Q. The conjugate base of $H P O_{4}^{2 -}$ is

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A

$P O_{4}^{3 -}$

38%

B

$H_{2} P O_{4}^{-}$

41%

C

$H_{3} P O_{4}$

16%

D

$H_{3} P O_{3}$

5%

Solution:

$H P O_{4}^{-} ⇌ H^{+} + P O_{4}^{3 -}$
Base is obtained by loss of $H^{+}$