A 73.5-mL sample of 0.18 M HNO₂ (K_a = 4.0 × 10^-4) is titrated with 0.12 M NaOH. What is the pH after 25.2 mL of NaOH has been added? A) 10.07 B) 7.00 C) 3.93 D) 2.76 E) 2.87

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After adding 25.2 mL of 0.12 M NaOH to 73.5 mL of 0.18 M HNO2, the solution is in the buffer region. Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), where [A-] and [HA] are the concentrations of the conjugate base and acid, respectively. Calculating these concentrations and applying the equation gives a pH of approximately 2.87.

A 735-ML Sample of 0