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(C) The given reaction is the hydrolysis of the formate ion $(HCOO^-)$,which is the conjugate base of a weak acid $(HCOOH)$.The equilibrium constant f

Vedclass · Accepted Answer

$5.56 	imes 10^{-11}$

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(C) The given reaction is the hydrolysis of the formate ion $(HCOO^-)$,which is the conjugate base of a weak acid $(HCOOH)$.The equilibrium constant for this hydrolysis reaction is denoted as $K_h$.The relationship between $K_h$,$K_w$ (ionic product of water),and $K_a$ (acid dissociation constant) is given by:$K_h = \frac{K_w}{K_a}$Given $K_w = 1.0 	imes 10^{-14}$ at $25^{\circ}C$ and $K_a = 1.8 	imes 10^{-4}$.$K_h = \frac{1.0 	imes 10^{-14}}{1.8 	imes 10^{-4}}$$K_h = 0.5556 	imes 10^{-10} = 5.56 	imes 10^{-11}$Thus,the correct option is $C$.

Find the equilibrium constant for the equilibrium reaction:
$HCOO^{-} + H_2O \rightleftharpoons HCOOH + OH^{-}$
Given that the dissociation constant of formic acid is $K_a(HCOOH) = 1.8 \times 10^{-4}$.

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Find the equilibrium constant for the equilibrium reaction:$HCOO^{-} + H_2O \rightleftharpoons HCOOH + OH^{-}$Given that the dissociation constant of formic acid is $K_a(HCOOH) = 1.8 \times 10^{-4}$.