We write the formula #H_3O^+# or #H^+#. We mean the characteristic cation of water. This might be a cluster of 3 or 4 or 5 water molecules with an extra proton, #H^+#, to make #H_7O_3^+# for instance. It is sufficient for calculations to write the ion product for water dissociation as:
#K_(H_2O)# #=# #[H_3O^+][OH^-]# #=# #10^(-14)#. Taking #-log_(10)# of both sides, we get:
#pH + pOH =14#. At neutrality, #pH# #=# #pOH# #=# #7#. (That is #[H_3O^+]# #=# #[OH^-]#). If #pH# #<# #7#, then the solution is acidic (because #[H_3O^+]# #># #[OH^-]#). And of course the solution is alkaline, #pH# #># #7#, when #[H_3O^+]# #<# #[OH^-]#.
I hope this is at the appropriate level.
