If 0.5 M methanol undergo selft dissociation like `CH_(3) OH hArr CH_(3) O^(-) + H^(+)` & if concentration of `H^(+)` is `2.5 xx 10^(-4) M` then calculate % dissociation of methanol

Answer

To solve the problem of calculating the percent dissociation of methanol, we can follow these steps:

### Step 1: Write the dissociation equation
The self-dissociation of methanol can be represented as:
\[ \text{CH}_3\text{OH} \rightleftharpoons \text{CH}_3\text{O}^- + \text{H}^+ \]

### Step 2: Set up the initial concentrations
Initially, we have:
- Concentration of CH₃OH = 0.5 M
- Concentration of CH₃O⁻ = 0 M
- Concentration of H⁺ = 0 M

### Step 3: Define the change in concentration
Let \( \alpha \) be the degree of dissociation. At equilibrium, the concentrations will be:
- CH₃OH = \( 0.5 - \alpha \)
- CH₃O⁻ = \( \alpha \)
- H⁺ = \( \alpha \)

### Step 4: Use the given concentration of H⁺
We are given that the concentration of H⁺ is \( 2.5 \times 10^{-4} \) M. Therefore,
\[ \alpha = 2.5 \times 10^{-4} \]

### Step 5: Express the degree of dissociation in terms of initial concentration
From the initial concentration of methanol (0.5 M), we can express \( \alpha \):
\[ \alpha = \frac{[\text{H}^+]}{[\text{CH}_3\text{OH}]_{\text{initial}}} \]

### Step 6: Substitute the values
Substituting the known values:
\[ \alpha = \frac{2.5 \times 10^{-4}}{0.5} \]

### Step 7: Calculate \( \alpha \)
Calculating \( \alpha \):
\[ \alpha = 5 \times 10^{-4} \]

### Step 8: Calculate percent dissociation
Percent dissociation is calculated as:
\[ \text{Percent dissociation} = \alpha \times 100 \]
Substituting the value of \( \alpha \):
\[ \text{Percent dissociation} = 5 \times 10^{-4} \times 100 = 0.05\% \]

### Final Answer
The percent dissociation of methanol is **0.05%**.

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