A monobasic weak acid dissociates 2 % in its | vedclass

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(B) For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degr

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$8 	imes 10^{-7}$

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(B) For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of dissociation.Given: $C = 0.002 \ M = 2 	imes 10^{-3} \ M$.Degree of dissociation $\alpha = 2 \% = 0.02 = 2 	imes 10^{-2}$.Substituting the values into the formula:$K_a = (2 	imes 10^{-3}) 	imes (2 	imes 10^{-2})^2$$K_a = (2 	imes 10^{-3}) 	imes (4 	imes 10^{-4})$$K_a = 8 	imes 10^{-7}$.

$A$ monobasic weak acid dissociates $2 \%$ in its $0.002 \ M$ solution. Calculate the dissociation constant of the weak acid.

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