For ortho phosphoric acid,
$\begin{array}{l} H_{3} {PO}_{4} (aq) + H_{2} O (aq) ⟶ H_{3} O^{+} (aq) + H_{2} {PO}_{4}^{-} (aq); K_{a_{1}} \end{array}$ $\begin{array}{l} H_{2} {PO}_{4}^{-} (aq) + H_{2} O (aq) ⟶ H_{3} O^{+} (aq) + {HPO}_{4}^{2 -} (aq); K_{a_{2}} \\ {HPO}_{4}^{2 -} (aq) + H_{2} O (aq) ⟶ H_{3} O^{+} (aq) + {PO}_{4}^{3 -} (aq); K_{a_{3}} \end{array}$
The correct order of $K_{a}$ values is

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$K_{a_{1}} > K_{a_{2}} < K_{a_{3}}$

$K_{a_{1}} < K_{a_{2}} < K_{a_{3}}$

$K_{a_{1}} > K_{a_{2}} > K_{a_{3}}$

$K_{a_{1}} < K_{a_{2}} > K_{a_{3}}$

answer is C.

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Detailed Solution

$K_{a_{1}} > K_{a_{2}} > K_{a_{3}}$ (because it is difficult to remove a proton from species from which already proton has been lost).

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