Question

Identify \(Base_2\) for the following equation according to Brønsted-Lowry theory:
\( HCl(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + Cl^-(aq) \)

• A. \( H_3O^+(aq) \)
• B. \( H_2O(l) \)
• C. \( Cl^-(aq) \)
• D. \( HCl(aq) \)

Hint:

In Brønsted-Lowry reactions:
• Species gaining \(H^+\) acts as a base.
• Species losing \(H^+\) acts as an acid. A quick trick: \[ \text{Gain } H^+ \Rightarrow \text{Base} \] \[ \text{Lose } H^+ \Rightarrow \text{Acid} \] Water can behave both as acid and base depending on the reaction, making it an amphoteric substance.

Answer 1

The Correct Option is B

Concept: According to the Brønsted-Lowry acid-base theory:
• A Brønsted-Lowry acid is a species that donates a proton \((H^+)\).
• A Brønsted-Lowry base is a species that accepts a proton \((H^+)\). Whenever an acid donates a proton, another species must accept that proton. Thus, acid-base reactions always involve proton transfer. The general representation is: \[ \text{Acid}_1 + \text{Base}_2 \rightleftharpoons \text{Acid}_2 + \text{Base}_1 \] where:
• \(Acid_1\) donates proton.
• \(Base_2\) accepts proton.
• \(Acid_2\) is the conjugate acid formed after proton acceptance.
• \(Base_1\) is the conjugate base formed after proton donation.

Step 1: Writing the given reaction clearly.}
The given reaction is: \[ HCl(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + Cl^-(aq) \] We now analyze the movement of the proton \((H^+)\).

Step 2: Identifying the proton donor.}
In the reaction, \(HCl\) changes into \(Cl^-\). \[ HCl \rightarrow Cl^- \] This transformation is possible only if \(HCl\) loses a proton \((H^+)\). Thus: \[ HCl \text{ donates } H^+ \] Therefore, \(HCl\) behaves as a Brønsted-Lowry acid. Hence, \[ HCl = Acid_1 \]

Step 3: Identifying the proton acceptor.}
Water molecule changes into hydronium ion: \[ H_2O \rightarrow H_3O^+ \] This occurs because water accepts the proton released by \(HCl\). \[ H_2O + H^+ \rightarrow H_3O^+ \] Since \(H_2O\) accepts a proton, it behaves as a Brønsted-Lowry base. Thus: \[ H_2O = Base_2 \]

Step 4: Identifying conjugate acid-base pairs.}
The reaction can now be classified as: \[ \underbrace{HCl}_{Acid_1} + \underbrace{H_2O}_{Base_2} \rightleftharpoons \underbrace{H_3O^+}_{Acid_2} + \underbrace{Cl^-}_{Base_1} \]
• \(HCl\) loses proton and forms \(Cl^-\), therefore \(Cl^-\) is the conjugate base.
• \(H_2O\) gains proton and forms \(H_3O^+\), therefore \(H_3O^+\) is the conjugate acid.

Step 5: Evaluating the options.}

• \(H_3O^+\): This is the conjugate acid formed after proton acceptance, not \(Base_2\).
• \(H_2O\): Correct, because it accepts proton from \(HCl\).
• \(Cl^-\): This is the conjugate base formed after proton donation by \(HCl\).
• \(HCl\): This acts as acid because it donates proton. Therefore, the correct answer is: \[ \boxed{H_2O(l)} \]