What mass of NaF must be added to 50.0 mL of a 0.500 M HF solution to achieve a pH of 3.25? For HF, K_a = 7.2 × 10^-4. A) 1.3 g B) 0.69 g C) 6.9 g D) 23 g E) 1.5 g

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The correct answer is A.The pH of a solution can be calculated using the Henderson-Hasselbalch equation:```pH = pKa + log([A-]/[HA])```where pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.In this case, we have a solution of HF with a pH of 3.25 and a pKa of 7.2 × 10^-4. We can use the Henderson-Hasselbalch equation to calculate the ratio of [A-]/[HA]:```3.25 = 7.2 × 10^-4 + log([A-]/[HA])``````log([A-]/[HA]) = -4.05``````[A-]/[HA] = 10^-4.05 = 8.9 × 10^-5```This means that the concentration of the conjugate base, NaF, is 8.9 × 10^-5 times the concentration of the acid, HF.We can use the concentration of HF and the volume of the solution to calculate the number of moles of HF:```moles of HF = (0.500 M) × (50.0 mL) = 0.0250 mol```We can then use the ratio of [A-]/[HA] to calculate the number of moles of NaF that must be added:```moles of NaF = (8.9 × 10^-5) × (0.0250 mol) = 2.23 × 10^-6 mol```Finally, we can convert the number of moles of NaF to grams:```grams of NaF = (2.23 × 10^-6 mol) × (41.99 g/mol) = 1.3 g```Therefore, the correct answer is A: 1.3 g.

What Mass of NaF Must Be Added to 50