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The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 x 10-4, 1.8 x 10-4 and 4.8 x 10-4 respectively. Calculate the ionization constants of the corresponding conjugate base.

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For a conjugate acid-base pair

KaKb = Kw

Acid Ka Conjugate Kb = Kw/Ka
HF 6.8 x 10-4 F- \(\frac{1\times 10^{-14}}{6.8\times 10^{-14}}\) = 1.5 x 10-11
HCOOH 1.8 x 10-4 HCOO- \(\frac{1\times 10^{-14}}{6.8\times 10^{-14}}\) = 5.6 x 10-11
HCN 4.8 x 10-4 CN- \(\frac{1\times 10^{-14}}{6.8\times 10^{-14}}\) = 2.08 x 10-11

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