The $\mathrm{pH}$ of pure water at $80°\mathrm{C}$ is

At  ${25}^{°}C$, the value of the ionic product of water is${10}^{-14}$. The ionic product of water is defined as 

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right] \\ \Rightarrow {10}^{-14}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right] \\ \mathrm{For} \mathrm{pure} \mathrm{water} \left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right] \\ \mathrm{So} \left[{\mathrm{H}}^{+}\right]={10}^{-7} \\ \mathrm{Thus} \mathrm{the} \mathrm{pH}=7 \end{gathered}$$

Since the degree of ionization is proportional to temperature, the value of the ionic product of water increases with increase in temperature. So, the ionic product will increase at a temperature more than $25°C$
Therefore, at $80°\mathrm{C}$, the ionic product will increase and so the hydrogen ion concentration will increase and thus, the pH will become less than 7.