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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $HCOO^{-}$?

Similar Questions

Which is the strongest of the following acids?

The successive equilibrium constants for the stepwise dissociation of a tribasic acid are $K_1$, $K_2$ and $K_3$, respectively. The equilibrium constant for the overall dissociation is

Which salt can furnish $H^{+}$ ion in its aqueous solution?

Acidity of $BF_3$ can be explained on which of the following concepts?

The conjugate base of $NH_2^-$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module