Comparing acidity using pH: A solution with pH = 5 is less acidic than a solution with pH = 2 by what factor? (Assume 25°C and aqueous solutions.)

Introduction / Context:The pH scale is logarithmic, not linear. Every one-unit change corresponds to a tenfold change in hydrogen ion concentration. Converting qualitative statements like “more acidic” into quantitative factors is a routine task in water chemistry, bioprocessing, and environmental engineering.

Given Data / Assumptions:

• Definition: pH = −log10[H+].
• Two solutions: pH1 = 5 and pH2 = 2.
• Temperature and activity corrections are neglected for this comparison (typical for textbook problems).

Concept / Approach:The acidity comparison factor is the ratio of hydrogen ion concentrations. Because pH is a base-10 logarithm, a difference of ΔpH corresponds to a factor of 10^ΔpH in [H+]. Lower pH means higher [H+] (more acidic). We compute the factor by evaluating 10^(pH_high − pH_low).

Step-by-Step Solution:

Verification / Alternative check:Compute [H+] explicitly: at pH 2, [H+] = 10^−2 M; at pH 5, [H+] = 10^−5 M. Ratio = (10^−2)/(10^−5) = 10^3 = 1000.

Why Other Options Are Wrong:

• 3: Mistakenly treats the scale as linear.
• 100: Only accounts for a 2-unit change, not the full 3 units.
• None of these / 30: Not supported by the logarithmic definition.

Common Pitfalls:Forgetting the logarithmic nature of pH; mixing up which solution is more acidic; confusing “times less acidic” with differences in pH units.

Final Answer:1000