Phosphoric acid ionizes in three steps with t | vedclass

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Question

(D) $H_3PO_4 \rightleftharpoons H^{+} + H_2PO_4^{-}; K_{a_1}$$H_2PO_4^{-} \rightleftharpoons H^{+} + HPO_4^{2-}; K_{a_2}$$HPO_4^{2-} \rightleftharpoon

Vedclass · Accepted Answer

$A, B$ and $C$ only

Vedclass · Answer

(D) $H_3PO_4 ightleftharpoons H^{+} + H_2PO_4^{-}; K_{a_1}$$H_2PO_4^{-} ightleftharpoons H^{+} + HPO_4^{2-}; K_{a_2}$$HPO_4^{2-} ightleftharpoons H^{+} + PO_4^{3-}; K_{a_3}$Adding these steps,the overall reaction is: $H_3PO_4 ightleftharpoons 3H^{+} + PO_4^{3-}$.The overall ionization constant is $K = K_{a_1} 	imes K_{a_2} 	imes K_{a_3}$.Taking $\log$ on both sides: $\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$. Thus,statement $A$ is true.For polybasic acids,the order of ionization constants is $K_{a_1} > K_{a_2} > K_{a_3}$ because as the negative charge on the anion increases,the electrostatic attraction for the $H^{+}$ ion increases,making it harder to remove subsequent protons. Thus,statement $C$ is true.Since $K_{a_1} > K_{a_2} > K_{a_3}$,$H_3PO_4$ is the strongest acid among the species,making statement $B$ true.Statement $D$ is incorrect as there is no such relationship.

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