A2.6 grams sample of solid naf is dissolved into 250.0 ml of a 0.75 m solution of hf(aq). assuming no significant change of volume, what is the ph of the resulting solution?

pH = 2.7

Explanation:

Please recognize this is a buffer solution question since we have the weak acid HF and its conjugate base F⁻ from the salt NaF.

For buffer solutions we utilize the Henderson-Hasselbach formula to calculate the pH:

pH = pKa + log [A⁻]/[HA]

where pKa  for the acid is given in reference tables, [A⁻] is the molarity of the weak base, and [HA] is the acid concentration.

pKa = 3.2 for HF

[F⁻] = mol Na / Vol (L) = (2.6 g/42 g/mol) / 0.250 L = 0.25 M

[HF] = 0.75 M

plugging our values into the equation:

pH = 3.2 + log (0.25/0.75) = 3.1 - 0.48 = 2.7

Note if you only had Ka for the acid, its pKa is the negative log of Ka.

12 is it do you think so