What is the pH of a buffer solution made by mixing 50.0 mL of 0.100 M potassium hydrogen phthalate with 13.6 mL of 0.100 M NaOH and diluting the mixture to 100.0 mL with water? The K_a2 for hydrogen phthalate is 3.1 × 10^-6. A)3.25 B)5.08 C)5.51 D)5.94

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First, calculate the moles of potassium hydrogen phthalate (KHP) and NaOH. Moles of KHP = 0.100 M * 0.050 L = 0.005 moles. Moles of NaOH = 0.100 M * 0.0136 L = 0.00136 moles. Since NaOH neutralizes KHP, the moles of remaining KHP = 0.005 - 0.00136 = 0.00364 moles. The moles of produced potassium phthalate = 0.00136 moles (equal to the moles of NaOH reacted). The total volume of the solution is 100.0 mL or 0.1 L. The concentrations are: [KHP] = 0.00364 moles / 0.1 L = 0.0364 M, [potassium phthalate] = 0.00136 moles / 0.1 L = 0.0136 M. Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa = -log(3.1 × 10^-6) = 5.51, [A-] = [potassium phthalate] = 0.0136 M, and [HA] = [KHP] = 0.0364 M. Substituting the values: pH = 5.51 + log(0.0136/0.0364) = 5.51 + log(0.3736) ≈ 5.51 - 0.428 = 5.08.

What Is the PH of a Buffer Solution Made by Mixing