If an aqueous solution at 25° Chas twice as many ${\mathrm{OH}}^{-}$as pure water its $\mathrm{pOH}$ will be :

$\left[{\mathrm{OH}}^{-}\right]$ for pure water at ${25}^{\circ }\mathrm{C}={10}^{-7}$

$\left[{\mathrm{OH}}^{-}\right]$ for the aqueous at  ${25}^{\circ }\mathrm{C}=2\times {10}^{-7}$

Hence, $\mathrm{pOH}-\log \left[{\mathrm{OH}}^{-}\right]=-\log \left(2\times {10}^{-7}\right)=7-\log 2=6.698$