Question

$ {{K}_{a}} $ for HF is $ 3.5\times {{10}^{-7}} $ calculate $ {{K}_{b}} $ for the fluoride ion.

• A. $ 3.5\times {{10}^{-4}} $
• B. $ 1.0\times {{10}^{-7}} $
• C. $ 2.85\times {{10}^{-11}} $
• D. $ 1.0\times {{10}^{-14}} $

Answer 1

Answer: (C)

For a conjugate acid-base pair: $ {{K}_{a}}\cdot {{K}_{b}}={{K}_{w}} $ $ {{K}_{b}}=\frac{{{K}_{w}}}{{{K}_{a}}}=\frac{{{10}^{-14}}}{3.5\times {{10}^{-4}}}=2.85\times {{10}^{-11}} $