# How can i dentify the nucleophile and the electrophile in H-Br + HO^-)hArr Br^-+H_2O acid–base reaction?

Nucleophiles tend to be negatively charged or have a lone pair of electrons. In this case it is easy to see that $O {H}^{-}$ has a negative charge as a reactant, and so it is the nucleophile. Thus, $H - B r$ is the electrophile.
The nucleophilic $O {H}^{-}$ wants a proton from $H B r$, to become water.
The $p K a$ of water is $15.7$, whereas the $p K a$ of $H B r$ is about $- 8$. Since the equilibrium lies on the side of the weaker acid, this equilibrium is skewed towards the products by about $24$ orders of magnitude.
(Remember that ${10}^{- p K a} = {K}_{a}$, thus the ${K}_{a}$'s are ${10}^{- 15.7}$ vs. ${10}^{8}$, water vs. $H B r$, and water hardly dissociates while $H B r$ does quite a bit)