By definition, #pH=-log_10[H_3O^+]#...and thus the LOWER (or more negative) the #pH# the GREATER the concentration of #H_3O^+#.
In aqueous solution, in which we ASSUME all these solvents are dissolved, the following equilibrium operates:
#2H_2Orightleftharpoons H_3O^+ + HO^-#
And, by careful measurement, the following equilibrium is established and quantified....
#K_w=10^-14=[H_3O^+][HO^-]#...under standard conditions...
And we take #log_10# of BOTH sides...
#underbrace(log_(10)10^-14)_(-14)=log_10([H_3O^+][HO^-])=log_10[H_3O^+]+log_10[HO^-]#
And so #-14=underbrace(log_10[H_3O^+])_(-pH)+underbrace(log_10[HO^-])_(-pOH)# ... or...
#14=pH+pOH#...
And thus THE GREATER #[H_3O^+]# the LOWER the #pH#. Acetic acid and #H_2S# are weak acids, however, the extent of equilibrium reduces the #pH<7#...
What are #pH# for #1.0*mol*L^-1# #HCl#, and for #10*mol*L^-1# #HCl#?
On the other hand, were we to have #1.0*mol*L^-1# #NaOH#..#[HO^-]=1.0*mol*L^-1#...and #pOH=-log_(10)1=0#...i.e. #pH=14#. Capisce?