We assess an AQUEOUS SOLUTION...in which the following equilibrium operates...
#underbrace(2H_2O(l) rightleftharpoons H_3O^+ + HO^-)_"the autoprotolysis rxn"#
And certainly this a measurable equilibrium, for which under a given set of standard conditions at #298*K# and near atmospheric pressure we can provide values...
#K_w=[H_3O^+][HO^-]=10^-14#...
How do you think #K_w# would evolve at HIGHER temperatures, given that the reaction as shown is A BOND-BREAKING reaction...?
And in the usual way we take #log_10# of both sides to get....
#log_10K_w=log_10([H_3O^+][HO^-])-=log_10(10^-14)#
And so....
#log_10(10^-14)=-14=log_10[H_3O^+]+log_10[HO^-][HO^-]#
And thus #+14=underbrace(-log_10[H_3O^+]-log_10[HO^-])_"by definition, pH+pOH"#...
And our defining relationship..#pH+pOH=14#
And suppose that we had #0.1*mol*L^-1# #H_3O^+#...#pH=-log_10(10^-1)=--1=+1#...and so #pOH=13#...
And should we take antilogs…#[HO^-]=10^(-pOH)=10^(-13)*mol*L^-1#...VERY LOW, but NON-ZERO...
The solution is ACIDIC because there are (many) more #H_3O^+# ions than #HO^-# ions....
