How much water must be added to `300 mL` of a `0.2M` solution of `CH_(3)COOH` for the degree of dissociation of the acid to double ? ( Assume `K_(a)` of acetic is of order of `10^(-5)M)`

Options

A. `900 ml`
B. `300ml`
C. `600 ml`
D. `1200ml`

Explanation

To solve the problem of how much water must be added to 300 mL of a 0.2 M solution of acetic acid (CH₃COOH) to double its degree of dissociation, we can follow these steps:

### Step 1: Understand the Initial Conditions
We start with a 0.2 M solution of acetic acid. The degree of dissociation (α) is the fraction of the acid that dissociates into ions. At equilibrium, the dissociation of acetic acid can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Write the Expression for Ka
The dissociation constant \( K_a \) for acetic acid is given as \( 10^{-5} \) M. The expression for \( K_a \) is:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]

### Step 3: Calculate Initial Degree of Dissociation
Let the initial degree of dissociation be \( \alpha \). At equilibrium, the concentrations will be:
- \([\text{CH}_3\text{COOH}] = 0.2(1 - \alpha)\)
- \([\text{CH}_3\text{COO}^-] = [\text{H}^+] = 0.2\alpha\)

Substituting these into the \( K_a \) expression gives:
\[ K_a = \frac{(0.2\alpha)(0.2\alpha)}{0.2(1 - \alpha)} \]
\[ K_a = \frac{0.04\alpha^2}{0.2(1 - \alpha)} \]
Assuming \( \alpha \) is small, we can neglect \( (1 - \alpha) \) and simplify:
\[ K_a \approx \frac{0.04\alpha^2}{0.2} \]
\[ K_a \approx 0.2\alpha^2 \]
\[ \alpha^2 = \frac{K_a}{0.2} \]
\[ \alpha = \sqrt{\frac{K_a}{0.2}} = \sqrt{\frac{10^{-5}}{0.2}} \]

### Step 4: Calculate the New Degree of Dissociation
To double the degree of dissociation, we need \( 2\alpha \). The new concentration \( C' \) after dilution will be:
\[ C' = \frac{K_a}{(2\alpha)^2} \]
Using the previous relation, we can find:
\[ C' = \frac{K_a}{4\alpha^2} \]
Since \( \alpha^2 = \frac{K_a}{0.2} \):
\[ C' = \frac{K_a}{4 \cdot \frac{K_a}{0.2}} = \frac{0.2}{4} = 0.05 \text{ M} \]

### Step 5: Use the Dilution Formula
Using the dilution formula \( C_1V_1 = C_2V_2 \):
- \( C_1 = 0.2 \) M (initial concentration)
- \( V_1 = 300 \) mL (initial volume)
- \( C_2 = 0.05 \) M (final concentration)

We need to find \( V_2 \):
\[ 0.2 \times 300 = 0.05 \times V_2 \]
\[ V_2 = \frac{0.2 \times 300}{0.05} = 1200 \text{ mL} \]

### Step 6: Calculate the Amount of Water to be Added
The volume of water to be added is:
\[ \text{Volume of water} = V_2 - V_1 = 1200 \text{ mL} - 300 \text{ mL} = 900 \text{ mL} \]

### Final Answer
Thus, **900 mL** of water must be added to double the degree of dissociation of the acetic acid solution.