The dissociation constant of a weak acid is 1 | vedclass

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Question

(C) The dissociation of a weak acid is given by: $HA ightleftharpoons H^{+} + A^{-}$,where $K_a = \frac{[H^{+}][A^{-}]}{[HA]} = 1.0 	imes 10^{-5}$

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$1.0 	imes 10^{9}$

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(C) The dissociation of a weak acid is given by: $HA ightleftharpoons H^{+} + A^{-}$,where $K_a = \frac{[H^{+}][A^{-}]}{[HA]} = 1.0 	imes 10^{-5}$ $(I)$.The reaction of a weak acid with a strong base is: $HA + OH^{-} ightleftharpoons A^{-} + H_2O$,with equilibrium constant $K = \frac{[A^{-}]}{[HA][OH^{-}]}$ $(II)$.We know that the auto-ionization of water is: $H_2O ightleftharpoons H^{+} + OH^{-}$,where $K_w = [H^{+}][OH^{-}] = 1.0 	imes 10^{-14}$ $(III)$.Dividing $(I)$ by $(III)$: $\frac{K_a}{K_w} = \frac{[H^{+}][A^{-}]}{[HA]} 	imes \frac{1}{[H^{+}][OH^{-}]} = \frac{[A^{-}]}{[HA][OH^{-}]} = K$.Therefore,$K = \frac{K_a}{K_w} = \frac{1.0 	imes 10^{-5}}{1.0 	imes 10^{-14}} = 1.0 	imes 10^{9}$.

The dissociation constant of a weak acid is $1.0 \times 10^{-5}$. The equilibrium constant for its reaction with a strong base is:

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