To solve the problem of how much water must be added to 300 mL of a 0.2 M solution of acetic acid (CH₃COOH) for the degree of dissociation of the acid to double, we will follow these steps:
### Step 1: Understand the Degree of Dissociation
The degree of dissociation (α) is defined as the fraction of the acid that dissociates into ions. If we denote the initial degree of dissociation as α₁, we want to find the concentration of acetic acid when the degree of dissociation becomes double, i.e., α₂ = 2α₁.
### Step 2: Use the Dissociation Constant
For a weak acid like acetic acid, the dissociation constant \( K_a \) is given by:
\[
K_a = \frac{[CH₃COO^-][H^+]}{[CH₃COOH]}
\]
Assuming the initial concentration of acetic acid is \( C \), at equilibrium:
- The concentration of dissociated acetic acid will be \( Cα \).
- The concentration of undissociated acetic acid will be \( C(1 - α) \).
Thus, we can express \( K_a \) as:
\[
K_a = \frac{(Cα)(Cα)}{C(1 - α)} = \frac{Cα^2}{1 - α}
\]
### Step 3: Simplify the Expression
Since \( K_a \) is very small (around \( 10^{-5} \)), we can assume \( α \) is small compared to 1, allowing us to neglect \( 1 - α \) in the denominator:
\[
K_a \approx Cα^2
\]
This gives us:
\[
α = \sqrt{\frac{K_a}{C}}
\]
### Step 4: Determine the Initial and Final Concentrations
Let:
- \( C_1 = 0.2 \, M \) (initial concentration)
- \( α_1 = \sqrt{\frac{K_a}{C_1}} \)
Now, for the final concentration when the degree of dissociation doubles:
\[
α_2 = 2α_1
\]
Using the same formula for \( C_2 \):
\[
α_2 = \sqrt{\frac{K_a}{C_2}}
\]
### Step 5: Set Up the Ratio
From the relationship between the degrees of dissociation:
\[
\frac{α_2}{α_1} = \frac{2α_1}{α_1} = 2 \implies \frac{C_1}{C_2} = 4
\]
Thus:
\[
C_2 = \frac{C_1}{4} = \frac{0.2}{4} = 0.05 \, M
\]
### Step 6: Use the Dilution Equation
Using the dilution equation:
\[
C_1V_1 = C_2V_2
\]
Where:
- \( V_1 = 300 \, mL \)
- \( C_1 = 0.2 \, M \)
- \( C_2 = 0.05 \, M \)
We can find \( V_2 \):
\[
0.2 \times 300 = 0.05 \times V_2
\]
\[
V_2 = \frac{0.2 \times 300}{0.05} = 1200 \, mL
\]
### Step 7: Calculate the Amount of Water to be Added
To find the volume of water to be added:
\[
\text{Volume of water} = V_2 - V_1 = 1200 \, mL - 300 \, mL = 900 \, mL
\]
### Final Answer
Thus, the amount of water that must be added is **900 mL**.
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