A weak acid,HA, has a K_a of 1.00 times 10^{- | vedclass

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(A) The dissociation of the weak acid is given by: $HA \rightleftharpoons H^{+} + A^{-}$At equilibrium,let the concentration of $H^{+}$ be $x$. Then $

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$1$

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(A) The dissociation of the weak acid is given by: $HA ightleftharpoons H^{+} + A^{-}$At equilibrium,let the concentration of $H^{+}$ be $x$. Then $[H^{+}] = [A^{-}] = x$ and $[HA] = 0.1 - x \approx 0.1$ (since $K_a$ is very small).$K_a = \frac{[H^{+}][A^{-}]}{[HA]} = \frac{x^{2}}{0.1} = 1.00 	imes 10^{-5}$$x^{2} = 1.00 	imes 10^{-6}$$x = 1.00 	imes 10^{-3} \ M$Percentage of dissociation $\alpha = \frac{x}{C} 	imes 100 = \frac{1.00 	imes 10^{-3}}{0.100} 	imes 100 = 1 \%$

$A$ weak acid,$HA,$ has a $K_a$ of $1.00 \times 10^{-5}.$ If $0.100 \ mol$ of this acid is dissolved in one litre of water,the percentage of acid dissociated at equilibrium is closest to $...\%$

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