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(A) For a weak acid,the dissociation constant $K_a$ is given by $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of ioniza

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$3.37$

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(A) For a weak acid,the dissociation constant $K_a$ is given by $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of ionization. Given $\alpha = 4.242 \% = 0.04242$ and $K_a = 1.8 	imes 10^{-5}$. $1.8 	imes 10^{-5} = C 	imes (0.04242)^2$. $C = \frac{1.8 	imes 10^{-5}}{0.001799} \approx 0.01 \ M$. The concentration of $H^+$ ions is $[H^+] = C \alpha = 0.01 	imes 0.04242 = 4.242 	imes 10^{-4} \ M$. The pH is calculated as $pH = -\log[H^+] = -\log(4.242 	imes 10^{-4})$. $pH = -(\log 4.242 + \log 10^{-4}) = -(0.6275 - 4) = 3.3725 \approx 3.37$.

At $25^{\circ} C$,the percentage of ionization of $x \ M$ acetic acid is $4.242$. What is the pH of the acetic acid solution?
$(\log 4.242=0.6275) ;(\log 0.04242=-1.372) \quad (K_a=1.8 \times 10^{-5})$

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At $25^{\circ} C$,the percentage of ionization of $x \ M$ acetic acid is $4.242$. What is the pH of the acetic acid solution?$(\log 4.242=0.6275) ;(\log 0.04242=-1.372) \quad (K_a=1.8 \times 10^{-5})$