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

Options

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

Explanation

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 for HF
HF is a weak acid and partially dissociates in water:
\[ \text{HF} \rightleftharpoons \text{H}^+ + \text{F}^- \]

### Step 2: Set up the expression for the acid dissociation constant (Ka)
The acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[\text{H}^+][\text{F}^-]}{[\text{HF}]} \]
Given that \( K_a = 2 \times 10^{-4} \).

### Step 3: Define initial concentrations and changes
Let the initial concentration of HF be \( C = 0.5 \, \text{M} \). At equilibrium, if \( x \) is the amount that dissociates:
- \([\text{HF}] = C - x = 0.5 - x\)
- \([\text{H}^+] = x\)
- \([\text{F}^-] = x\)

### Step 4: Substitute into the \( K_a \) expression
Substituting the equilibrium concentrations into the \( K_a \) expression gives:
\[ K_a = \frac{x \cdot x}{0.5 - x} = \frac{x^2}{0.5 - x} \]

### Step 5: Assume \( x \) is small compared to \( C \)
Since HF is a weak acid, we can assume that \( x \) is small compared to 0.5 M. Therefore, \( 0.5 - x \approx 0.5 \):
\[ K_a = \frac{x^2}{0.5} \]

### Step 6: Solve for \( x \)
Rearranging gives:
\[ x^2 = K_a \cdot 0.5 \]
\[ x^2 = (2 \times 10^{-4}) \cdot 0.5 \]
\[ x^2 = 1 \times 10^{-4} \]
\[ x = \sqrt{1 \times 10^{-4}} = 1 \times 10^{-2} \, \text{M} \]

### Step 7: Calculate pH
The concentration of \( \text{H}^+ \) ions is \( x \):
\[ [\text{H}^+] = 1 \times 10^{-2} \, \text{M} \]
Now, 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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