Example 1

Solution Although a free proton is never actually produced in solution, it is often convenient to break the proton-transfer process into two hypothetical steps: (1) the loss of a proton by the acid, and (2) the gain of a proton by H₂O.

a) When HClO₄ loses a proton, H⁺, the valence electron originally associated with the H atom is left behind, producing a negative ion, ClO₄^–. The proton can then be added to a water molecule in the second hypothetical step. Summing the two steps gives the overall proton transfer:

    `HClO_4 -> H^(+) + ClO_4^(-)`               step 1

    `cancelH^(+) + H_2O -> H_3O^(+)`                   step 2
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`HClO_4 + H_2O -> H_3O^(+) + ClO_4^(-)`      overall

b) Proceeding as in part a, we have

          `HBr -> H^(+) + Br^(-)`          step 1

  `cancelH^(+) + H_2O -> H_3O^(+)`              step 2
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`HBr + H_2O -> H_3O^(+) + Br^(-)`      overall

With practice, you should be able to write overall proton transfers without having to write steps 1 and 2 every time.