Water undergoes self-ionization:
#H_2O rightleftharpoons H^+ + OH^-# (i)
Alternatively,
#2H_2O rightleftharpoons H_3O^+ + OH^-#(ii)
Both (i) and (ii) are equivalent representations; (ii) tends to be more common, and characterizes the cation of water as #H_3O^+#.
This is an equilibrium reaction, whose extent may be very accurately measured.
Under standard conditions, #[H_3O^+][OH^-]# #=# #10^(-14)#. Taking logarithms (base 10) of both sides, and multiplying by #-1#, we get:
#pH + pOH# #=# #14.# That is #pH# #=# #-log_10[H^+]#, and #pOH# #=# #-log_10[OH^-]#
This relationship is always obeyed. At high concentrations of #H_3O^+#, there are low concentrations of #OH^-#, at low concentrations of #H^+#, there are high concentrations of #OH^-#. But the sum of #pH + pOH# always equals #14.#
Students tend to have problems with the logarithmic function. When I write #log_ab = c#, I am asking to what power I raise the base #a# to get #b#; in other words, #a^c = b#. So #log_(10) 100 = 2#, #log_(10) 10 = 1#, and #log_(10) 0.1 = -1#.
So now, assuming complete ionization of #HCl# (reasonable!), can you tell me the #pH# of #0.1#, #1.0#, and #10# #mol# #L^-1# solutions?
#HCl(g) + H_2O(l) rightarrow H_3O^+ + Cl^-#.
