(i) H_3PO_{4text{(aq)}} rightleftharpoons H^+ | vedclass

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Question

(D) The given reactions are:
(i) $H_3PO_{4	ext{(aq)}} ightleftharpoons H^+{_{	ext{(aq)}}} + H_2PO_4^-{_{	ext{(aq)}}} ; \ K_1 = \frac{[H^+][H_2PO

Vedclass · Accepted Answer

$K_1 K_2 K_3$

Vedclass · Answer

(D) The given reactions are:
(i) $H_3PO_{4	ext{(aq)}} ightleftharpoons H^+{_{	ext{(aq)}}} + H_2PO_4^-{_{	ext{(aq)}}} ; \ K_1 = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]}$
(ii) $H_2PO_4^-{_{	ext{(aq)}}} ightleftharpoons H^+{_{	ext{(aq)}}} + HPO_4^{2-}{_{	ext{(aq)}}} ; \ K_2 = \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]}$
(iii) $HPO_4^{2-}{_{	ext{(aq)}}} ightleftharpoons H^+{_{	ext{(aq)}}} + PO_4^{3-}{_{	ext{(aq)}}} ; \ K_3 = \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$
Adding these three reactions gives the net reaction:
$H_3PO_{4	ext{(aq)}} ightleftharpoons 3H^{+}{_{	ext{(aq)}}} + PO_4^{3-}{_{	ext{(aq)}}}$
When reactions are added,their equilibrium constants are multiplied.
Therefore,the equilibrium constant $K$ for the net reaction is:
$K = K_1 	imes K_2 	imes K_3$

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