The pKa of hydrofluoric acid, HF is 3.17. The pKb of sodium fluoride, NaF, is
The correct answer and explanation is:
The pKa of hydrofluoric acid (HF) is 3.17. Sodium fluoride (NaF) is a salt derived from the weak acid HF and the strong base NaOH. When NaF dissolves in water, it dissociates into Na⁺ and F⁻ ions. The fluoride ion (F⁻) is the conjugate base of HF.
The relationship between pKa and pKb for a conjugate acid-base pair is given by:
pKa+pKb=14pK_a + pK_b = 14
This equation assumes the solution is at 25°C, where the ion product of water (Kw) is 1×10−141 \times 10^{-14}.
Given:
- pKapK_a of HF = 3.17
To find the pKb of the fluoride ion (F⁻), use:
pKb=14−pKapK_b = 14 – pK_a
Substitute the given value:
pKb=14−3.17=10.83pK_b = 14 – 3.17 = 10.83
Therefore, the pKb of the fluoride ion (from NaF) is 10.83.
Explanation
Hydrofluoric acid (HF) is a weak acid with a pKa of 3.17. This means it only partially dissociates in water to release H⁺ ions. The fluoride ion (F⁻) is its conjugate base, formed when HF loses a proton. Since HF is weak, its conjugate base F⁻ is relatively strong, but not a strong base; it can accept protons but does so weakly.
pKa and pKb measure the strength of acids and bases. The lower the pKa, the stronger the acid; the lower the pKb, the stronger the base. Because HF is a weak acid, its conjugate base has a high pKb value, indicating weak basicity.
The sum of pKa and pKb equals 14 at 25°C, reflecting the ionization balance in water. This allows one to calculate the pKb of a conjugate base if the pKa of its acid is known.
Knowing the pKb of fluoride ion helps predict the behavior of NaF solutions. Because the fluoride ion is a weak base, NaF solutions tend to be slightly basic in water, affecting pH and equilibrium in buffer solutions involving HF/NaF pairs. This concept is essential for understanding buffer chemistry and acid-base equilibria in aqueous systems.