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 will follow these steps:

### Step 1: Calculate the number of moles of CH₃COOH
Given:
- Mass of CH₃COOH = 10 g
- Molecular weight of CH₃COOH = 60 g/mol (calculated as follows: C(2) = 2×12 + H(4) = 4×1 + O(2) = 2×16 = 60 g/mol)

Using the formula for moles:
\[ \text{Moles of CH₃COOH} = \frac{\text{Mass}}{\text{Molecular weight}} = \frac{10 \text{ g}}{60 \text{ g/mol}} = \frac{1}{6} \text{ mol} \]

### Step 2: Calculate the concentration of CH₃COOH in both vessels
1. **For the vessel with 6 L of water:**
\[ \text{Concentration} (C_1) = \frac{\text{Moles}}{\text{Volume}} = \frac{\frac{1}{6} \text{ mol}}{6 \text{ L}} = \frac{1}{36} \text{ mol/L} \approx 0.0278 \text{ mol/L} \]

2. **For the vessel with 12 L of water:**
\[ \text{Concentration} (C_2) = \frac{\text{Moles}}{\text{Volume}} = \frac{\frac{1}{6} \text{ mol}}{12 \text{ L}} = \frac{1}{72} \text{ mol/L} \approx 0.0139 \text{ mol/L} \]

### Step 3: Use the dissociation constant (Kₐ) to find the degree of dissociation (α)
The dissociation of acetic acid can be represented as:
\[ \text{CH₃COOH} \rightleftharpoons \text{CH₃COO}^- + \text{H}^+ \]

At equilibrium:
- Initial concentration = C
- Change in concentration = -Cα for CH₃COOH, +Cα for CH₃COO⁻ and H⁺

The equilibrium concentrations are:
- CH₃COOH: \( C - Cα = C(1 - α) \)
- CH₃COO⁻ and H⁺: \( Cα \)

The expression for the dissociation constant (Kₐ) is:
\[ K_a = \frac{[CH₃COO^-][H^+]}{[CH₃COOH]} = \frac{(Cα)(Cα)}{C(1 - α)} = \frac{Cα^2}{C(1 - α)} = \frac{Cα^2}{1 - α} \]

Assuming α is small (1 - α ≈ 1):
\[ K_a \approx Cα^2 \]

### Step 4: Calculate α for both concentrations
1. **For the vessel with 6 L:**
\[ K_a = 1.8 \times 10^{-5}, \quad C = 0.0278 \text{ mol/L} \]
\[ α = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.8 \times 10^{-5}}{0.0278}} \approx 0.0255 \]

2. **For the vessel with 12 L:**
\[ K_a = 1.8 \times 10^{-5}, \quad C = 0.0139 \text{ mol/L} \]
\[ α = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.8 \times 10^{-5}}{0.0139}} \approx 0.0431 \]

### Step 5: Compare the degrees of dissociation
From the calculations:
- Degree of dissociation in 6 L vessel (α₁) ≈ 0.0255
- Degree of dissociation in 12 L vessel (α₂) ≈ 0.0431

### Conclusion
The degree of dissociation is greater in the vessel with 12 L of water.