What is the molar concentration of acetic aci | vedclass

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(B) Given: $K_{a} = 1.8 	imes 10^{-5}$ and $\alpha = 0.02$. For a weak monobasic acid,the relationship between dissociation constant $(K_{a})$,degree

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$4.5 	imes 10^{-2} \ M$

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(B) Given: $K_{a} = 1.8 	imes 10^{-5}$ and $\alpha = 0.02$. For a weak monobasic acid,the relationship between dissociation constant $(K_{a})$,degree of dissociation $(\alpha)$,and molar concentration $(c)$ is given by the formula: $K_{a} = \alpha^2 c$ Rearranging the formula to solve for concentration $(c)$: $c = \frac{K_{a}}{\alpha^2}$ Substituting the given values: $c = \frac{1.8 	imes 10^{-5}}{(0.02)^2} = \frac{1.8 	imes 10^{-5}}{4 	imes 10^{-4}} = 0.45 	imes 10^{-1} \ M = 4.5 	imes 10^{-2} \ M$.

What is the molar concentration of acetic acid if the value of its dissociation constant is $1.8 \times 10^{-5}$ and the degree of dissociation is $0.02$?

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