Calculate the pH after 0.16 mol of NaOH is added to 1.00 L of the solution that is 0.55 M HF and 1.08 M KF.?

A)
Refer to: http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf
Ka for HF = 7.2 × 10⁻⁴ and pKa for HF = 3.14
(Different pKa values would lead to different answers.)

After addition of 0.16 M NaOH, HF reacts with NaOH to form NaF. Then,
[F⁻] = (1.08 + 0.16) M = 1.24 M
[HF] = (0.55 - 0.16) M = 0.39 M

Consider the dissociation of HF :
HF(aq) ⇌ H⁺(aq) + F⁻(aq)

pH = pKa + log([F⁻]/[HF]) = 3.14 + log(1.24/0.39) = 3.64 ≈ 3.6 (to 2 sig. fig.)


B)
After addition of 0.30 HCl, KF reacts with HCl to form HF. Then,
[F⁻] = (1.08 - 0.30) M = 0.78 M
[HF] = (0.55 + 0.30) M = 0.85 M

Consider the dissociation of HF :
HF(aq) ⇌ H⁺(aq) + F⁻(aq)

pH = pKa + log([F⁻]/[HF]) = 3.14 + log(0.78/0.85) = 3.10 ≈ 3.1 (to 2 sig. fig.)