For HF,pK_a = 3.45.What is the pH of an aqueous buffer solution that is 0.100 M HF(aq)and 0.300 M KF(aq)?

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Quizplus · Accepted Answer

The dissociation of HF in water can be represented by the equation: HF(aq) + H2O(l) ⇌ H3O+(aq) + F-(aq) The equilibrium expression for this reaction is: Ka = [H3O+][F-]/[HF] We know that pK_a = 3.45, which means that: pKa = -log(Ka) = -log([H3O+][F-]/[HF]) = 3.45 Therefore, we can solve for [H3O+] using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) pH = 3.45 + log([KF]/[HF]) pH = 3.45 + log(0.300/0.100) = 3.93 Therefore, the pH of the buffer solution is 3.93, which is closest to option D (3.93).

For HF,pKa = 3

For HF,pK_a = 3