Water is known to undergo autoprotolysis according to the following reaction:
2H_2O(l) rightleftharpoons H_3O^+ + HO^-2H2O(l)⇌H3O++HO−
This equilibrium has been exhaustively studied and we write the equilibrium in the normal way,
K_"eq"=([H_3O^+][HO^-])/([H_2O])Keq=[H3O+][HO−][H2O]
Because [H_2O][H2O] is LARGE, it can be assumed to be constant, and removed from the expression to give:
K_w=[H_3O^+][HO^-]=10^-14Kw=[H3O+][HO−]=10−14 at 298*K298⋅K
A temperature is specified, because the extent of reaction depends upon temperature, especially for a bond-breaking reaction. Now this is a mathematical expression, the which we can divide, multiply etc., provided that we do it to BOTH sides of the expression. One think we can do is to take log_10log10 of both sides for reasons that will become apparent later:
log_10K_w=log_10{[H_3O^+][HO^-]}=log_10{10^-14}log10Kw=log10{[H3O+][HO−]}=log10{10−14}
And thus, log_10[H_3O^+]+log_10[HO^-]=log_10{10^-14}log10[H3O+]+log10[HO−]=log10{10−14}
But log_10{10^-14}=-14log10{10−14}=−14 by definition, and we can rearrange the given expression to give:
14=-log_10[H_3O^+]-log_10[HO^-]14=−log10[H3O+]−log10[HO−]
Of course, by definition -log_10[H_3O^+]=pH−log10[H3O+]=pH and -log_10[HO^-]=pOH−log10[HO−]=pOH
So for water at 298K298K, pH+pOH=14pH+pOH=14. This is the defining expression for acid base behaviour in water, and it is one with which you will get very familiar.
So to answer your question (finally!), pHpH is a quantitative measure of the concentration of H_3O^+H3O+ (in water).
I apologize for going on so long, but you will need the given background if you don't know it already.
