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Ionization Enthalpy

The ionizatio...

Question

The ionization constant of HF, HCOOH and HCN at 298K 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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Solution

It is known that,

Given,

K_a of HF = 6.8 × 10^–4

Hence, K_b of its conjugate base F^–

Given,

K_a of HCOOH = 1.8 × 10^–4

Hence, K_b of its conjugate base HCOO^–

Given,

K_a of HCN = 4.8 × 10^–9

Hence, K_b of its conjugate base CN^–

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Q. The ionization constant of

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6.8 \times 10^{- 4}, 1.8 \times 10^{- 4} and 4.8 \times 10^{- 9}

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Q. Equimolar solutions of

H F, H C O O H

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6.8 \times 10^{- 4}, 1.8 \times 10^{- 4}

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25^{o} C

, the dissociation constant of

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of its conjugate base if the ionization constant of acid is

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?

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.The ionization constant for the conjugate base of the acid will be (consider molarity

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