According to Bronsted Lowry concept, in given reaction, water will behave as :
`CH_3COOH+H_2O hArr CH_3COO^(ᶱ)+H_3O^(o+)`

Options

A. weak acid
B. weak base
C. strong acid
D. strong base

Answer

To determine how water behaves in the given reaction according to the Bronsted-Lowry concept, we can analyze the reaction step by step.

### Step 1: Identify the reactants and products
The reaction provided is:
\[ \text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+ \]

Here, we have acetic acid (CH₃COOH) and water (H₂O) as reactants, and acetate ion (CH₃COO⁻) and hydronium ion (H₃O⁺) as products.

### Step 2: Identify the acid and base
According to the Bronsted-Lowry theory:
- An acid is a substance that donates a proton (H⁺).
- A base is a substance that accepts a proton (H⁺).

In this reaction:
- Acetic acid (CH₃COOH) donates a proton to water (H₂O).
- Water (H₂O) accepts the proton and is converted into hydronium ion (H₃O⁺).

### Step 3: Determine the role of water
Since water is accepting a proton (H⁺) from acetic acid, it behaves as a base in this reaction.

### Conclusion
Thus, according to the Bronsted-Lowry concept, in the reaction:
\[ \text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+ \]
water (H₂O) behaves as a **Bronsted-Lowry base**.