The dissociation constant of a weak monobasic | vedclass

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(B) For a weak monobasic acid $HA$,the dissociation equilibrium is: $HA_{(aq)} \rightleftharpoons H^+_{(aq)} + A^-_{(aq)}$ At equilibrium,the concentr

Vedclass · Accepted Answer

$0.03$

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(B) For a weak monobasic acid $HA$,the dissociation equilibrium is: $HA_{(aq)} ightleftharpoons H^+_{(aq)} + A^-_{(aq)}$ At equilibrium,the concentration of $H^+$ and $A^-$ is $C\alpha$ and the concentration of $HA$ is $C(1-\alpha)$,where $C$ is the initial concentration and $\alpha$ is the degree of dissociation. The dissociation constant $K_a$ is given by: $K_a = \frac{[H^+][A^-]}{[HA]} = \frac{C\alpha^2}{1-\alpha}$ Since the acid is weak,$\alpha \ll 1$,so $1-\alpha \approx 1$. Thus,$K_a = C\alpha^2$. Rearranging to solve for $C$: $C = \frac{K_a}{\alpha^2} = \frac{2.7 	imes 10^{-5}}{(3 	imes 10^{-2})^2} = \frac{2.7 	imes 10^{-5}}{9 	imes 10^{-4}} = 0.3 	imes 10^{-1} = 0.03 \ M$.

The dissociation constant of a weak monobasic acid is $2.7 \times 10^{-5}$. If the degree of dissociation of the acid is $3 \times 10^{-2}$,what is the concentration of the acid (in $M$)?

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