At 25^{circ} C,K_{a} of formic acid is 1.8 ti | vedclass

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Question

(C) The relationship between the dissociation constant of a weak acid $(K_{a})$ and the dissociation constant of its conjugate base $(K_{b})$ is given

Vedclass · Accepted Answer

$5.55 	imes 10^{-11}$

Vedclass · Answer

(C) The relationship between the dissociation constant of a weak acid $(K_{a})$ and the dissociation constant of its conjugate base $(K_{b})$ is given by the equation: $K_{a} 	imes K_{b} = K_{w}$.At $25^{\circ} C$,the ionic product of water is $K_{w} = 1.0 	imes 10^{-14}$.Given $K_{a} = 1.8 	imes 10^{-4}$ for formic acid $(HCOOH)$.Therefore,$K_{b} = \frac{K_{w}}{K_{a}} = \frac{1.0 	imes 10^{-14}}{1.8 	imes 10^{-4}}$.$K_{b} = 0.555 	imes 10^{-10} = 5.55 	imes 10^{-11}$.

At $25^{\circ} C$,$K_{a}$ of formic acid is $1.8 \times 10^{-4}$. What is the $K_{b}$ of $HCOO^{-}$?

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