If 20 mL of 0.10 M NaOH is added to 40 mL of 0.20 M HC₂H₃O₂,what is the pH of the resulting solution at 25°C? K_a for HC₂H₃O₂ is 1.8 × 10^-5 at 25°C. A) 9.8 B) 4.1 C) 5.2 D) 2.7 E) 4.3

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First, calculate the moles of NaOH and HC₂H₃O₂ (which seems to be a generic representation for an acid). For NaOH: moles = 0.02 L * 0.10 M = 0.002 moles. For the acid: moles = 0.04 L * 0.20 M = 0.008 moles. After reaction, 0.002 moles of the acid will be neutralized, leaving 0.006 moles of the acid unreacted in a total volume of 60 mL (0.06 L). The concentration of the acid is then 0.006 moles / 0.06 L = 0.1 M. Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa = -log(1.8 × 10^-5) and assuming the concentration of A- (the conjugate base) is approximately equal to the amount of NaOH added since it's a strong base and fully dissociates, we find that the pH is close to the pKa value due to the ratio of [A-]/[HA] being close to 1. The pKa calculated from the given Ka is approximately 4.74. Given the options, the closest value considering slight inaccuracies in the rough calculation and the effect of dilution and neutralization would be 4.3, as it's the only option reasonably close to this calculated pKa value.

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