A weak monobasic acid is 4% dissociated in 0. | vedclass

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(A) For a weak acid, the degree of dissociation $\alpha$ is related to concentration $C$ by the relation $\alpha \approx \sqrt{\frac{K_a}{C}}$, which

Vedclass · Accepted Answer

$2.83$

Vedclass · Answer

(A) For a weak acid, the degree of dissociation $\alpha$ is related to concentration $C$ by the relation $\alpha \approx \sqrt{\frac{K_a}{C}}$, which implies $\alpha \propto \frac{1}{\sqrt{C}}$.Given: $\alpha_1 = 4\% = 0.04$, $C_1 = 0.05 	ext{ M}$, $C_2 = 0.1 	ext{ M}$.Using the ratio: $\frac{\alpha_2}{\alpha_1} = \sqrt{\frac{C_1}{C_2}}$.$\frac{\alpha_2}{0.04} = \sqrt{\frac{0.05}{0.1}} = \sqrt{0.5} \approx 0.707$.$\alpha_2 = 0.04 	imes 0.707 = 0.02828$.Percent dissociation = $0.02828 	imes 100 = 2.828\% \approx 2.83\%$.

$A$ weak monobasic acid is $4\%$ dissociated in $0.05 \text{ M}$ solution. What is the percent dissociation in $0.1 \text{ M}$ solution (in $\%$)?

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