Why do strong acids have a low pH?

1 Answer
Mar 18, 2018

Because of the way we express the #p# function....

Explanation:

By definition, #pH=-log_10[H_3O^+]#. 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 #HCl# at MAXIMUM concentration, approx. #10.6*mol*L^-1#, which is conceived to ionize completely in aqueous solution, we gots...

#HCl(aq) + H_2O(l) rarr H_3O^+ +Cl^-#

Now here, #[H_3O^+]=10.6*mol*L^-1#....

And so #pH=-log_10[H_3O^+]=-log_10{10.6}=-(+1.03)=-1.03#..

And thus for stronger acid #[H_3O^+]# GIVE A MORE NEGATIVE #pH#....

For background...

Just to note that in aqueous solution under standard conditions, the ion product...

#K_w=[H_3O^+][HO^-]=10^(-14)#...

And we can take #log_10# of both sides to give....

#log_(10)K_w=log_(10)10^(-14)=log_10[H_3O^+]+log_10[HO^-]#.

And thus.... #-14=log_(10)[H_3O^+]+log_(10)[HO^-]#

Or.....

#14=-log_(10)[H_3O^+]-log_(10)[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#.