Sodium amide, $\mathrm{NaNH}{}_2$, is dissolved in water. The strongest base actually present in the solution is

$\mathrm{NH}{}_2{}^{-}$ is a much stronger base than $\mathrm{OH}{}^{-}$ (the pKa of $\mathrm{NH}{}_3$ is 38 and of water 15.7), so it takes a proton from water completely: $\mathrm{NH}{}_2{}^{-}+\mathrm{H}{}_2\mathrm{O}\to \mathrm{NH}{}_3+\mathrm{OH}{}^{-}$. In water no base stronger than $\mathrm{OH}{}^{-}$ can exist (levelling effect).