#NaOH(aq)#, #KOH(aq)#, #Na_2CO_3(aq)#
Water undergoes the following equilibrium.....the #"autoprotolysis reaction"#:
#2H_2O(l) rightleftharpoonsH_3O^+ + HO^-#
And the given equilibrium has been precisely measured under standard conditions....
#[H_3O^+][HO^-]=10^-14#
If #[HO^-]>10^-7*mol*L^-1#, then the solution is basic.......and if #[HO^-]<10^-7*mol*L^-1#, then the solution is acidic.....
We can introduce the #pH# measurement by taking #-log_10# of both sides......#[H_3O^+][HO^-]=10^-14#
Thus #-log_10[H_3O^+]-log_10[HO^-]=underbrace(-log_10(10^-14))_14#
And thus our defining relationship, which you will use shortly.....
#underbrace(-log_10[H_3O^+])_(pH)underbrace(-(log_10[HO^-]))_(pOH)=underbrace(-log_10(10^-14))_14#
p H+ p OH = 14
So actual examples.....#NaOH(aq)#, #Na_2CO_3(aq)#, #NH_3(aq)#, #KOH(aq)#, #NaF#..........
