At 80^ C , distilled water has H^ + ions concentration equal to 1 ×...

\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=1 \times 10^{-6} \mathrm{~mol} \mathrm{~L} \mathrm{~L}^{-1}$or$\left[\mathrm{H}^{+}\right]=1 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$As for distilled water$\left[\mathrm{H}^{+}\right]=$\left[\mathrm{OH}^{-}\right]$ \begin{aligned} & {\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]=1 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}} \\ & \mathrm{~K}_w=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-12} \mathrm{~mol}^2 \mathrm{~L}^{-2} \end{aligned}