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Q. Which one of the following is conjugate acid of water in the reaction? $ {{H}_{2}}S{{O}_{4}}+{{H}_{2}}O{{H}_{3}}{{O}^{+}}+HSO_{4}^{-} $

Rajasthan PMTRajasthan PMT 2009

Solution:

$ \underset{aci{{d}_{1}}}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,+\underset{bas{{e}_{2}}}{\mathop{{{H}_{2}}O}}\,\underset{aci{{d}_{2}}}{\mathop{{{H}_{3}}{{O}^{+}}}}\,+\underset{bas{{e}_{1}}}{\mathop{HSO_{4}^{-}}}\, $ In the above reaction, $ {{H}_{2}}S{{O}_{4}} $ lose a proton $ ({{H}^{+}}) $ which is gain by $ {{H}_{2}}O $ , hence $ {{H}_{2}}S{{O}_{4}} $ is an acid and $ {{H}_{2}}O $ is base. If this reaction is reversible $ {{H}_{3}}{{O}^{+}} $ losses a proton which is accepted by $ HSO_{4}^{-} $ , so $ {{H}_{3}}{{O}^{+}} $ is an acid and $ HSO_{4}^{-} $ is base. So, conjugate base of $ {{H}_{2}}S{{O}_{4}} $ is $ HSO_{4}^{-} $ and conjugate acid of $ {{H}_{2}}O $ is $ {{H}_{3}}{{O}^{+}} $ .