If an aqueous solution at ${25}^0\mathrm{c}$  has twice as many ${\mathrm{OH}}^{-}$ as pure water ,its ${\mathrm{p}}^{\mathrm{OH}}$ will be

 $\left[{\mathrm{OH}}^{-}\right] \mathrm{of} \mathrm{pure} \mathrm{water} = 1\times 1{\mathrm{O}}^{-7}$

$\left[{\mathrm{OH}}^{-}\right] \mathrm{given} \mathrm{solution} = 2 \times {10}^{-7}$

$$\begin{gathered} {\mathrm{p}}^{\mathrm{OH}}=-\log \left[{\mathrm{OH}}^{-}\right]=-\log \left[2\times {10}^{-7}\right] \\ =7-\log 2=7-0.3010=6.990 \end{gathered}$$