The dihydrogen phosphate ion undergoes these reactions in water.H2PO4−(aq)+H2O(l)⟶HPO42−(aq)+H3O+(aq)K=6.2×10−8H2PO4−(aq)+H2O(l)⟶H3PO4(aq)+OH−(aq)K=1.6×10−7What 1s the conjugate base of H2PO4− ?

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The dihydrogen phosphate ion undergoes these reactions in water.
$\begin{array}{l} H_{2} {PO}_{4}^{-} (a q) + H_{2} O (l) ⟶ {HPO}_{4}^{2 -} (a q) + H_{3} O^{+} (a q) \\ K = 6.2 \times 10^{- 8} \end{array}$
$\begin{array}{l} H_{2} {PO}_{4}^{-} (a q) + H_{2} O (l) ⟶ H_{3} {PO}_{4} (a q) + {OH}^{-} (a q) \\ K = 1.6 \times 10^{- 7} \end{array}$
What 1s the conjugate base of $H_{2} {PO}_{4}^{-}$ ?

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$\begin{aligned} H_{2} O (l) \end{aligned}$

${OH}^{-} (a q)$

$H_{3} {PO}_{4} (a q)$

$\begin{aligned} {HPO}_{4}^{2 -} (a q) \end{aligned}$

answer is A.

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$H_{2} {PO}_{4}^{-} \to {HPO}_{4}^{- 2} + H^{+}$

Hence, conjugate base of $H_{2} {PO}_{4}^{-}$ is ${HPO}_{4}^{- 2}$ .

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