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The ionization constants of HF, HCOOH and HCN at 298 K are 6.8 × 10-4 , 1.8 × 10-4 and 4.8 × 10-9 respectively. Calculate the ionization constants of the corresponding conjugate base. |
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Answer» The relation between ionization constant of an acid and that of its conjugate base is Ka x Kb= Kw Ka x Kb = Kw ∴ Kb = \(\frac{K_w}{K_a}\) The conjugate base of HF is F- , ∴ Kb(F-) = \(\frac{K_w}{K_{a(HF)}}\) = \(\frac{1\times10^{-14}}{6.8\times10^{-4}}\) = 1.5 x 10-11 The conjugate base of HCOOH is HCOO-. ∴ Kb(HCOO-) = \(\frac{K_w}{K_{a(HCOOH)}}\) = \(\frac{1\times10^{-14}}{1.8\times10^{-4}}\) = 5.6 x 10-11 The conjugate base of HCN is CN-. ∴ Kb(CN-) = \(\frac{K_w}{K_{a(HCN)}}\) = \(\frac{1\times10^{-14}}{4.8\times10^{-9}}\) = 2.1 x 10-9 |
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