A) \[PO_{4}^{3-}\]
B) \[{{P}_{2}}{{O}_{5}}\]
C) \[{{H}_{3}}P{{O}_{4}}\]
D) \[HPO_{4}^{2-}\]
Correct Answer: D
Solution :
\[{{H}_{3}}P{{O}_{4}}\]is a tribasic acid, thus ionising in three steps I. \[{{H}_{2}}P{{O}_{4}}{{H}^{+}}+{{H}_{2}}PO_{4}^{-}\] II. \[{{H}_{2}}PO_{4}^{-}\,{{H}^{+}}\,+HPO_{4}^{2-}\] III. \[HPO_{4}^{2-}{{H}^{+}}+PO_{4}^{3-}\] .Conjugate base is formed when an acid loses its proton. Thus,\[HPO_{4}^{2-}\]is the conjugate base of\[{{H}_{2}}PO_{4}^{-}\](which is an acid in step II, but is the conjugate base of \[{{H}_{3}}P{{O}_{4}}\] in step I).
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