pH is short for "pouvoir hydrogen" i.e. "the power of hydrogen" or "hydrogen potential". This is applied to aqueous solutions in the following way.
In water, the following reaction takes place, and this is referred to as "autoprotolysis":
2H_2O rightleftharpoonsH_3O^+ + HO^-
As with any equilibrium, we can write the following equilibrium expression:
([H_3O^+][HO^-])/([H_2O]^2) = K.
Because [H_2O] is large and effectively constant, this simplifies to:
[H_3O^+][HO^-] = K_w.
This equilibrium has been measured meticulously, and at 298 K, we can give a value for K_w:
K_w = 10^(-14) = [H_3O^+][HO^-]
We can take log_10 of both sides to give:
log_10(10^(-14)) = log_10[H_3O^+]+ log_10[HO^-]
But log_10(10^(-14)) = -14 by definition of the log function.
And so, -14=log_10[H_3O^+]+ log_10[HO^-] or
14=-log_10[H_3O^+]- log_10[HO^-]
Now if we define -log_10[H_3O^+]=pH and - log_10[HO^-]=pOH, then:
14=pH+pOH.
And this given a value of pH, I can find [H_3O^+] or [HO^-] by taking antilogarithms.
What is the pH of a solution that is 0.5 mol*L^-1 with respect to H_3O^+? What is the corresponding pOH?
