The degree of ionization of 0.4 M acetic aci | vedclass

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(A) For a weak acid like acetic acid,the dissociation constant $K_{a}$ is related to the degree of ionization $\alpha$ and concentration $C$ by the fo

Vedclass · Accepted Answer

$6.71 	imes 10^{-3}$

Vedclass · Answer

(A) For a weak acid like acetic acid,the dissociation constant $K_{a}$ is related to the degree of ionization $\alpha$ and concentration $C$ by the formula: $K_{a} = C \alpha^{2}$.Given $K_{a} = 1.8 	imes 10^{-5}$ and $C = 0.4 \ M$.Substituting the values: $1.8 	imes 10^{-5} = 0.4 	imes \alpha^{2}$.$\alpha^{2} = \frac{1.8 	imes 10^{-5}}{0.4} = 4.5 	imes 10^{-5} = 45 	imes 10^{-6}$.$\alpha = \sqrt{45 	imes 10^{-6}} \approx 6.71 	imes 10^{-3}$.

The degree of ionization of $0.4 \ M$ acetic acid will be $(K_{a} = 1.8 \times 10^{-5})$.

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