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The ionization constant of $HF$, $HCOOH$ and $HCN$ at $298\, K$ are $6.8 \times 10^{-4}, 1.8 \times 10^{-4}$ and $4.8 \times 10^{-9}$ respectively. Calculate the ionization constants of the corresponding conjugate base.
Solution
It is known that,
$K_{b}=\frac{K_{w}}{K_{a}}$
Given $K_{a}$ of $HF =6.8 \times 10^{-4}$
Hence, $K_{b}$ of its conjugate base $F^{-}$
$=\frac{K_{w}}{K_{a}}$
$=\frac{10^{-14}}{6.8 \times 10^{-4}}$
$=1.5 \times 10^{-11}$
Given,
$K_{a}$ of $HCOOH =1.8 \times 10^{-4}$
Hence, $K_{b}$ of its conjugate base $HCOO ^{-}$
$=\frac{K_{w}}{K_{a}}$
$=\frac{10^{-14}}{1.8 \times 10^{-4}}$
$=5.6 \times 10^{-11}$
Given,
$K_{a}$ of $HCN =4.8 \times 10^{-9}$
Hence, $K_{b}$ of its conjugate base $CN ^{-}$
$=\frac{K_{w}}{K_{a}}$
$=\frac{10^{-14}}{4.8 \times 10^{-9}}$
$=2.08 \times 10^{-6}$
