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(A) For a diprotic acid ${H_2}A$,the dissociation steps are:$1$. ${H_2}A ightleftharpoons H^+ + HA^-$; $K_{a1} = 1.0 	imes 10^{-5}$$2$. $HA^- igh

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$5.0 	imes 10^{-15}$

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(A) For a diprotic acid ${H_2}A$,the dissociation steps are:$1$. ${H_2}A ightleftharpoons H^+ + HA^-$; $K_{a1} = 1.0 	imes 10^{-5}$$2$. $HA^- ightleftharpoons H^+ + A^{2-}$; $K_{a2} = 5.0 	imes 10^{-10}$The overall reaction is the sum of these two steps:${H_2}A ightleftharpoons 2H^+ + A^{2-}$The overall dissociation constant $K_a$ is the product of the individual dissociation constants:$K_a = K_{a1} 	imes K_{a2}$$K_a = (1.0 	imes 10^{-5}) 	imes (5.0 	imes 10^{-10}) = 5.0 	imes 10^{-15}$

The first and second dissociation constants of a diprotic acid ${H_2}A$ are $1.0 \times 10^{-5}$ and $5.0 \times 10^{-10}$ respectively. The overall dissociation constant of the acid ${H_2}A$ is .......

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