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The ionization constant of HF, HCOOH and...

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.

Text Solution

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If `K_(a)` is the ionization constant of a weak acid (HA) and `K_(cb)` is the ionization constant of its conjugate base `(A^(-))` then `K_(a).K_(cb)=K_(w)`
`K_(b)(F^(-))=(1xx10^(-14))/(K_(a)(HF)) = (1xx10^(-14))/(6.8xx10^(-4))=147xx10^(-11)`
`K_(b)(HCOO^(-))=(1xx10^(-14))/(K_(a)(HCOOH))=(1xx10^(-14))/(1.8xx10^(-4))=5.56xx10^(-11)`
`K_(b)(CN^(-))=(1xx10^(-14))/(K_(a)(HCN))=(1xx10^(-14))/(4.8xx10^(-9)`=`2.08xx10^(-6)`
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