By definition, pH=-log_10[H_3O^+]pH=−log10[H3O+]. And the use of the logarithmic function dates back to pre-electronic calculator days, when students, and engineers, and scientists, used logarithmic tables for more complex calculations, the which a modern calculator, available for a dollar or so, would EAT today....
For a strong acid, say HClHCl at MAXIMUM concentration, approx. 10.6*mol*L^-110.6⋅mol⋅L−1, which is conceived to ionize completely in aqueous solution, we gots...
HCl(aq) + H_2O(l) rarr H_3O^+ +Cl^-HCl(aq)+H2O(l)→H3O++Cl−
Now here, [H_3O^+]=10.6*mol*L^-1[H3O+]=10.6⋅mol⋅L−1....
And so pH=-log_10[H_3O^+]=-log_10{10.6}=-(+1.03)=-1.03pH=−log10[H3O+]=−log10{10.6}=−(+1.03)=−1.03..
And thus for stronger acid [H_3O^+][H3O+] GIVE A MORE NEGATIVE pHpH....
For background...
Just to note that in aqueous solution under standard conditions, the ion product...
K_w=[H_3O^+][HO^-]=10^(-14)Kw=[H3O+][HO−]=10−14...
And we can take log_10log10 of both sides to give....
log_(10)K_w=log_(10)10^(-14)=log_10[H_3O^+]+log_10[HO^-]log10Kw=log1010−14=log10[H3O+]+log10[HO−].
And thus.... -14=log_(10)[H_3O^+]+log_(10)[HO^-]−14=log10[H3O+]+log10[HO−]
Or.....
14=-log_(10)[H_3O^+]-log_(10)[HO^-]14=−log10[H3O+]−log10[HO−]
14=underbrace(-log_10[H^+])_(pH)underbrace(-log_10[OH^-])_(pOH)
14=pH+pOH
By definition, -log_10[H^+]=pH, -log_10[HO^-]=pOH.
