A) \[{{H}_{3}}{{O}^{+}}<NH_{4}^{+}<HF<O{{H}^{-}}<{{H}_{2}}O\]
B) \[NH_{4}^{+}<HF<{{H}_{3}}{{O}^{+}}<{{H}_{2}}O<O{{H}^{-}}\]
C) \[O{{H}^{-}}<{{H}_{2}}O<NH_{4}^{+}<HF<{{H}_{3}}{{O}^{+}}\]
D) \[{{H}_{3}}{{O}^{+}}>HF>{{H}_{2}}O>NH_{4}^{+}>O{{H}^{-}}\]
Correct Answer: C
Solution :
Among the given species, correct order of increasing acidic strength is: \[O{{H}^{-}}<{{H}_{2}}O<NH_{4}^{+}<HF<{{H}_{3}}{{O}^{+}}\] \[\text{O}{{\text{H}}^{-}}\] is basic {i.e., it tends to gain a proton) and hence is least acidic. \[{{\text{H}}_{\text{2}}}\text{O}\] is neutral species. \[{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\] is most acidic as it readily loose proton.
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