The expression pH = -log [H+] = -log [H3O+] does not work for very dilute acids.

The expression pH = -log [H+] = -log [H3O+] does not work for very dilute acids.

• A. Very dilute acids
• B. Very basic solutions
• C. High temperatures
• D. Strong bases

Answer: (A)

The key idea is that pH depends on hydrogen ion activity, not just the acid’s concentration. The formula pH = -log[H+] works well when the hydrogen ions in solution come almost entirely from the added acid and when activity roughly matches concentration. But in very dilute solutions, water’s autoionization becomes a significant source of H+. That means total [H+] is not simply the acid’s concentration, and the activity is not simply the concentration either. Because of that, using -log of the acid concentration gives an inaccurate pH. A concrete example helps: if you add a very small amount of strong acid, say 1×10^-8 M, water already contributes about 1×10^-7 M H+ through autoionization. Solving with Kw shows the total hydrogen ion concentration is around 1×10^-7 M, giving a pH close to 6.98 (slightly acidic), rather than pH = -log(1×10^-8) = 8. So the simple relationship fails for very dilute acids because you must account for both water autoionization and the difference between activity and concentration.

The key idea is that pH depends on hydrogen ion activity, not just the acid’s concentration. The formula pH = -log[H+] works well when the hydrogen ions in solution come almost entirely from the added acid and when activity roughly matches concentration. But in very dilute solutions, water’s autoionization becomes a significant source of H+. That means total [H+] is not simply the acid’s concentration, and the activity is not simply the concentration either.

Because of that, using -log of the acid concentration gives an inaccurate pH. A concrete example helps: if you add a very small amount of strong acid, say 1×10^-8 M, water already contributes about 1×10^-7 M H+ through autoionization. Solving with Kw shows the total hydrogen ion concentration is around 1×10^-7 M, giving a pH close to 6.98 (slightly acidic), rather than pH = -log(1×10^-8) = 8. So the simple relationship fails for very dilute acids because you must account for both water autoionization and the difference between activity and concentration.