The ionization constant of `HF,HCOOH` and `HCN` at `298 K` are `6.8xx10^(-4), 1.8xx10^(-4)` and `4.8xx10^(-9)` respectively. Calculate the ionization constant of the corresponding conjugate base.

Answer

`F^(-) = 1.5 xx 10^(-11), HCOO^(-) = 5.6 xx 10^(-11), CN^(-) = 2.08 xx 10^(-6)`

Explanation

It is given that,
`K_(b) = (K_(w))/(K_(a))`
Given, `K_(a)` of `HF = 6.8 xx 10^(-4)`
Hence, `K_(b)` of its conjugates 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 xx 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)` is given 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)`