The successive equilibrium constants for the | vedclass

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Question

(D) For a tribasic acid like $H_3PO_4$, there are three ionisation constants. The overall ionisation constant $(K)$ is the product of the ionisation c

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$K_1 	imes K_2 	imes K_3$

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(D) For a tribasic acid like $H_3PO_4$, there are three ionisation constants. The overall ionisation constant $(K)$ is the product of the ionisation constants of the three individual steps.Step $1$: $H_3PO_4 ightleftharpoons H^+ + H_2PO_4^-$; $K_1 = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]}$ $(i)$Step $2$: $H_2PO_4^- ightleftharpoons H^+ + HPO_4^{2-}$; $K_2 = \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]}$ $(ii)$Step $3$: $HPO_4^{2-} ightleftharpoons H^+ + PO_4^{3-}$; $K_3 = \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$ $(iii)$Adding the three equations gives the overall reaction: $H_3PO_4 ightleftharpoons 3H^+ + PO_4^{3-}$.The overall equilibrium constant $K$ is given by $K = \frac{[H^+]^3[PO_4^{3-}]}{[H_3PO_4]}$.Multiplying the three stepwise equilibrium constants:$K_1 	imes K_2 	imes K_3 = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} 	imes \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]} 	imes \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$$K_1 	imes K_2 	imes K_3 = \frac{[H^+]^3[PO_4^{3-}]}{[H_3PO_4]} = K$Therefore, $K = K_1 	imes K_2 	imes K_3$.

The successive equilibrium constants for the stepwise dissociation of a tribasic acid are $K_1$, $K_2$ and $K_3$, respectively. The equilibrium constant for the overall dissociation is

Similar Questions

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