In water, we invoke the following equilibrium...
#2H_2O(l) rightleftharpoonsH_3O^+ + HO^-#
And we can write an equilibrium expression, to quantify this dissociation.......
#underbrace(K_w=[HO^-][H_3O^+]=10^-14)_"specified under standard conditions of 298 K and near 1 atm"#
And we take #log_10# OF BOTH SIDES....
#log_10K_w=log_10[HO^-]+log_10[H_3O^+]=log_(10)10^-14#
But #log_(10)10^-14=-14#
And so #+14=underbrace(-log_10[H_3O^+])_"pH by definition"underbrace(-log_10[HO^-])_"pOH by definition"#
And thus our working relationship...#14=pH+pOH#
And thus if we gots #0.5*mol*L^-1# #HCl(aq)#, we gots #[H_3O^+]=0.50*mol*L^-1#...#pH=-log_10(0.50)=-(-0.301)=+0.301#..
And so #pOH=14-0.301=13.7#...
Conc. #HCl(aq)# is #10.6*mol*L^-1# out of the bottle. What is #pH# here?
