A student is asked to determine the value of K a for hydrofluoric acid by titrat
ID: 1030092 • Letter: A
Question
A student is asked to determine the value of Ka for hydrofluoric acid by titration with potassium hydroxide.
The student begins titrating a 34.2 mL sample of a 0.461 M aqueous solution of hydrofluoric acid with a 0.290 M aqueous potassium hydroxide solution, but runs out of standardized base before reaching the equivalence point. The student's last observation is that when 35.3milliliters of potassium hydroxide have been added, the pH is 3.382.
What is Ka for hydrofluoric acid based on the student's data?
_________
Explanation / Answer
HF + KOH -------------> KF + H2O
millimoles of HF = 34.2 x 0.461 = 15.77
millimoles of KOH added = 35.3 x 0.290 = 10.34
15.77 - 10.34 = 5.43 millimoles HF left
10.34 millimoles KF formed
total volume = 34.2 + 35.3 = 69.5 mL
[HF] = 5.43 / 69.5 = 0.078 M
[KF] = 10.34 / 69.5 = 0.149 M
mixture of KF and HF act as acidic buffer
pH = pKa + log [KF] / [HF]
3.382 = pKa + log [0.149] / [0.078]
3.382 = pKa + 0.281
pKa = 3.101
Ka = 10-pKa = 10-3.101
Ka = 7.92 x 10-4
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.