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 1: Understand the dissociation of NH₃
Ammonia (NH₃) in water can be represented as:
\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]

### Step 2: Set up the initial concentration
The initial concentration of NH₃ is given as 0.5 M. At equilibrium, if α is the degree of dissociation, the concentrations will be:
- [NH₃] = 0.5 - 0.5α
- [NH₄⁺] = 0.5α
- [OH⁻] = 0.5α

### Step 3: Calculate pOH from pH
We know that:
\[ \text{pH} + \text{pOH} = 14 \]
Given that pH = 12, we can find pOH:
\[ \text{pOH} = 14 - 12 = 2 \]

### Step 4: Calculate the concentration of OH⁻
Using the pOH, we can find the concentration of hydroxide ions:
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2} \, \text{M} = 0.01 \, \text{M} \]

### Step 5: Relate OH⁻ concentration to α
From the dissociation equation, we have:
\[ [\text{OH}^-] = 0.5α \]
Setting this equal to the concentration we calculated:
\[ 0.5α = 0.01 \]
Now, solve for α:
\[ α = \frac{0.01}{0.5} = 0.02 \]

### Step 6: Calculate percentage dissociation
To find the percentage dissociation, we multiply α by 100:
\[ \text{Percentage dissociation} = α \times 100 = 0.02 \times 100 = 2\% \]

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

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