To what volume of 10 litre of 0.5 M `CH_(3)COOH` `(K_(a)=1.8xx10^(-5))` be diluted in order to double the hydroxide ion concentration :
(a) 20L
(b) 30L
(c) 40L
(d) None of these

Options

A. 20 L
B. 30 L
C. 40 L
D. None of these

Answer

To solve the problem of how much volume of a 0.5 M acetic acid solution needs to be diluted to double the hydroxide ion concentration, we can follow these steps:

### Step 1: Calculate the initial concentration of H⁺ ions
We start with the dissociation constant \( K_a \) for acetic acid, which is given as \( 1.8 \times 10^{-5} \). The concentration of acetic acid \( C \) is 0.5 M.

Using the formula for the concentration of H⁺ ions:
\[ [H^+] = \sqrt{K_a \times C} \]
Substituting the known values:
\[ [H^+] = \sqrt{1.8 \times 10^{-5} \times 0.5} \]
Calculating this gives:
\[ [H^+] = \sqrt{9.0 \times 10^{-6}} \approx 3.0 \times 10^{-3} \, \text{M} \]

### Step 2: Calculate the initial concentration of OH⁻ ions
Using the relationship between \( H^+ \) and \( OH^- \) ions:
\[ K_w = [H^+][OH^-] \]
Where \( K_w = 1.0 \times 10^{-14} \) at 25°C. We can rearrange this to find \( [OH^-] \):
\[ [OH^-] = \frac{K_w}{[H^+]} \]
Substituting the calculated \( [H^+] \):
\[ [OH^-] = \frac{1.0 \times 10^{-14}}{3.0 \times 10^{-3}} \approx 3.33 \times 10^{-12} \, \text{M} \]

### Step 3: Determine the new concentration of OH⁻ ions
To double the hydroxide ion concentration:
\[ [OH^-]_{new} = 2 \times 3.33 \times 10^{-12} \approx 6.66 \times 10^{-12} \, \text{M} \]

### Step 4: Calculate the new concentration of H⁺ ions
Using the \( K_w \) relationship again:
\[ [H^+]_{new} = \frac{K_w}{[OH^-]_{new}} = \frac{1.0 \times 10^{-14}}{6.66 \times 10^{-12}} \approx 1.50 \times 10^{-3} \, \text{M} \]

### Step 5: Relate the new concentration of H⁺ ions to the new volume
Using the formula again:
\[ [H^+]_{new} = \sqrt{K_a \times C_{new}} \]
Setting this equal to our new \( [H^+] \):
\[ 1.50 \times 10^{-3} = \sqrt{1.8 \times 10^{-5} \times C_{new}} \]
Squaring both sides:
\[ (1.50 \times 10^{-3})^2 = 1.8 \times 10^{-5} \times C_{new} \]
Calculating \( (1.50 \times 10^{-3})^2 \):
\[ 2.25 \times 10^{-6} = 1.8 \times 10^{-5} \times C_{new} \]
Solving for \( C_{new} \):
\[ C_{new} = \frac{2.25 \times 10^{-6}}{1.8 \times 10^{-5}} \approx 0.125 \, \text{M} \]

### Step 6: Use the dilution equation to find the new volume
Using the dilution equation \( C_1 V_1 = C_2 V_2 \):
\[ 0.5 \, \text{M} \times 10 \, \text{L} = 0.125 \, \text{M} \times V_2 \]
Solving for \( V_2 \):
\[ V_2 = \frac{0.5 \times 10}{0.125} = 40 \, \text{L} \]

### Conclusion
The volume to which the acetic acid solution should be diluted to double the hydroxide ion concentration is **40 L**.