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The ionization constant of HF,HCOOH and ...

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.

Text Solution

Verified by Experts

For conjugate acid-base pair : `K_(a) xx K_(b) =K_(w)`
`K_(b)" of "F^(-)=(10^(-14))/(6.8 xx 10^(-4))=1.47 xx 10^(-11)`
`K_(b)" of "HCOO^(-)=(10^(-14))/(1.8 xx 10^(-4))=5.6 xx 10^(-11)`
`K_(b)" of "CN^(-)=(10^(-14))/(4.8 xx 10^(-8))=2.08 xx 10^(-7)`
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