A 120.0 mL sample of 0.20 M HF is titrated with 0.10 M NaOH. Determine the pH of the solution after the addition of 240.0 mL of NaOH. The Ka of HF is 6.8 × 10-4.

Answer 1

See my answer to your previous question about HF and NaOH.

initial mols HF = 0.120 L x 0.20 mol/L = 0.024 mol

initial mols OH- = 0.240 L x 0.10 mol/L = 0.024 mol

Henderson Hasselbalch: pH = pKa + log [conj. base[ / [ acid]

EDITED RESPONSE

HF + OH- ==> F- + H2O

0.024..0.024.....0..........Initial

-0.024..-0.024..+0.024..Change

0.........0..........0.024....Equilibrium

At equilibrium all of the HF has been neutralized by the OH- (equivalence point) so there is no longer a buffer (no weak acid is present). The pH of the solution will be dependent on the hydrolysis of F- acting as a conjugate base. This occurs as follows:

F- + H2O ==> HF + OH-

Kb for F- = 1x10^-14 / 6..8x10^-10 since KaKb = Kw

Kb = 1.47x10^-11

1.47x10^-11 = [HF][OH-] / [F-] = (x)(x) / 0.0667-x (0.0667 comes from 0.024 mols / 0.360 L)

ignore x in the denominator assuming it is small relative to 0.0667

1.47x10^-11 = x² / 0.0667

x² = 9.81x10^-13

x = 9.91x10^-7 = [OH-] (small relative to 0.0667 so assumption was valid)

pOH = -log 9.91x10-7 = 6.00

pH = 14 - 6.00

pH = 8.00