At 25^{circ} C,the percentage of ionization o | vedclass

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(D) For a weak acid,the degree of ionization $\alpha$ is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$,where $C$ is the molar concentration '$x

Vedclass · Accepted Answer

$0.01$

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(D) For a weak acid,the degree of ionization $\alpha$ is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$,where $C$ is the molar concentration '$x$'.Given,percentage of ionization $= 4.242\%$,so $\alpha = \frac{4.242}{100} = 0.04242$.Given $K_a = 1.8 	imes 10^{-5}$.Using the formula $\alpha^2 = \frac{K_a}{C}$,we get $C = \frac{K_a}{\alpha^2}$.$C = \frac{1.8 	imes 10^{-5}}{(0.04242)^2}$.$(0.04242)^2 \approx 0.0018$.$C = \frac{1.8 	imes 10^{-5}}{1.8 	imes 10^{-3}} = 10^{-2} = 0.01 \ M$.Therefore,the value of $x$ is $0.01$.

At $25^{\circ} C$,the percentage of ionization of '$x$' $M$ acetic acid is $4.242$. What is the value of $x$? $(K_a = 1.8 \times 10^{-5})$

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