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(D) For a polyprotic acid $H_2A$,the dissociation steps are:$H_2A ightleftharpoons H^{+} + HA^{-}$; $K_1 = \frac{[H^{+}][HA^{-}]}{[H_2A]} = 1.0 	im

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$5.0 	imes 10^{-15}$

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(D) For a polyprotic acid $H_2A$,the dissociation steps are:$H_2A ightleftharpoons H^{+} + HA^{-}$; $K_1 = \frac{[H^{+}][HA^{-}]}{[H_2A]} = 1.0 	imes 10^{-5}$$HA^{-} ightleftharpoons H^{+} + A^{2-}$; $K_2 = \frac{[H^{+}][A^{2-}]}{[HA^{-}]} = 5.0 	imes 10^{-10}$The overall dissociation reaction is the sum of these two steps:$H_2A ightleftharpoons 2H^{+} + A^{2-}$The overall dissociation constant $K$ is the product of the individual dissociation constants:$K = K_1 	imes K_2$$K = (1.0 	imes 10^{-5}) 	imes (5.0 	imes 10^{-10})$$K = 5.0 	imes 10^{-15}$

The first and second dissociation constants of an acid $H_2A$ are $1.0 \times 10^{-5}$ and $5.0 \times 10^{-10}$ respectively. The overall dissociation constant of the acid will be

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