Calculate the percentage dissociation of 0.5 M `NH_(3)` at `25^(@)C` in a solution of pH = 12.

Answer

To calculate the percentage dissociation of 0.5 M `NH₃` at `25°C` in a solution with a pH of 12, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**:
Ammonia (`NH₃`) in water can be represented as:
\[ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \]
Here, `NH₃` dissociates into ammonium ions (`NH₄⁺`) and hydroxide ions (`OH⁻`).

2. **Determine the pH and pOH**:
Given that the pH of the solution is 12, we can calculate the pOH:
\[ pOH = 14 - pH = 14 - 12 = 2 \]

3. **Calculate Hydroxide Ion Concentration**:
The concentration of hydroxide ions can be calculated using the formula:
\[ [OH^-] = 10^{-pOH} = 10^{-2} = 0.01 \, M \]

4. **Set Up the Initial Concentration**:
Let the initial concentration of `NH₃` be `C = 0.5 M`. At equilibrium, if `α` is the degree of dissociation, the concentrations will be:
- `[NH₃] = C(1 - α) = 0.5(1 - α)`
- `[NH₄⁺] = Cα = 0.5α`
- `[OH⁻] = 0.01 M` (from the pH calculation)

5. **Equilibrium Expression**:
At equilibrium, the concentration of `OH⁻` produced from the dissociation of `NH₃` will be equal to the concentration of `OH⁻` in the solution:
\[ [OH^-] = 0.5α \]
Setting this equal to the hydroxide concentration we calculated:
\[ 0.5α = 0.01 \]

6. **Solve for α**:
Rearranging the equation gives:
\[ α = \frac{0.01}{0.5} = 0.02 \]

7. **Calculate Percentage Dissociation**:
The percentage dissociation is given by:
\[ \text{Percentage Dissociation} = α \times 100 = 0.02 \times 100 = 2\% \]

### Final Answer:
The percentage dissociation of 0.5 M `NH₃` at `25°C` in a solution of pH 12 is **2%**.

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