this temperature?

Because acid base dissociation is a bond breaking reaction, and you have increased the temperature (thus favoring bond-breaking), $p H$ should be less than $7$.
${H}_{2} O \rightarrow {H}^{+} + O {H}^{-}$; ${K}_{w}$ $=$ $\left[{H}^{+}\right] \left[O {H}^{-}\right]$ $=$ $2.7 \times {10}^{- 14}$ at $310$ $K$ (this is almost 3 times the value at $298$ $K$).
By definition $p H = - {\log}_{10} \left[{H}^{+}\right]$ $=$ $- {\log}_{10} \left\{\sqrt{2.7 \times {10}^{- 14}}\right\}$. $p H$ should be less than $7$, because the higher temperature facilitates the breaking of the $H - O H$ bond.