For a 0.1 M solution of a weak acid HA (K_a | vedclass

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(A) For a weak acid $HA$,the dissociation is given by $HA \rightleftharpoons H^+ + A^-$. The degree of dissociation $\alpha$ is given by $\alpha = \sq

Vedclass · Accepted Answer

Percentage of dissociation = $1.18 \%$,$pH = 2.93$

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(A) For a weak acid $HA$,the dissociation is given by $HA ightleftharpoons H^+ + A^-$. The degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$. Given $K_a = 1.4 	imes 10^{-5}$ and $C = 0.1 \ M$. $\alpha = \sqrt{\frac{1.4 	imes 10^{-5}}{0.1}} = \sqrt{1.4 	imes 10^{-4}} = 1.18 	imes 10^{-2}$. Percentage of dissociation = $\alpha 	imes 100 = 1.18 	imes 10^{-2} 	imes 100 = 1.18 \%$. The concentration of $H^+$ ions is $[H^+] = C 	imes \alpha = 0.1 	imes 1.18 	imes 10^{-2} = 1.18 	imes 10^{-3} \ M$. $pH = -\log[H^+] = -\log(1.18 	imes 10^{-3}) = 3 - \log(1.18) = 3 - 0.0719 = 2.9281 \approx 2.93$.

For a $0.1 \ M$ solution of a weak acid $HA$ $(K_a = 1.4 \times 10^{-5})$ in $2 \ L$ of solution,calculate the percentage of dissociation and the $pH$ of the solution.

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