The dissociation constant of a weak monobasic | vedclass

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(D) Given: Dissociation constant $K_a = 3.2 	imes 10^{-4}$ and concentration $c = 0.04 \ M$. For a weak monobasic acid,the degree of dissociation $\a

Vedclass · Accepted Answer

$0.089$

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(D) Given: Dissociation constant $K_a = 3.2 	imes 10^{-4}$ and concentration $c = 0.04 \ M$. For a weak monobasic acid,the degree of dissociation $\alpha$ is given by the formula: $\alpha = \sqrt{\frac{K_a}{c}}$ Substituting the values: $\alpha = \sqrt{\frac{3.2 	imes 10^{-4}}{0.04}} = \sqrt{\frac{320 	imes 10^{-6}}{4 	imes 10^{-2}}} = \sqrt{80 	imes 10^{-4}} = \sqrt{8 	imes 10^{-3}} \approx 0.0894$ Thus,the degree of dissociation is approximately $0.089$.

The dissociation constant of a weak monobasic acid is $3.2 \times 10^{-4}$. Calculate the degree of dissociation in its $0.04 \ M$ solution.

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