A) \[{{H}_{2}}O\]
B) \[{{H}_{3}}{{O}^{+}}\]
C) \[SO_{4}^{2-}\]
D) \[HSO_{4}^{-}\]
Correct Answer: B
Solution :
\[\underset{acid{{\,}_{1}}}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,+\underset{bas{{e}_{2}}}{\mathop{{{H}_{2}}O}}\,\underset{aci{{d}_{2}}}{\mathop{{{H}_{3}}{{O}^{+}}}}\,+\underset{bas{{e}_{1}}}{\mathop{HSO_{4}^{-}}}\,\] 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 \[{{H}_{3}}{{O}^{+}}\]
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