Two samples of `CH_3COOH` each of 10 g were taken separately in two vessels containing water of 6 litre and 12 litre respectively at `27^@C` . The degree of dissociation of `CH_3COOH` will be

Options

A. More in 12 litre vessel
B. More in 6 litre vessel
C. Equal in both vessels
D. Half in 6 litre vessel than in 12 litre vessel

Answer

To solve the problem of determining the degree of dissociation of acetic acid (CH₃COOH) in two different volumes of water, we can follow these steps:

### Step 1: Understand the dissociation of acetic acid
Acetic acid (CH₃COOH) is a weak acid that dissociates in water according to the following equilibrium reaction:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Define the degree of dissociation
Let α be the degree of dissociation of acetic acid. At equilibrium, if we start with an initial concentration \( C \) of acetic acid, the concentrations at equilibrium will be:
- Concentration of CH₃COOH: \( C(1 - \alpha) \)
- Concentration of CH₃COO⁻: \( C\alpha \)
- Concentration of H⁺: \( C\alpha \)

### Step 3: Write the expression for the dissociation constant (Kₐ)
The dissociation constant \( K_a \) for acetic acid can be expressed as:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} \]

### Step 4: Simplify the expression
Assuming \( \alpha \) is small (which is typical for weak acids), we can approximate \( 1 - \alpha \approx 1 \):
\[ K_a \approx \frac{C\alpha^2}{C} = C\alpha^2 \]

### Step 5: Solve for α
Rearranging gives:
\[ \alpha = \sqrt{\frac{K_a}{C}} \]

### Step 6: Analyze the concentration in both vessels
We have two vessels:
1. **Vessel 1**: 10 g of CH₃COOH in 6 L of water
2. **Vessel 2**: 10 g of CH₃COOH in 12 L of water

The number of moles of CH₃COOH in both vessels is the same:
\[ \text{Moles of CH}_3\text{COOH} = \frac{10 \text{ g}}{60 \text{ g/mol}} = \frac{1}{6} \text{ mol} \]

Now, we calculate the concentration \( C \) in each vessel:
- For **Vessel 1** (6 L):
\[ C_1 = \frac{\frac{1}{6} \text{ mol}}{6 \text{ L}} = \frac{1}{36} \text{ mol/L} \]

- For **Vessel 2** (12 L):
\[ C_2 = \frac{\frac{1}{6} \text{ mol}}{12 \text{ L}} = \frac{1}{72} \text{ mol/L} \]

### Step 7: Compare the concentrations
Since \( C_1 > C_2 \), we can see that:
- Vessel 1 has a higher concentration of acetic acid than Vessel 2.

### Step 8: Determine the degree of dissociation
From the derived formula \( \alpha = \sqrt{\frac{K_a}{C}} \), we can conclude:
- The degree of dissociation \( \alpha \) is inversely proportional to the square root of the concentration \( C \). Therefore:
\[ \alpha_1 \propto \frac{1}{\sqrt{C_1}} \]
\[ \alpha_2 \propto \frac{1}{\sqrt{C_2}} \]

Since \( C_1 > C_2 \):
- \( \alpha_1 < \alpha_2 \)

### Conclusion
The degree of dissociation of acetic acid will be higher in the vessel with 12 L of water compared to the vessel with 6 L of water.

### Final Answer
The degree of dissociation of CH₃COOH will be more in the 12 L vessel.

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