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(C) For a weak acid $HA$,the dissociation is $HA ightleftharpoons H^{+} + A^{-}$,so $K_a = \frac{[H^{+}][A^{-}]}{[HA]} = 1.0 	imes 10^{-5}$.The rea

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$1.0 	imes 10^9$

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(C) For a weak acid $HA$,the dissociation is $HA ightleftharpoons H^{+} + A^{-}$,so $K_a = \frac{[H^{+}][A^{-}]}{[HA]} = 1.0 	imes 10^{-5}$.The reaction of a weak acid with a strong base is $HA + OH^{-} ightleftharpoons A^{-} + H_2O$.The equilibrium constant $K$ for this reaction is $K = \frac{[A^{-}]}{[HA][OH^{-}]}$.We know that $K_w = [H^{+}][OH^{-}] = 1.0 	imes 10^{-14}$.Multiplying the numerator and denominator of $K$ by $[H^{+}]$,we get $K = \frac{[A^{-}][H^{+}]}{[HA][OH^{-}][H^{+}]} = \frac{K_a}{K_w}$.Substituting the values: $K = \frac{1.0 	imes 10^{-5}}{1.0 	imes 10^{-14}} = 1.0 	imes 10^9$.

If the dissociation constant of a weak acid is $1.0 \times 10^{-5}$,then the equilibrium constant for its reaction with a strong base will be......

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