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: $K_a\cdot K_b=K_w$ $K_b=\frac{K_w}{K_a}=\frac{{10}^{-14}}{3.5\times {10}^{-4}}=2.85\times {10}^{-11}$