An aqueous solution of ammonium carbonate is

Options

A. Weakly acidic
B. Weakly basic
C. Strongly acidic
D. Neither acidic nor basic

Explanation

To determine the nature of an aqueous solution of ammonium carbonate, we need to analyze its components and their properties. Here’s a step-by-step solution:

### Step 1: Identify the Components
Ammonium carbonate \((NH_4)_2CO_3\) dissociates in water to form ammonium ions \((NH_4^+)\) and carbonate ions \((CO_3^{2-})\).

### Step 2: Determine the Nature of the Ions
- **Ammonium ion \((NH_4^+)\)**: This is the conjugate acid of ammonia \((NH_3)\), which is a weak base. Therefore, \(NH_4^+\) can act as a weak acid.
- **Carbonate ion \((CO_3^{2-})\)**: This is the conjugate base of bicarbonate \((HCO_3^-)\) and can act as a weak base.

### Step 3: Compare the Strengths of the Acid and Base
To determine the overall nature of the solution, we need to compare the strengths of the weak acid \((NH_4^+)\) and the weak base \((CO_3^{2-})\). This can be done using their dissociation constants:
- The dissociation constant for ammonium ion (weak acid) is \(K_a\).
- The dissociation constant for carbonate ion (weak base) is \(K_b\).

### Step 4: Use Known Values
From literature, we know:
- \(K_a\) for \(NH_4^+\) is approximately \(5.6 \times 10^{-10}\).
- \(K_b\) for \(CO_3^{2-}\) can be derived from the \(K_w\) (the ion-product constant of water) and \(K_a\) of \(HCO_3^-\). The \(K_b\) for \(CO_3^{2-}\) is approximately \(2.1 \times 10^{-4}\).

### Step 5: Compare \(K_a\) and \(K_b\)
Now, we compare the two values:
- \(K_a (NH_4^+) = 5.6 \times 10^{-10}\)
- \(K_b (CO_3^{2-}) = 2.1 \times 10^{-4}\)

Since \(K_b > K_a\), the base (carbonate ion) is stronger than the acid (ammonium ion).

### Step 6: Conclusion
Since the base is stronger than the acid, the solution of ammonium carbonate will be basic.

### Final Answer
An aqueous solution of ammonium carbonate is basic.

---