What is the molar concentration of a weak mon | vedclass

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(B) For a weak monobasic acid $HA$,the dissociation equilibrium is $HA \rightleftharpoons H^+ + A^-$.Let $C$ be the initial molar concentration and $x

Vedclass · Accepted Answer

$0.002$

Vedclass · Answer

(B) For a weak monobasic acid $HA$,the dissociation equilibrium is $HA ightleftharpoons H^+ + A^-$.Let $C$ be the initial molar concentration and $x$ be the degree of dissociation.$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(Cx)(Cx)}{C(1-x)} = \frac{Cx^2}{1-x}$.Since the acid is weak,$x \ll 1$,so $1-x \approx 1$,and $K_a \approx Cx^2$.Given $K_a = 5 	imes 10^{-8}$ and $x = 0.5 \% = 0.005 = 5 	imes 10^{-3}$.$C = \frac{K_a}{x^2} = \frac{5 	imes 10^{-8}}{(5 	imes 10^{-3})^2}$.$C = \frac{5 	imes 10^{-8}}{25 	imes 10^{-6}} = \frac{5}{25} 	imes 10^{-2} = 0.2 	imes 10^{-2} = 0.002 \ M$.

What is the molar concentration of a weak monobasic acid if its dissociation constant is $5 \times 10^{-8}$ and it undergoes $0.5 \%$ dissociation (in $M$)?

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