Answer:
0.0562 moles
Step-by-step explanation:
Based on the H-H equation, for the HF buffer:
pH = pKa + log [F⁻] / [HF]
Where pH is the pH of the buffer = 3.630
pKa is pKa of HF buffer: 3.17
And [] could be taken as the moles of each reactant
3.630 = 3.17 + log [F⁻] / [HF]
0.46 = log [F⁻] / [HF]
2.884 = [F⁻] / [HF] (1)
The initial moles of HF are:
0.250L * (0.303mol / L) = 0.07575 moles
The HF reacts with NaOH as follows:
HF + NaOH → NaF + H₂O
That means the moles of HF will be the initial moles of HF - Moles NaOH
And moles NaF = F⁻ = Moles of NaOH added
That means:
0.07575 moles = [F⁻] +[HF] (2)
Replacing (2) in (1):
2.884 = [F⁻] / 0.07575 moles - [F⁻]
0.218 - 2.884[F⁻] = [F⁻]
0.218 = 3.884[F⁻]
[F⁻] = 0.0562 moles
That means the moles of NaOH that must be added are:
0.0562 moles