Starting with 0.750L of a buffer solution containing 0.30 M benzoic acid (C₆H₅COOH) and 0.35 M sodium benzoate (C₆H₅COONa), what will the pH of the solution be after the addition of 340.0 mL of 0.350M HCl (K_a (C₆H₅COOH) = 6.5 x 10^-5)

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The addition of HCl will result in the reaction: C₆H₅COOH (aq) + H3O+ (aq) ⇌ C₆H₅COOH2+ (aq) which will consume some of the benzoic acid and shift the equilibrium of the buffer system. To calculate the pH after the addition of HCl, we need to first calculate the concentrations of the species in the buffer solution before the addition of HCl. Using the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]) where pKa is the dissociation constant of benzoic acid (6.5 x 10^-5), [A-] is the concentration of sodium benzoate, and [HA] is the concentration of benzoic acid. Since we know the volumes and concentrations of the components in the buffer solution, we can calculate the moles of each component: moles of benzoic acid = (0.750L)(0.30 mol/L) = 0.225 mol moles of sodium benzoate = (0.750L)(0.35 mol/L) = 0.263 mol The total volume of the buffer solution is 0.750L + 0.340L = 1.09L Using the moles of each component and the total volume, we can calculate the concentrations: [HA] = 0.225 mol / 1.09 L = 0.206 M [A-] = 0.263 mol / 1.09 L = 0.241 M Now we can plug these values into the Henderson-Hasselbalch equation: pH = 6.5 x 10^-5 + log (0.241/0.206) = 4.17 So the initial pH of the buffer solution is 4.17. Next, we need to consider the effect of adding 0.340 L of 0.350M HCl. This will add moles of HCl to the solution, which will react with the benzoic acid to form benzoic acid protonated: C₆H₅COOH (aq) + H3O+ (aq) → C₆H₅COOH2+ (aq) We need to calculate the moles of HCl added: moles of HCl = (0.340 L)(0.350 mol/L) = 0.119 mol Since the concentration of benzoic acid is 0.206 M and one mole of HCl reacts with one mole of benzoic acid to form benzoic acid protonated, we can calculate the moles of benzoic acid that are converted to its protonated form: moles of benzoic acid protonated = 0.119 mol This means that the new concentration of benzoic acid is: [HA] = (0.225 mol - 0.119 mol) / (0.750 L + 0.340 L) = 0.0896 M The new concentration of [A-] remains the same as before the addition of HCl because NaC6H5COO does not react with HCl. Once again using the Henderson-Hasselbalch equation with the new values of [HA] and [A-]: pH = 6.5 x 10^-5 + log (0.241/0.0896) = 3.81 So the final pH of the solution after the addition of HCl is 3.81, which is option C.