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

Answer 1
Nucleophiles tend to be negatively charged or have a lone pair of electrons. In this case it is easy to see that #OH^(-)# has a negative charge as a reactant, and so it is the nucleophile. Thus, #H-Br# is the electrophile.
The nucleophilic #OH^(-)# wants a proton from #HBr#, to become water.
The #pKa# of water is #15.7#, whereas the #pKa# of #HBr# 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^(-pKa) = K_a#, thus the #K_a#'s are #10^(-15.7)# vs. #10^8#, water vs. #HBr#, and water hardly dissociates while #HBr# does quite a bit)
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Answer 2

In the acid-base reaction between H-Br and HO^-, the nucleophile is HO^- and the electrophile is H in H-Br.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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