1.

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.

Answer»

For a conjugate acid-base pair

KaKb = Kw

AcidKaConjugateKb = Kw/Ka
HF6.8 x 10-4F-\(\frac{1\times 10^{-14}}{6.8\times 10^{-14}}\) = 1.5 x 10-11
HCOOH1.8 x 10-4HCOO-\(\frac{1\times 10^{-14}}{6.8\times 10^{-14}}\) = 5.6 x 10-11
HCN4.8 x 10-4CN-\(\frac{1\times 10^{-14}}{6.8\times 10^{-14}}\) = 2.08 x 10-11


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