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The equilibrium constants for the reactions are:
$H_{3} {PO}_{4} ⇌ H^{+} + H_{2} {PO}_{4}^{-}; K_{1}$
$H_{2} {PO}_{4} ⇌ H^{+} + H_{2} {PO}_{4}^{2 -}; K_{2}$
${HPO}_{4}^{2 -} ⇌ H^{+} + {PO}_{4}^{3 -}; K_{3}$
The equilibrium constant for, $H_{3} {PO}_{4} ⇌ 3 H^{+} + {PO}_{4}^{3 -} will be :$

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$K_{1} / K_{2} K_{3}$

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

$K_{2} / K_{1} K_{3}$

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

answer is B.

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Detailed Solution

$K_{1} = \frac{[H^{+}] [H_{2} {PO}_{4}^{-}]}{[H_{3} {PO}_{4}]};$

$K_{2} = \frac{[H^{+}] [{HPO}_{4}^{2 -}]}{[H_{2} {PO}_{4}^{-}]}$

$K_{3} = \frac{[H^{+}] [{PO}_{4}^{3 -}]}{[{HPO}_{4}^{2 -}]}$

Multiplying these three $K_{1} \times K_{2} \times K_{3} = \frac{{[H^{+}]}^{3} [{PO}_{4}^{3 -}]}{[H_{3} {PO}_{4}]}$

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