NH3(aq) + HCl(aq) → NH4+(aq) + Cl-(aq) The Brønsted-Lowry bases in the reaction represented above ar — Acids and Bases Chemistry Question

STEPS:

1. Define the Brønsted^-Lowry Concept: Recall that according to the Brønsted^-Lowry theory, an acid is a proton ($H^+$) donor and a base is a proton ($H^+$) acceptor.
2. Analyze the Forward Reaction: Look at the reactants and determine which species is accepting a proton.
* $NH_3(aq)$ becomes $NH_4^+(aq)$. To do this, it must accept a proton ($H^+$) from $HCl$. Therefore, $NH_3$ is a Brønsted^-Lowry base.
3. Identify the Acid and its Conjugate:
* $HCl(aq)$ becomes $Cl^-(aq)$ by donating a proton to $NH_3$. This makes $HCl$ the Brønsted^-Lowry acid.
* The species that remains after an acid has donated its proton is called the conjugate base. In this reaction, $Cl^-$ is the conjugate base of $HCl$.
4. Analyze the Reverse Reaction: Consider what would happen if the reaction proceeded in the opposite direction.
* To reform the reactants, $Cl^-(aq)$ would have to accept a proton from $NH_4^+(aq)$ to become $HCl$. Since $Cl^-$ acts as a proton acceptor in the reverse process, it is also classified as a Brønsted^-Lowry base.
5. Conclusion: The two species in this equilibrium system that act as proton acceptors (bases) are $NH_3(aq)$ and $Cl^-(aq)$, which corresponds to option B.

WHY_OTHERS_WRONG:

• $NH_4^+(aq)$ (Found in A and D): This is the conjugate acid of $NH_3$. In the reverse reaction, it would donate a proton to $Cl^-$ to reform $NH_3$; therefore, it is an acid, not a base.
• $HCl(aq)$ (Found in C, D, and E): $HCl$ is a strong acid that dissociates to donate a proton to the base ($NH_3$). By definition, a proton donor cannot be a Brønsted^-Lowry base in the same reaction.