In the following reaction `HCO_3^(-)+H_2OhArrCO_3^(2-)+H_3O^+` which two substances are Bronsted base ?

Options

A. `CO_3^(2-) and H_3O^+`
B. `HCO_3^- and H_3O^+`
C. `HCO_3^- and CO_3^(2-)`
D. `CO_3^(2-) and H_2O`

Answer

To determine which two substances are Bronsted bases in the reaction:

\[ \text{HCO}_3^{-} + \text{H}_2\text{O} \rightleftharpoons \text{CO}_3^{2-} + \text{H}_3\text{O}^{+} \]

we can follow these steps:

### Step 1: Understand the Bronsted-Lowry Theory
A Bronsted base is defined as a substance that can accept protons (H⁺ ions).

**Hint:** Recall the definition of Bronsted bases and acids.

### Step 2: Identify the Reactants and Products
In this reaction, we have:
- Reactants: HCO₃⁻ (bicarbonate) and H₂O (water)
- Products: CO₃²⁻ (carbonate) and H₃O⁺ (hydronium ion)

**Hint:** Write down the reactants and products clearly to visualize the reaction.

### Step 3: Analyze the Role of Each Substance
- **H₂O (water)**: In the product side, H₂O donates a proton (H⁺) and becomes H₃O⁺. This means H₂O is acting as an acid in this reaction, but it can also act as a base in other contexts.
- **HCO₃⁻ (bicarbonate)**: It can accept a proton to form H₂CO₃ (carbonic acid), but in this reaction, it is acting as an acid because it donates a proton to form CO₃²⁻.
- **CO₃²⁻ (carbonate)**: This ion can accept a proton to form HCO₃⁻. Therefore, it acts as a Bronsted base in this reaction.
- **H₃O⁺ (hydronium ion)**: This is a proton donor and acts as a Bronsted acid.

**Hint:** Determine which substances are gaining protons and which are losing them.

### Step 4: Identify the Bronsted Bases
From the analysis:
- **CO₃²⁻** accepts a proton to become HCO₃⁻, making it a Bronsted base.
- **H₂O** can also accept a proton to become H₃O⁺, thus it is also a Bronsted base.

### Conclusion
The two substances that are Bronsted bases in the given reaction are:
1. **CO₃²⁻ (carbonate ion)**
2. **H₂O (water)**

**Final Answer:** CO₃²⁻ and H₂O are the Bronsted bases in the reaction.

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