7.43. The ionization constant of HF, HCOOH and HCN at 298 K are is 6.8 x 10-4, 1.8x 10-4 and 4.8 x 10-9 respectively, calculate the ionization constant of the corresponding conjugate base.
7.43. The ionization constant of HF, HCOOH and HCN at 298 K are is 6.8 x 10-4, 1.8x 10-4 and 4.8 x 10-9 respectively, calculate the ionization constant of the corresponding conjugate base.
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1 Answer
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For F–, Kb =Kw/Ka= 10-14/ (6.8 x 10-4)
= 1.47 x 10-11 = 1.5 x 10-11 .
For HCOO-, Kb = 10-14/ (1.8 x 10-4)
= 5.6 x 10-11
For CN–, Kb= 10-14/ (4.8 X 10-9)
= 2.08 x 10-6
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