How do we get this expression?
Well, in water solution, the following equilibrium takes place:
2H_2O(l) rightleftharpoonsH_3O^+ + HO^-
AS with any equilibrium, we can write the equilibrium expression:
K_w = [H_3O^+][HO^-]
Now K_w has been extensively and carefully measured under a variety of circumstances. At 298 K,
K_w = [H_3O^+][HO^-] = 10^(-14).
As with any equation, we can divide it, multiply it, etc. PROVIDED that we do it to both the left hand side and right hand side of the equation. We can take log_10 of BOTH sides to give:
log_10K_w = log_(10)10^(-14) = log_10[H_3O^+] + log[HO^-]
But log_(10)10^(-14) = -14 BY DEFINITION of the logarithmic function. (i.e. log_(b)b^(c)=c)
Thus -14 = log_10[H_3O^+] + log_(10)[HO^-]
OR
+14 = -log_10[H_3O^+] - log_(10)[HO^-]
And again by DEFINITION, -log_10[H_3O^+]=pH, and likewise, -log_10[HO^-]=pOH
And thus, pH+pOH=14
So in most cases, given an actual HO^- concentration we can get both pH and pOH very rapidly and straightforwardly.
So you must remember pH+pOH=14
Note that this relation is valid under standard conditions, aqueous solution, 1 atm pressure, and 298*K. Under non-standard conditions, say at 398*K, how would expect pH to evolve? Remember that the acid base dissociation is a BOND BREAKING reaction.
