Self-ionisation of liquid ammonia occurs as `2NH_(3) hArr NH_(4)^(+) +NH_(2)^(-), K = 10^(-10)` . In this solvent , an acid might be

Options

A. `NH_(4)^(+)`
B. `NH_(3)`
C. Any species that will form `NH_(4)^(+)`
D. All of these

Explanation

To determine what an acid might be in the context of the self-ionization of liquid ammonia, we can follow these steps:

### Step 1: Understand the Self-Ionization of Liquid Ammonia
The self-ionization of liquid ammonia can be represented by the equation:
\[ 2NH_3 \rightleftharpoons NH_4^+ + NH_2^- \]
This indicates that two molecules of ammonia (NH₃) can react to form one ammonium ion (NH₄⁺) and one amide ion (NH₂⁻).

### Step 2: Analyze the Equilibrium Constant (K)
The equilibrium constant (K) for this reaction is given as:
\[ K = 10^{-10} \]
This value indicates that the concentration of the products (NH₄⁺ and NH₂⁻) is much lower than that of the reactants (NH₃) at equilibrium, suggesting that in pure liquid ammonia, the self-ionization is not significant.

### Step 3: Identify the Acid in This Context
An acid is defined as a substance that can donate a proton (H⁺). In the context of liquid ammonia, the ammonium ion (NH₄⁺) can act as an acid because it can donate a proton to form ammonia (NH₃):
\[ NH_4^+ \rightleftharpoons NH_3 + H^+ \]
Thus, NH₄⁺ is a Brønsted acid in this solvent.

### Step 4: Consider Other Possible Acids
In addition to NH₄⁺, other substances that can donate protons in liquid ammonia could also be considered acids. For example, any substance that can release H⁺ ions when dissolved in ammonia could be classified as an acid.

### Conclusion
In the solvent of liquid ammonia, an acid might be the ammonium ion (NH₄⁺), as it can donate a proton to form ammonia (NH₃).

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