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

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$\sqrt[3]{K_{1} + K_{2} + K_{3}}$

$K_{1} + K_{2} + K_{3}$

$K_{1} \times K_{2} \times K_{3}$

${(K_{1} \times K_{2} \times K_{3})}^{3}$

answer is D.

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Detailed Solution

Tribasic acid = H₃A

$H_{3} A ⇌ H^{+} + H_{2} A^{-}, K_{c} = K_{1} H_{2} A^{-} ⇌ H^{+} + {HA}^{- 2}, K_{c} = K_{2} {HA}^{- 2} ⇌ H^{+} + A^{- 3}, K_{c} = K_{3}$

Net reaction: $H_{3} A ⇌ 3 H^{+} + A^{- 3}, K_{c} = K_{1} \times K_{2} \times K_{3}$

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