If 25.0 mL of 0.100 M HCl is titrated with 0.150 M NaOH, what is the pH at the equivalence point? (Assume a strong acid–base reaction)

• A. pH = 7
• B. pH = 4
• C. pH = 2
• D. pH = 9

Answer: (C)

The key idea is that in a titration between a strong acid and a strong base, the equivalence point yields a neutral solution. When HCl (a strong acid) reacts completely with NaOH (a strong base), the products are Na+ and Cl− in water. Both ions are spectator ions in terms of acid-base chemistry: neither hydrolyzes to produce extra H+ or OH−, so the solution is essentially neutral and has pH around 7.

Working with the numbers: moles of HCl present are 0.0250 L × 0.100 M = 2.50×10−3 mol. You need the same amount of NaOH to reach equivalence, which is 2.50×10−3 mol / 0.150 M = 0.0167 L = 16.7 mL of NaOH. The total volume at equivalence is 25.0 mL + 16.7 mL = 41.7 mL. The resulting solution contains only Na+ and Cl−, which do not affect the pH, so pH ≈ 7.

Thus, the pH at the equivalence point for this strong acid–strong base titration is about 7, not 2, 4, or 9.