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^-.
