At $20.0° C$ water has ${\mathrm{K}}_{\mathrm{\omega }}=6.807\times {10}^{-15}$ . What is the pH of pure water at this temperature

Given,

${\mathrm{K}}_{\mathrm{\omega }}=6.807\times {10}^{-15}$ at $20.0° C$.

Value of pH of water at this temperature,

$$\begin{gathered} \mathrm{pH}=\frac{\mathrm{pk\omega }}{2} \\ \mathrm{pH}=\frac{1}{2}[15-\log 6.807] \\ \mathrm{pH}=7.084 \end{gathered}$$