The ionization constant of HF, HCOOH and HCN at 298K are `6.8 xx 10^(-4), 1.8 xx 10^(-4)` and `4.8 xx 10^(-9)` respectively. Calculate the ionization constants of the corresponding conjugate base.

Explanation

The conjugate base of HF is` F^(-)`. It will react with water as follows
` F^(-) +H_2O hArr HF+OH^(-)`
Obviously, this is the equation of hydrolysis of a salt of HF with a strong base. Hence,
` " " K_(F^(-))= K_h= (K_w)/(K_a) = ( 1.0 xx 10^(-14))/(6.8 xx 10 ^(-4)) = 1.5 xx 10^(-11)`
(ii) Similarly , for HCOOH
` " "K_(HCOO^(-) ) = J_h = (K_w)/(K_a) = (1.0 xx 10 ^(-14))/(1.8 xx 10 ^(-4))= 1.5 xx 10 ^(-11)`
(iii) For HCN,
` " " K_(CN) - = K_h = (K_w)/(K_a) = (1.0 xx 10^(-14))/(4.8 xx 10^(-9))= 2.1 xx 10 ^(-6)`