Answer:
pH = 2.40
Explanation:
It is possible to answer this question using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [A⁻] / [HA]
For HF / F⁻ buffer, [HA] is weakacid, HF, and F⁻ is conjugate base, [A⁻].
pKa of this buffer is -log Ka → 3.456
As molarity of NaF is 0.102M and molarity of HF is: 2.45mol / 2.1L = 1.167M. pH of resulting solution is:
pH = 3.456 + log₁₀ [0.102] / [1.167]
pH = 2.40
I hope it helps!
