How many moles of sodium hydroxide would have to be added to 250 mL of a 0.303 M hydrofluoric acid solution, in order to prepare a buffer with a pH of 3.630?

0.0562 moles

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:

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