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(B) For a weak monobasic acid $(HA)$,the dissociation is given by: $HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$. The dissociation consta

Vedclass · Accepted Answer

$7.17$

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(B) For a weak monobasic acid $(HA)$,the dissociation is given by: $HA_{(aq)} ightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$. The dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1-\alpha}$. Since the acid is weak,$\alpha \ll 1$,so $1-\alpha \approx 1$,and $K_a \approx C \alpha^2$. For condition $1$: $C_1 = 0.05 \ M$,$\alpha_1 = 0.10$. $K_a = 0.05 	imes (0.1)^2 = 0.05 	imes 0.01 = 0.0005$. For condition $2$: $C_2 = 0.10 \ M$,$\alpha_2 = ?$. $K_a = C_2 \alpha_2^2 \implies 0.0005 = 0.10 	imes \alpha_2^2$. $\alpha_2^2 = \frac{0.0005}{0.10} = 0.005$. $\alpha_2 = \sqrt{0.005} \approx 0.0707$. Percentage dissociation $= 0.0707 	imes 100 = 7.07 \%$. The closest option is $7.17 \%$.

$A$ weak monobasic acid is $10 \%$ dissociated in $0.05 \ M$ solution. What is its percentage dissociation in $0.10 \ M$ solution (in $\%$)?

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