The `pH` of `0.5 M` aqueous solution of `HF`
`(K_(a)=2xx10^(-4))` is

Options

A. 2
B. 4
C. 6
D. 10

Answer

To find the pH of a 0.5 M aqueous solution of HF (hydrofluoric acid), we can follow these steps:

### Step 1: Write the dissociation equation
HF dissociates in water as follows:
\[ \text{HF} \rightleftharpoons \text{H}^+ + \text{F}^- \]

### Step 2: Set up the initial concentrations
Let the initial concentration of HF be \( C = 0.5 \, \text{M} \). At equilibrium, let \( \alpha \) be the degree of dissociation. The concentrations at equilibrium will be:
- \([\text{HF}] = C(1 - \alpha) = 0.5(1 - \alpha)\)
- \([\text{H}^+] = C\alpha = 0.5\alpha\)
- \([\text{F}^-] = C\alpha = 0.5\alpha\)

### Step 3: Write the expression for the acid dissociation constant \( K_a \)
The expression for the dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[\text{H}^+][\text{F}^-]}{[\text{HF}]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(0.5\alpha)(0.5\alpha)}{0.5(1 - \alpha)} \]

### Step 4: Substitute the given value of \( K_a \)
Given \( K_a = 2 \times 10^{-4} \):
\[ 2 \times 10^{-4} = \frac{(0.5\alpha)(0.5\alpha)}{0.5(1 - \alpha)} \]
This simplifies to:
\[ 2 \times 10^{-4} = \frac{0.25\alpha^2}{0.5(1 - \alpha)} \]
\[ 2 \times 10^{-4} = \frac{0.25\alpha^2}{0.5} \cdot \frac{1}{(1 - \alpha)} \]
\[ 2 \times 10^{-4} = 0.5\alpha^2 \cdot \frac{1}{(1 - \alpha)} \]

### Step 5: Assume \( \alpha \) is small
Since \( K_a \) is small, we can assume \( \alpha \) is small compared to 1, so \( 1 - \alpha \approx 1 \):
\[ 2 \times 10^{-4} \approx 0.5\alpha^2 \]
\[ \alpha^2 = \frac{2 \times 10^{-4}}{0.5} = 4 \times 10^{-4} \]
\[ \alpha = \sqrt{4 \times 10^{-4}} = 2 \times 10^{-2} \]

### Step 6: Calculate \([\text{H}^+]\)
Now, we can find the concentration of \([\text{H}^+]\):
\[ [\text{H}^+] = C\alpha = 0.5 \times 2 \times 10^{-2} = 1 \times 10^{-2} \, \text{M} \]

### Step 7: Calculate the pH
Finally, we can calculate the pH:
\[ \text{pH} = -\log[\text{H}^+] = -\log(1 \times 10^{-2}) = 2 \]

### Final Answer
The pH of the 0.5 M aqueous solution of HF is **2**.

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