If `pK_b` for fluoride ion at ` 25 ^(@) C ` is 10.83 the ionisation constant of hydrofluoric acid in water at this temperature is

Options

A. ` 1.74 xx 10 ^(-5)`
B. ` 3.52 xx 10 ^(-3)`
C. ` 6. 75 xx 10 ^(-4)`
D. ` 5.38 xx 10 ^(-2)`

Answer

To find the ionization constant of hydrofluoric acid (HF) in water at 25 °C, given that the pK_b for the fluoride ion (F⁻) is 10.83, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between K_a, K_b, and K_w**:
The ionization constant of hydrofluoric acid (K_a) and the base ionization constant of fluoride ion (K_b) are related through the ion product of water (K_w). The equation is:
\[ K_w = K_a \times K_b \]
At 25 °C, \( K_w = 1.0 \times 10^{-14} \).

2. **Convert pK_b to K_b**:
We are given pK_b = 10.83. To find K_b, we use the relationship:
\[ K_b = 10^{-pK_b} = 10^{-10.83} \]
Calculating this gives:
\[ K_b \approx 1.47 \times 10^{-11} \]

3. **Calculate K_a using K_w and K_b**:
Now we can rearrange the first equation to find K_a:
\[ K_a = \frac{K_w}{K_b} \]
Substituting the known values:
\[ K_a = \frac{1.0 \times 10^{-14}}{1.47 \times 10^{-11}} \approx 6.80 \times 10^{-4} \]

4. **Convert K_a to pK_a**:
Finally, we can find pK_a using:
\[ pK_a = -\log(K_a) \]
Calculating this gives:
\[ pK_a \approx -\log(6.80 \times 10^{-4}) \approx 3.17 \]

### Final Answer:
The ionization constant (K_a) of hydrofluoric acid (HF) in water at 25 °C is approximately:
\[ K_a \approx 6.80 \times 10^{-4} \]