The dissociation constant of a weak acid HA i | vedclass

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Question

(A) Step $1$: Calculate the molar concentration $C$ of the solution.$C = \frac{n}{V} = \frac{0.2 	ext{ mol}}{2 	ext{ L}} = 0.1 	ext{ M}$.Step $2$:

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$1.22$

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(A) Step $1$: Calculate the molar concentration $C$ of the solution.$C = \frac{n}{V} = \frac{0.2 	ext{ mol}}{2 	ext{ L}} = 0.1 	ext{ M}$.Step $2$: Use the formula for the degree of dissociation $\alpha$ for a weak acid.$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.5 	imes 10^{-5}}{0.1}} = \sqrt{1.5 	imes 10^{-4}} = 1.2247 	imes 10^{-2}$.Step $3$: Convert the degree of dissociation to percent dissociation.$	ext{Percent dissociation} = \alpha 	imes 100 = 1.2247 	imes 10^{-2} 	imes 100 = 1.2247 \% \approx 1.22 \%$.

The dissociation constant of a weak acid $HA$ is $1.5 \times 10^{-5}$. Find the percent dissociation of a solution containing $0.2 \text{ moles}$ of $HA$ in $2 \text{ liters}$ of solution. (in $\%$)

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