Stop this qualitative technobabble, it's time for quantitative calculations! pH is mathematically simple, if the activity coefficients are about 1, i.e. assuming an ideal-dilute solution. And yes, pH can be negative or over 14. The 'p' is shorthand for "power of", to denote very small concentrations; pH 2 is ten times diluted from pH 1, and so on. That is, pH is -1 times tenth base logarithm of the ionic activity of hydrogen ions: *pH = -log _{10} a_{H+}*.

The ionic activity is the "effective concentration" of the protons, which is different from the stoichiometric concentration ("how much we added") since ions tend to cluster in solutions. This actually requires a fair bit of calculations for pH measurement in concentrated or salt-laden solutions, such as volcanic waters. In dilute solutions, we can assume that the activity is the same as the stoichiometric concentration. (In 0.1 to about 0.01 M we have to calculate the activity; see Debye-Hückel.)

So, we're left with this: *pH = -log _{10} a_{H+}*. (The square brackets mean "the concentration of" in this case, in mol/l or moles per litre as always; this is essentially the count of ions per litre.)

Water dissociates even on when nothing else is present, although the extent of ionization is very small. Water dissociation is an equilibrium reaction, so we can measure an equilibrium constant, which is called water dissociation constant K_{W}.

The water is a solvent, so its activity is 1, aThe reaction:H_{2}O <-> H^{+}+ OH^{-}

K = (a_{H+}a_{OH-}) / a_{H2O}

_{H2O}= 1. One way of looking at this is that we could say that the water is of a different phase - it's not aqueous ions like H

^{+}and OH

^{-}but liquid - and when a reaction has one reactant in a different phase, this reactant is left out of the function for the position of the equilibrium. The environment in an dilute aqueous solution is completely saturated with water, which that means we can assume that [H

_{2}O] is constant. Dissolving small amounts of substances to water will not affect the volume considerably or change the number of water molecules. Thus, [H

_{2}O] is left out, and the constant is approximated such that activities are assumed to be the same as concentrations:

KIt can be measured that the constant has the value K_{W}= [H^{+}][OH^{-}]

_{W}= 10

^{-14}at 298 K (room temperature). This is an actual experimental value, not a definition. An implication is that pH is not a "scale" in the same way centigrade is, for example; it is a dimensionless number based on real physical quantities.

Neutral pure water has a pH of 7, as you will learn in a chemistry class. Here is the calculation:

What happens to hydrochloric acid in water?In pure water, [H^{+}] = [OH^{-}], so

[H^{+}][OH^{-}] = [H^{+}]^{2}= K_{W}

==> [H^{+}] = (10^{-14})^{0.5}= 10^{-7}

By definition,pH = -log_{10}10^{-7}

==> pH = 7.

an activity of 1 mol HClIn this case, we ignore the original dissociation of water, because the acid gives ten million times more Hto1 dm^{3}water

HCl -> H^{+}+ Cl^{-}(dissociation)

1 mol H^{+}==> pH = -log 1

pH = 0.

^{+}ions than the water does. This assumption is valid for (not too dilute) solutions with only acid and water. Of course, when concentrations of ions from the acid and from the water are closer, i.e. at 10

^{-7}mol dm

^{-3}of acid, you have to take the original dissociation into account. As you can see, if we put more than 1 mol of active protons, we get a negative pH. The result is not accurate for concentration, though; using the Debye-Hückel formula gives the result that an activity of 1 mol/l corresponds to 2 mol/l stoichiometric concentration, although Debye-Hückel probably doesn't work at this high a concentration.

When we remove or add hydrogen ions to the solution, the reaction shifts, because (surprise surprise!) the dissociation constant stays constant. (See Le Chatelier's principle.) This way we can understand how the pH can be over 7. When we increase the [OH^{-}] part in the reaction, the [H^{+}] part has to decrease.

So when we put, for instance, NaOH into the solution, it dissociates into OH^{-}, which in turn decreases the number of hydrogen ions in the water. As in adding H^{+}, we can neglect the small amount of OH^{-} from the water in the calculations and assume that all OH^{-} comes from the base.

KAn example:_{W}= [H^{+}][OH^{-}]

[H^{+}] = K_{W}/[OH^{-}]

2 mol dmThere you see how the pH is not always from 0 to 14. As you may have noticed, a handy shortcut for the pH is to take the logarithm of the whole K^{-3}of NaOH

[H^{+}] = 10^{-14}/(2 mol dm^{-3})

pH = 14.3

_{W}expression. Because the numerical value of K

_{W}is 10

^{-14}, its tenth-base logarithm is -14.

KpH is simply fourteen minus the logarithm of the concentration of the strong acid or base. In this case, log 2 is about -0.3, so pH = 14 - (-0.3) = 14.3. (Notice that in this concentrated solutions, this analysis breaks down, because activity no longer the same as the concentration. Use the Debye-Hückel theory in these cases.) If you want to apply this to weak bases, calculate the concentration of hydroxide ions first. In general:_{W}= [H^{+}][OH^{-}](-log_{10}both sides)

14 =-log- log_{10}[H^{+}]_{10}[OH^{-}](There's the pH!)

pH = 14 - log_{10}[OH^{-}]

pH = 14 - pOH

Not cut-n-paste. No source would have come up with this level of clarity and possibly some highly creative errors.