0.01 mol of a weak acid HA (K_{a} = 2.0 time | vedclass

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(C) The dissociation equilibrium of the weak acid $HA$ is given by: $HA \rightleftharpoons H^{+} + A^{-}$.Initial concentrations: $[HA] = 0.01 \ M$,$[

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$2$

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(C) The dissociation equilibrium of the weak acid $HA$ is given by: $HA ightleftharpoons H^{+} + A^{-}$.Initial concentrations: $[HA] = 0.01 \ M$,$[H^{+}] = 0.1 \ M$ (from $HCl$),$[A^{-}] = 0$.At equilibrium,let the concentration of $A^{-}$ be $x \ M$. Then $[HA] = (0.01 - x) \approx 0.01 \ M$ and $[H^{+}] = (0.1 + x) \approx 0.1 \ M$ (since $\alpha The acid dissociation constant expression is $K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}$.Substituting the values: $2.0 	imes 10^{-6} = \frac{0.1 	imes x}{0.01}$.Solving for $x$: $x = \frac{2.0 	imes 10^{-6} 	imes 0.01}{0.1} = 2.0 	imes 10^{-7} \ M$.The degree of dissociation $\alpha = \frac{x}{C} = \frac{2.0 	imes 10^{-7}}{0.01} = 2.0 	imes 10^{-5}$.Thus,the value is $2 	imes 10^{-5}$.

$0.01 \ mol$ of a weak acid $HA$ $(K_{a} = 2.0 \times 10^{-6})$ is dissolved in $1.0 \ L$ of $0.1 \ M$ $HCl$ solution. The degree of dissociation of $HA$ is ............. $\times 10^{-5}$ (Round off to the Nearest Integer). [Neglect volume change on adding $HA$. Assume degree of dissociation $<< 1$]

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