A handbook states that to prepare a particular buffer solution mix 63.0 mL of 0.200 M aqueous HC₂H₃O₂ with 37.0 mL of 0.200 M aqueous NaC₂H₃O₂.What is the pH of this buffer? (K_a = 1.8 × 10^-5) A)4.74 B)4.51 C)4.98 D)5.33 E)7.00

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To calculate the pH of a buffer solution, we need to use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) where pKa is the acid dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid. In this case, HC₂H₃ is a weak acid and NaC₂H₃ is its conjugate base. The acid dissociation constant for HC₂H₃ (K_a) is given as 1.8 × 10^-5. To use the Henderson-Hasselbalch equation, we need to calculate the concentrations of HC₂H₃ and its conjugate base NaC₂H₃. First, we need to convert the given volumes and concentrations of the solutions to moles: moles of HC₂H₃ = 63.0 mL x (0.200 mol/L) = 12.6 mmol moles of NaC₂H₃ = 37.0 mL x (0.200 mol/L) = 7.4 mmol Next, we need to calculate the concentrations: [HC₂H₃] = moles of HC₂H₃ / total volume of solution = 12.6 + 37.0 mL = 2500 mL = 0.00504 M [NaC₂H₃] = moles of NaC₂H₃ / total volume of solution = 7.4 + 63.0 mL = 2500 mL = 0.00296 M Now we can plug these values into the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) pH = 4.74 + log(0.00296/0.00504) pH = 4.51 Therefore, the pH of this buffer is 4.51, which is closest to answer choice B.

A Handbook States That to Prepare a Particular Buffer Solution