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.

#### Solution 1

It is known that,

`K_b = K_w/K_a`

Given,

*K*_{a} of HF = 6.8 × 10^{–4}

Hence, *K*_{b} of its conjugate base F^{–}

^{= `K_w/K_a`}

`= 10^(-14)/(6.8 xx 10^(-4))`

`= 1.5 xx 10^(-11)`

Given,

*K*_{a} of HCOOH = 1.8 × 10^{–4}

Hence, *K*_{b} of its conjugate base HCOO^{–}

= `K_w/K_a`

`= 10^(-14)/(1.8 xx 10^(-4))`

`= 5.6 xx 10^(-11)`

Given

`K_a of HCN = 4.8 xx 10^(-9)`

Hence , K_b of its conjugate base `CN^(-)`

`= K_w/K_(a)`

`= 10^(-14)/(4.8 xx 10^(-9))`

`= 2.08 xx 10^(-6)`

#### Solution 2

For F^{– }, K_{b} =K_{w}/K_{a}= 10^{-14}/(6.8 x 10^{-4}) = 1.47 x 10^{-11} = 1.5 x 10^{-11} .

For HCOO-, K_{b} = 10^{-14}/(1.8 x 10^{-4}) = 5.6 x 10^{-11 }

For CN^{–}, K_{b}= 10^{-14}/(4.8 X 10^{-9}) = 2.08 x 10^{-6}