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)`

Explanation

To find the ionization constant of hydrofluoric acid (HF) in water at 25°C, we can use the relationship between the pKa of an acid and the pKb of its conjugate base. Here are the steps to solve the problem:

### Step 1: Understand the relationship between pKa, pKb, and pKw.
The relationship is given by the equation:
\[ pKa + pKb = pKw \]
At 25°C, the value of \( pKw \) is 14.

### Step 2: Substitute the known values into the equation.
We know that \( pKb \) for the fluoride ion (F⁻) is given as 10.83. We can substitute this value into the equation:
\[ pKa + 10.83 = 14 \]

### Step 3: Solve for pKa.
Rearranging the equation to find \( pKa \):
\[ pKa = 14 - 10.83 \]
\[ pKa = 3.17 \]

### Step 4: Calculate the ionization constant (Ka) from pKa.
The ionization constant \( Ka \) can be calculated using the formula:
\[ Ka = 10^{-pKa} \]
Substituting the value of \( pKa \):
\[ Ka = 10^{-3.17} \]

### Step 5: Calculate the numerical value of Ka.
Using a calculator, we find:
\[ Ka \approx 6.76 \times 10^{-4} \]

### Final Answer:
The ionization constant of hydrofluoric acid (HF) in water at 25°C is approximately \( 6.76 \times 10^{-4} \).

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