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

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First, let's find the moles of each component of the buffer: moles HC₂H₃ = (0.200 mol/L)(0.0630 L) = 0.0126 mol moles NaC₂H₃ = (0.200 mol/L)(0.0370 L) = 0.0074 mol Next, let's calculate the concentrations of each component in the buffer solution: [HC₂H₃] = moles HC₂H₃ / total volume of buffer solution = 0.0126 mol / (0.0630 L + 0.0370 L) = 0.216 M [NaC₂H₃] = moles NaC₂H₃ / total volume of buffer solution = 0.0074 mol / (0.0630 L + 0.0370 L) = 0.127 M Now we can use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) where pKa is the acid dissociation constant of the acid (in this case, HC₂H₃, with pKa = 1.8 × 10^-5) [A-] is the concentration of the conjugate base (in this case, NaC₂H₃) [HA] is the concentration of the acid (in this case, HC₂H₃) pH = 4.51 (rounded to two decimal places)

A Handbook States That to Prepare a Particular Buffer Solution