\( \mathrm{K}_{\mathrm{a}} \) for \( \mathrm{HF} \) is \( 3.5 \time...
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0\( \mathrm{K}_{\mathrm{a}} \) for \( \mathrm{HF} \) is \( 3.5 \times 10^{-4} \). Calculate \( \mathrm{K}_{\mathrm{b}} \) tor the fluoride
P ion.
W
(1) \( 3.5 \times 10^{-4} \)
(2) \( 1.0 \times 10^{-7} \)
(3) \( 2.9 \times 10^{-11} \)
(4) \( 1.0 \times 10^{-14} \)
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