Question

If \(pK_{b}\) value for fluoride ion at \(25^{\circ }\)C is \(10.83\), calculate the ionisation constant of hydrofluoric acid in water at this temperature.

• A. \(1.74\times 10^{-5}\)
• B. \(3.52\times 10^{-3}\)
• C. \(6.75\times 10^{-4}\)
• D. \(5.38\times 10^{-2}\)

Hint:

For a conjugate pair, pKa plus pKb equals 14 at 25 degrees Celsius. Convert pKa to Ka.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a conjugate acid-base pair, \(pK_a + pK_b = pK_w = 14\) at \(25^{\circ}\text{C}\).

Step 2: Find \(pK_a\)
\(pK_a = 14 - 10.83 = 3.17\)

Step 3: Find \(K_a\)
\(K_a = 10^{-3.17} = 10^{-4} \times 10^{0.83} \approx 6.75\times10^{-4}\)

Final Answer:
The ionisation constant of HF is about \(6.75\times10^{-4}\). \[ \boxed{\text{(C) }6.75\times10^{-4}} \]