A 600.0 mL sample of 0.20 M HF is titrated with 0.10 M NaOH.Determine the pH of the solution after the addition of 600.0 mL of NaOH.The K_a of HF is 3.5 × 10^-4. A) 2.08 B) 3.15 C) 4.33 D) 3.46 E) 4.15

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Quizplus · Accepted Answer

First, calculate the moles of HF and NaOH. Moles of HF = 0.600 L * 0.20 M = 0.12 mol. Moles of NaOH = 0.600 L * 0.10 M = 0.06 mol. Since NaOH is half the concentration of HF and equal volume is added, all NaOH will react with HF, leaving half of the initial HF unreacted. The reaction converts HF to F- (fluoride ion), so we have 0.06 mol of F- and 0.06 mol of unreacted HF in 1.2 L solution. The HF dissociation equation and its equilibrium constant (Ka = 3.5 × 10^-4) can be used to calculate the [H+], considering the concentrations of HF and F- at equilibrium. Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa = -log(3.5 × 10^-4) ≈ 3.46, [A-] = [F-] = 0.06 mol / 1.2 L, and [HA] = [HF] = 0.06 mol / 1.2 L, the ratio [A-]/[HA] = 1, log(1) = 0, so pH ≈ pKa = 3.46.

A 6000 ML Sample of 0