A) \[HCl+NaOH\]
B) \[C{{H}_{3}}COOH+NaOH\]
C) \[{{H}_{2}}S{{O}_{4}}+NaOH\]
D) \[HCl{{O}_{4}}+KOH\]
Correct Answer: B
Solution :
\[\Delta {{H}_{ionizaton}}\] for \[{{H}_{2}}O\] means \[{{H}_{2}}O\to {{H}^{+}}+O{{H}^{-}}\] i.e. reverse of neutalization of strong acid with strong base. For neutralization of weak acid heat evolved is less.
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