If for fluoride ion at 25°C is 10.3, the ionization constant of hydrofluoric acid in water at this t — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

For a conjugate acid-base pair at 25°C, $pK_a + pK_b = 14$. Therefore, $pK_a(\text{HF}) = 14 - pK_b(\text{F}^-) = 14 - 10.3 = 3.7$. The ionization constant is $K_a = 10^{-3.7} = 10^{0.3} \times 10^{-4}$. Since $\log 2 = 0.3$, $10^{0.3} = 2$. Thus, $K_a = 2 \times 10^{-4}$.