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pH the Power of H

The ionizatio...

Question

The ionization constant of HF, HCOOH and HCN at 298K are $6.8 \times 10^{- 4}, 1.8 \times 10^{- 4}$ and $4.8 \times 10^{- 9}$ respectively. Calculate the ionization constants of the corresponding conjugate base.

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Solution

It is known that,

$K_{b} = \frac{K_{w}}{K_{a}}$
Given,
$K_{a}$ of HF= $6.8 \times 10^{- 4}$
Hence, $K_{b}$ of its conjugate base $F^{-}$
$= \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{6.8 \times 10^{- 4}} = 1.5 \times 10^{- 11}$
Guven,
$K_{a}$ of $H C O O H = 1.8 \times 10^{- 4}$
Hence, $K_{b}$ of its conjugate base $H C O O^{-}$
$= \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{1.8 \times 10^{- 4}} = 5.6 \times 10^{- 11}$
Guven,
$K_{a}$ of $H C N = 4.8 \times 10^{- 9}$
Hence, $K_{b}$ of its conjugate base $C N^{-}$
$= \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{4.8 \times 10^{- 9}} = 2.08 \times 10^{- 6}$

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Similar questions

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.

Q. The ionization constant of

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Q. Equimolar solutions of

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pH the Power of H

Standard XII Chemistry

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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.

The ionization constant of HF, HCOOH and HCN at 298K are $6.8 \times 10^{- 4}, 1.8 \times 10^{- 4}$ and $4.8 \times 10^{- 9}$ respectively. Calculate the ionization constants of the corresponding conjugate base.