If pK_{a} = -log K_{a} = 4 for a weak acid HX | vedclass

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(C) Given $pK_{a} = 4$,so $K_{a} = 10^{-4}$.Given the dissociation equation for weak acid $HX$: $HX \rightleftharpoons H^+ + X^-$.The degree of dissoc

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$1.10$

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(C) Given $pK_{a} = 4$,so $K_{a} = 10^{-4}$.Given the dissociation equation for weak acid $HX$: $HX ightleftharpoons H^+ + X^-$.The degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_{a}}{C}}$.Substituting the values: $\alpha = \sqrt{\frac{10^{-4}}{0.01}} = \sqrt{\frac{10^{-4}}{10^{-2}}} = \sqrt{10^{-2}} = 0.1$.The Van't Hoff factor $i$ for a dissociation process is $i = 1 + (n-1)\alpha$,where $n$ is the number of ions produced per molecule.For $HX$,$n = 2$ ($H^+$ and $X^-$).$i = 1 + (2-1) 	imes 0.1 = 1 + 0.1 = 1.1$.

If $pK_{a} = -\log K_{a} = 4$ for a weak acid $HX$ and $K_{a} = C\alpha^{2}$,then the Van't Hoff factor when $C = 0.01 \ M$ is

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