A weak monobasic acid is 3.0 % dissociated in | vedclass

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Question

(B) For a weak monobasic acid $HA$,the dissociation equilibrium is: $HA_{(aq)} \rightleftharpoons H^+_{(aq)} + A^-_{(aq)}$Given concentration $C = 0.0

Vedclass · Accepted Answer

$3.6 	imes 10^{-5}$

Vedclass · Answer

(B) For a weak monobasic acid $HA$,the dissociation equilibrium is: $HA_{(aq)} ightleftharpoons H^+_{(aq)} + A^-_{(aq)}$Given concentration $C = 0.04 \ M$ and degree of dissociation $\alpha = 3.0 \% = 0.03$.The dissociation constant $K_a$ is given by the formula: $K_a = \frac{C \alpha^2}{1 - \alpha}$.Since the acid is weak,$\alpha \ll 1$,so $1 - \alpha \approx 1$.Thus,$K_a \approx C \alpha^2$.Substituting the values: $K_a = 0.04 	imes (0.03)^2$.$K_a = 0.04 	imes 0.0009 = 3.6 	imes 10^{-5}$.

$A$ weak monobasic acid is $3.0 \%$ dissociated in its $0.04 \ M$ solution. What is the dissociation constant of the acid?

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