25.0 mL of a hydrofluoric acid solution of unknown concentration is titrated with 0.200 M NaOH. After 20.0 mL of the base solution has been added, the pH in the titration flask is 3.00. What was the concentration of the original hydrofluoric acid solution. [K_a(HF) = 7.1 * 10^-4]

Question

Quizplus · Accepted Answer

The correct answer is A because the concentration of the original hydrofluoric acid solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). At the equivalence point, [A-] = [HA] and pH = pKa. Before the equivalence point, [A-] < [HA] and pH < pKa. After the equivalence point, [A-] > [HA] and pH > pKa. In this case, the pH is 3.00, which is less than the pKa of HF (3.15). Therefore, we are before the equivalence point and [A-] < [HA]. We can use the Henderson-Hasselbalch equation to calculate the ratio of [A-] to [HA]: 3.00 = 3.15 + log([A-]/[HA]). Solving for [A-]/[HA], we get [A-]/[HA] = 0.182. We know that the total volume of the solution is 25.0 mL + 20.0 mL = 45.0 mL. Therefore, the total number of moles of HF is 0.0250 L * [HF] = 0.0250 L * x mol. The number of moles of NaOH added is 0.0200 L * 0.200 mol/L = 0.00400 mol. Since NaOH is a strong base, it will completely react with HF to form NaF and H2O. Therefore, the number of moles of NaF formed is also 0.00400 mol. The number of moles of HF remaining is 0.0250 L * x mol - 0.00400 mol. The concentration of HF remaining is [HF] = (0.0250 L * x mol - 0.00400 mol) / 0.0450 L. We can use the ratio of [A-] to [HA] to calculate the value of x: 0.182 = [NaF] / [HF]. Substituting the expressions for [NaF] and [HF], we get 0.182 = (0.00400 mol) / ((0.0250 L * x mol - 0.00400 mol) / 0.0450 L). Solving for x, we get x = 0.39 M. Therefore, the concentration of the original hydrofluoric acid solution was 0.39 M.

250 ML of a Hydrofluoric Acid Solution of Unknown Concentration