What concentration of Ac^{-} ions will reduce | vedclass

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(D) The dissociation of $HAc$ is given by: $HAc ightleftharpoons H^{+} + Ac^{-}$.Given $[H^{+}] = [H_3O^{+}] = 2 	imes 10^{-4} \ M$ and $[HAc] \app

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$0.036$

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(D) The dissociation of $HAc$ is given by: $HAc ightleftharpoons H^{+} + Ac^{-}$.Given $[H^{+}] = [H_3O^{+}] = 2 	imes 10^{-4} \ M$ and $[HAc] \approx 0.40 \ M$ (assuming negligible dissociation).The expression for the acid dissociation constant is: $K_a = \frac{[H^{+}][Ac^{-}]}{[HAc]}$.Substituting the given values: $1.8 	imes 10^{-5} = \frac{(2 	imes 10^{-4}) 	imes [Ac^{-}]}{0.40}$.Solving for $[Ac^{-}]$: $[Ac^{-}] = \frac{1.8 	imes 10^{-5} 	imes 0.40}{2 	imes 10^{-4}}$.$[Ac^{-}] = \frac{0.72 	imes 10^{-5}}{2 	imes 10^{-4}} = 0.36 	imes 10^{-1} = 0.036 \ M$.

What concentration of $Ac^{-}$ ions will reduce $H_3O^{+}$ ion concentration to $2 \times 10^{-4} \ M$ in a $0.40 \ M$ solution of $HAc$ (in $M$)? Given $K_a (HAc) = 1.8 \times 10^{-5}$.

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