The ionisation constant of acetic acid is 1.8 | vedclass

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(B) For a weak acid,the dissociation constant $K_a$ is related to the degree of dissociation $\alpha$ and concentration $C$ by the formula: $K_a = C\a

Vedclass · Accepted Answer

$0.045$

Vedclass · Answer

(B) For a weak acid,the dissociation constant $K_a$ is related to the degree of dissociation $\alpha$ and concentration $C$ by the formula: $K_a = C\alpha^2 / (1 - \alpha)$.Given $K_a = 1.8 	imes 10^{-5}$ and $\alpha = 2\% = 0.02$.Since $\alpha$ is very small,we can approximate $1 - \alpha \approx 1$.Thus,$K_a \approx C\alpha^2$.Substituting the values: $1.8 	imes 10^{-5} = C 	imes (0.02)^2$.$1.8 	imes 10^{-5} = C 	imes 0.0004$.$C = (1.8 	imes 10^{-5}) / (4 	imes 10^{-4}) = 0.45 	imes 10^{-1} = 0.045 \ M$.

The ionisation constant of acetic acid is $1.8 \times 10^{-5}$. The concentration at which it will be dissociated to $2\%$,is (in $M$)

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