Ncert Exercises & Solutions

7.43 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.

NEET SOLUTION:

It is known that,

$K_{b} = \frac{K_{w}}{K_{a}}$

Given,

$K_{a} of HF = 6.8 \times 10^{- 4}$

$H_{a} of HCOOH = 1.8 \times 10^{- 4} Hence, K_{b} of its conjugate base {HCOO}^{-} = \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{1.8 \times 10^{- 4}} = 5.6 \times 10^{- 11}$

Given,

$K_{a} of HCN = 4.8 \times 10^{- 9} Hence, K_{b} of its conjugate base {CN}^{-} = \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{4.8 \times 10^{- 9}} = 2.08 \times 10^{- 6}$

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