A) \[{{H}_{2}}PO_{2}^{-}\]
B) \[HPO_{3}^{2-}\]
C) \[HPO_{4}^{2-}\]
D) All of these
Correct Answer: C
Solution :
\[HPO_{4}^{2-}+{{H}_{2}}O\overset{{}}{leftrightarrows}{{H}_{2}}PO_{4}^{2-}+O{{H}^{-}}\] \[HPO_{4}^{2-}+{{H}_{2}}O\overset{{}}{leftrightarrows}PO_{4}^{3-}+{{H}_{3}}{{O}^{+}}\] \[{{H}_{2}}PO_{2}^{-}\] is a conjugate base of \[{{H}_{3}}P{{O}_{2}}\] (a monobasic I acid) and does not give \[{{H}^{+}}\] further, \[HPO_{3}^{2-}\] is a conjugate base of \[{{H}_{2}}PO_{3}^{-}\] and does not ionize further, since \[{{H}_{3}}P{{O}_{3}}\] is a dibasic acid.
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